In this classic example of momentum conservation we’ll see why a rifle recoils when it is fired. A marksman holds a 3.00 kg rifle loosely, so that we can ignore any horizontal external forces acting on the rifle–bullet system. He fires a bullet of mass 5.00 g horizontally with a speed vbullet=300m/s . What is the recoil speed vrifle of the rifle? What are the final kinetic energies of the bullet and the rifle?

Question:

The same rifle fires a bullet with mass 7.7 g at the same speed as before. For the same idealized model, find the ratio of the final kinetic energies of the bullet and rifle.

The wavelength of red light in the glass would be 466.67 nm. The following is an explanation of how to get there:

We know that the wavelength of light changes as it moves from one medium to another. This change in the wavelength of light is described by the equation:

λ1/λ2 = n2/n1

where λ1 is the wavelength of light in the first medium, λ2 is the wavelength of light in the second medium, n1 is the refractive index of the first medium and n2 is the refractive index of the second medium.

In this case, the red light of wavelength 700 nm is moving from air (where its refractive index is 1.0) to glass (where its refractive index is 1.5). So, we can use the above equation to calculate the wavelength of light in the glass.

λ1/λ2

= n2/n1700/λ2

= 1.5/1.0λ2

= (700 nm x 1.0) / 1.5

λ2 = 466.67 nm

Therefore, the wavelength of the red light in the glass is 466.67 nm.

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The complete motion of the particle is linear in all the quadrants of the coordinate plane.

Given Vmax is the maximum speed, the particle is never stopped. A particle is said to have changed its direction when its velocity vector changes direction. Hence, the particle can reverse direction if the velocity vector becomes negative.

Let's discuss the particle's motion in each quadrant of a coordinate plane.

1. Quadrant I: In this quadrant, the x-component of the velocity vector is positive, and the y-component is also positive. Hence, the velocity vector lies in this quadrant. Therefore, the particle moves in this direction. Hence, the particle's motion is linear in this quadrant.

2. Quadrant II: In this quadrant, the x-component of the velocity vector is negative, but the y-component is positive. The velocity vector lies in this quadrant. Therefore, the particle moves in this direction. Hence, the particle's motion is linear in this quadrant.

3. Quadrant III: In this quadrant, the x-component of the velocity vector is negative, and the y-component is also negative. The velocity vector lies in this quadrant. Therefore, the particle moves in this direction. Hence, the particle's motion is linear in this quadrant.

4. Quadrant IV: In this quadrant, the x-component of the velocity vector is positive, but the y-component is negative. The velocity vector lies in this quadrant. Therefore, the particle moves in this direction. Hence, the particle's motion is linear in this quadrant.

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The displacement function is x(t) = 0.4 cos(3πt - 0.93) m, expressed in the given form. Determination of amplitude: In the given form of the displacement function x(t), the amplitude 'a' is given by the coefficient of the cosine function. Therefore, a = 0.4 m.

Determination of phase angle: The phase angle 'γ' can be determined by comparing the given function with the standard cosine function in the form of [tex]x(t) = a cos(ωt + γ).[/tex]

Here, we need to note that in the given function, the argument of the cosine function is (ωt - γ).

Therefore, [tex]γ = (ωt - arc cos (x/a))[/tex]

We know that [tex]cos(γ) = x/a[/tex]

∴ arc cos(x/a)

= γ= arc cos(0.4/0.6)

= 0.93 rad (approx)

Hence, the phase angle is γ = 0.93 rad.

Expressing displacement in the given form: Given that the displacement function is

x(t) = 0.4 cos(3πt - 0.93)

The angular frequency is ω = 3π rad/s and the phase angle is γ = 0.93 rad. Thus, the displacement function is x(t) = 0.4 cos(3πt - 0.93) m, expressed in the given form.

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According to the first law of thermodynamics, the internal energy of a system changes as the work is done on or by the system, or as heat is transferred to or from the system. The internal energy of a system is the sum of the kinetic and potential energies of its atoms and molecules.

δuint is the change in internal energy when objects a, b, and c are defined as separate systems. Hence, it is represented by the formula:δuint = q + w Where q is the heat absorbed or released, and w is the work done on or by the system. If the values of q and w are negative, the internal energy of the system decreases, and if they are positive, the internal energy of the system increases. The internal energy change is independent of the process by which it occurs, and only depends on the initial and final states of the system. Expressing the answer in Joules as an integer: δuint (J) = q(J) + w(J)

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. It can only be transformed from one form to another or transferred from one object to another. The total amount of energy in a closed system remains constant.

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The equilibrium constant (K) at 298 K for this reaction is 1.25 × 10¹⁰ mol⁻².

To calculate the equilibrium constant (K) at 298 K, we will need to utilize the equilibrium expression of the given chemical reaction.

The equilibrium constant (K) is defined as the ratio of the concentration of products raised to their stoichiometric coefficients to the concentration of reactants raised to their stoichiometric coefficients.

It is given as:K = [C]c[D]d / [A]a[B]b where A, B, C, and D are the chemical species present in the chemical reaction, and a, b, c, and d are the stoichiometric coefficients of A, B, C, and D respectively.

Also, [A], [B], [C], and [D] are the molar concentrations of A, B, C, and D at equilibrium, respectively.

Given reaction:N2(g) + 3H2(g) ⇌ 2NH3(g)In this reaction, a mole of nitrogen reacts with three moles of hydrogen to form two moles of ammonia.

Therefore, the equilibrium constant expression for this reaction is given as:K = [NH3]² / [N2][H2]³

The equilibrium constant (K) at 298 K for this reaction can be calculated by plugging the concentration of NH3, N2, and H2 at equilibrium in the above expression and solving for K.

Example:Suppose the concentration of NH3, N2, and H2 at equilibrium is found to be 0.2 M, 0.4 M, and 0.2 M respectively, then the equilibrium constant (K) at 298 K for this reaction will be:K = [NH3]² / [N2][H2]³K = (0.2)² / (0.4)(0.2)³K = 1.25 × 10¹⁰ mol⁻²

The equilibrium constant (K) at 298 K for this reaction is 1.25 × 10¹⁰ mol⁻².

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In this case, the object distance (u) is given as 50.0 cm and the lens to retina distance is given as 2.00 cm. We need to find the focal length (f) to calculate the power.

Since the eye is a complex optical system, we can consider it as a single thin lens. The lens to retina By substituting the calculated focal length (f) into the equation, we can determine the power of the eye when viewing an object 50.0 cm away.In this case, the lens to retina distance is given as 2.00 cm. Since the lens to retina distance represents the image distance (v), we need to find the object distance (u) to calculate the focal length (f).

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3.15 x [tex]10^-^3^4[/tex] J of energy are there in one photo. of orange light whose

wavelength is 630x[tex]10^9[/tex]m.

To calculate the energy of a photon, we can use the equation:

E = hc / λ

where E is the energy of the photon, h is Planck's constant (6.626 x [tex]10^-^3^4[/tex] J*s), c is the speed of light (3.0 x [tex]10^8[/tex] m/s), and λ is the wavelength of the light in meters.

Given the wavelength of the orange light as 630 x [tex]10^9[/tex]m, we can substitute the values into the equation to calculate the energy of one photon:

E = (6.626 x [tex]10^-^3^4[/tex]J*s * 3.0 x [tex]10^8[/tex] m/s) / (630 x [tex]10^9[/tex] m)

Simplifying the equation:

E = (1.988 x [tex]10^-^2^5[/tex]J*m) / (630 x[tex]10^9[/tex]m)

E = 3.15 x 10[tex]10^-^3^4[/tex] J

It's important to note that the energy of a single photon is very small due to its quantum nature. In practical applications, the energy of photons is often measured in terms of the number of photons rather than individual photon energy.

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The average force on the first ball is 0 N. The average force on the second ball is 0 N.

To solve this problem, we can use the principles of conservation of momentum and energy. Let's start by calculating the velocity of the second ball after the collision using the conservation of momentum:

Initial momentum = Final momentum
(mass_1 * velocity_1) + (mass_2 * velocity_2) = 0
(0.28 kg * 5.8 m/s) + (0.28 kg * velocity_2) = 0
velocity_2 = -(0.28 kg * 5.8 m/s) / 0.28 kg
velocity_2 = -5.8 m/s. The negative sign indicates that the second ball is moving in the opposite direction to the first ball. Now, we can calculate the change in kinetic energy of the first ball using the conservation of energy: Initial kinetic energy - Final kinetic energy = Work done by the force
(0.5 * mass_1 * velocity_1^2) - 0 = Average force * distance.
0.5 * 0.28 kg * (5.8 m/s)^2 = Average force * 0.
Average force on the first ball = 0 N
Since the first ball comes to rest, there is no change in kinetic energy, and therefore, no average force is exerted on it.
Next, we can calculate the change in kinetic energy of the second ball:
Initial kinetic energy - Final kinetic energy = Work done by the force
(0.5 * mass_2 * velocity_2^2) - 0 = Average force * distance

0.5 * 0.28 kg * (-5.8 m/s)^2 = Average force * 0
Average force on the second ball = 0 N.
Similarly, since the second ball flies off, there is no change in kinetic energy, and therefore, no average force is exerted on it. In conclusion:

A) The average force on the first ball is 0 N.

B) The average force on the second ball is 0 N.

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Option c) UNICEF is not an NGO, while options a) CARE, b) Red Cross, d) World Vision, and e) Oxfam are all NGOs.

The paragraph presents a question regarding non-governmental organizations (NGOs) and requires the identification of the option that is not an NGO.

NGOs are typically independent organizations that operate on a non-profit basis to address social, humanitarian, and environmental issues. They often work alongside governments and other entities to provide assistance and advocate for various causes.

Among the options provided, the United Nations International Children's Emergency Fund (UNICEF) is not considered an NGO.

UNICEF is a specialized agency of the United Nations (UN) and operates as a program within the UN system. It focuses specifically on child rights and well-being worldwide, collaborating with governments and other partners to fulfill its mandate.

On the other hand, CARE, Red Cross, World Vision, and Oxfam are all recognized NGOs that work on a range of issues such as poverty alleviation, disaster response, healthcare, and advocacy.

Therefore, option c) UNICEF is not an NGO, while options a) CARE, b) Red Cross, d) World Vision, and e) Oxfam are all NGOs.

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The Salem Witch Trials were the consequence of religious disputes within the Puritan community, widespread anxiety over wars with Indians, and fear and hatred of women who were perceived as different or challenging societal norms.

The Salem Witch Trials were influenced by religious disputes, anxiety over wars with Indians, and fear and prejudice towards women who deviated from societal norms.

The Salem Witch Trials of 1692 in colonial Massachusetts were primarily fueled by religious tensions within the Puritan community. Puritan beliefs and practices were deeply ingrained in the society, and any deviation from their strict religious doctrines was seen as a threat. The trials were fueled by a fear of witchcraft and the belief that Satan was actively working to corrupt the community.

Additionally, the ongoing conflicts between English colonists and Native American tribes during the time created a climate of widespread anxiety and fear. The fear of Indian attacks and the uncertainty of the frontier amplified the existing anxieties within the community, leading to a heightened sense of paranoia and the scapegoating of individuals as witches.

Furthermore, the trials were marked by a pervasive fear and prejudice against women who were seen as different or challenging the established norms. Many of the accused were women who didn't conform to the traditional roles and expectations placed upon them. Women who displayed independence, assertiveness, or unconventional behavior were viewed with suspicion and often targeted as witches.

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The baby will not succeed in moving the toy box with a horizontal force of 46N.

To determine if the baby will succeed in moving the toy box, we need to compare the force exerted by the baby (46N) with the maximum frictional force.

The maximum static frictional force can be calculated by multiplying the coefficient of static friction (0.8) by the normal force. The normal force is equal to the weight of the toy box, which is given by the formula:

weight = mass x gravity.

weight = 15 kg x 9.8 m/s^2 = 147 N

Maximum static frictional force = 0.8 x 147 N = 117.6 N

Since the force exerted by the baby (46N) is less than the maximum static frictional force (117.6 N), the toy box will not move. The static friction will be greater than the force applied, causing the toy box to remain stationary.

Therefore, the baby will not succeed in moving the toy box with a horizontal force of 46N.

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The amount of energy (in kJ/mole) that was converted to heat is 345 kJ/mol (rounded to three significant figures).

To find the energy that is converted to heat, we need to compare the energy of the absorbed light to the energy of the emitted light. The absorbed light has a wavelength of 320 nm = 320 × 10⁻⁹ m.

So:

E = hc/λ E = (6.626 × 10⁻³⁴ J·s) (3.00 × 10⁸ m/s) / (320 × 10⁻⁹ m) E = 1.85 × 10⁻¹⁸ J

The absorbed light has less energy than the emitted light. The difference in energy is converted to heat.

So:

ΔE = 3.81 × 10⁻¹⁷ J – 1.85 × 10⁻¹⁸ J

ΔE = 3.63 × 10⁻¹⁷ J

This is the energy that is converted to light. To convert this to energy per mole, we need to know the number of photons in one mole of the mineral. This can be calculated using Avogadro’s number:

N = 6.02 × 10²³ photons/mol

So the energy per mole is:

ΔE/mol = (3.63 × 10⁻¹⁷ J) (6.02 × 10²³ photons/mol) ΔE/mol = 2.19 × 10⁷ J/mol

To convert this to kJ/mol, we divide by 1000:

ΔE/mol = 2.19 × 10⁴ kJ/mol

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The energy that was not converted to light is converted to heat. If the mineral has absorbed energy with a wavelength of 320 nm, the amount of energy (in kJ/mole) that was converted to heat is 109 kJ/mole.

A fluorescent mineral absorbs "black light" from a mercury lamp. It then emits visible light with a wavelength 520 nm.

The energy not converted to light is converted into heat.

The energy absorbed by the mineral = 320 nm

We know that the frequency of the energy absorbed by the mineral is given by the formula: c = λv

Where:

c = speed of light (3.0 × 10⁸ m/s)

λ = wavelength of energy (in meters)

v = frequency of energy (in Hertz)

Therefore:

v = c/λ = 3.0 × 10⁸ m/s / 320 × 10⁻⁹ m = 9.375 × 10¹⁴ Hz

Now, the energy absorbed by the mineral (E) is given by the formula: E = hv

Where:

h = Planck's constant (6.626 × 10⁻³⁴ J s)v = frequency of energy (in Hertz)

Therefore:

E = hv = 6.626 × 10⁻³⁴ J s × 9.375 × 10¹⁴ Hz = 6.22 × 10⁻¹⁸ J/molecule

The mineral then emits visible light with a wavelength of 520 nm. The frequency of the emitted light is given by the formula: v = c/λ = 3.0 × 10⁸ m/s / 520 × 10⁻⁹ m = 5.769 × 10¹⁴ Hz

The energy emitted as light is given by the formula: E = hv = 6.626 × 10⁻³⁴ J s × 5.769 × 10¹⁴ Hz = 3.82 × 10⁻¹⁸ J/molecule

Therefore, the energy converted to heat is:ΔE = Energy absorbed - Energy emitted

ΔE = (6.22 - 3.82) × 10⁻¹⁸ J/moleculeΔE = 2.4 × 10⁻¹⁸ J/molecule

Now, to calculate the energy converted to heat in kJ/mol:2.4 × 10⁻¹⁸ J/molecule × (6.02 × 10²³ molecules/mol) / (1000 J/kJ) = 1.44 × 10⁻⁴ kJ/mole

Therefore, the amount of energy (in kJ/mole) that was converted to heat is 109 kJ/mole.

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The concentration of all species in a 0.100 M H₃PO₄ solution is as follows: [H₃PO₄] = 0.100 M, [H₂PO₄⁻] = 0.045 M, [HPO₄²⁻] = 0.0049 M, and [PO₄³⁻] = 1.0 x 10^-7 M.

Phosphoric acid, also known as orthophosphoric acid, is a triprotic acid with the chemical formula H₃PO₄. In water, the acid disassociates into H⁺ and H₂PO₄⁻. The second dissociation of H₂PO₄⁻⁻ results in the formation of H⁺ and HPO₄²⁻. Finally, the dissociation of HPO₄²⁻ produces H⁺ and PO₄³⁻. The following equations show the dissociation of H₃PO₄:
H₃PO₄ → H⁺ + H₂PO₄⁻
H₂PO₄⁻ → H⁺ + HPO₄²⁻
HPO₄²⁻ → H⁺ + PO₄³⁻
Using the dissociation constants of phosphoric acid, one can calculate the concentrations of all species in a 0.100 M H₃PO₄ solution. [H₃PO₄] = 0.100 M, [H₂PO₄⁻] = 0.045 M, [HPO₄²⁻] = 0.0049 M, and [PO₄³⁻] = 1.0 x 10^-7 M.

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The man does 480 J of work while loading the appliance across the ramp from bottom to top.

To solve this problem, we can use the equation for work:

Work = Force * Distance

We know that the force is equal to the weight of the appliance, which is 120 kg * 9.8 m/s² = 1176 N.

We also know that the distance is equal to the length of the ramp, which we can calculate using the Pythagorean theorem:

Length of ramp = √(4.0 m² + 4.0 m²) = 4.24 m

Plugging these values into the equation for work, we get:

Work = 1176 N * 4.24 m = 480 J

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Complete question :

A man loads 120kg appliance onto a truck across a ramp (sloped surface). The side opposite the ramps angle is 4.0 m in height. How much work does the man do while loading the appliance across the ramp from bottom to top

the depth of the pool is 3.08 meters.

Given:

Width of the swimming pool = 5.0 mThe pool is filled to the top.

The bottom of the pool becomes completely shaded in the afternoon when the sun is 23° above the horizon

We can solve the given question using Trigonometry.

ABC,cot 23° = AB/BCEquation (1)

But, AB + BC = 5.0 m

Equation (2)Also, AB^2 + BC^2 = AC^2

[Applying Pythagoras theorem in triangle ABC]  Equation (3)

From equation (2), we have BC = 5 - AB

Substituting it in equation (3),

we get:

AB^2 + (5 - AB)^2 = AC^2

Expanding and simplifying the above equation:

2AB^2 - 10AB + 25 = AC^2But, we know that AB/BC

Equation (1) => AB = BC × cot 23° => AB = (5 - AB) × cot 23°

Solving the above equation, we get AB = 1.92 m

Hence, the depth of the pool is BC = 5 - AB = 5 - 1.92 = 3.08 meters.

So, the depth of the pool is 3.08 meters.

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(a) The proton's average acceleration av in unit vector notation is (-2.50)i + (197)j + (6.70)k m/s^2.

(b) The magnitude of the proton's average acceleration av is 198 m/s^2.

(c) The angle between the average acceleration av and the positive direction of the x-axis is approximately 95.4 degrees.

Explanation to the above given short answers are written below,

(a) To find the average acceleration av, we need to calculate the change in velocity and divide it by the time interval. The change in velocity is given by
Δv = v_f - v_i,
where v_f is the final velocity and
v_i is the initial velocity.

Subtracting the initial velocity from the final velocity, we get
Δv = (7.50 - 18.0)i + (-4.90 - (-490))j + (13.0 - (-18.0))k = (-10.5)i + (485.1)j + (31.0)k.

Dividing Δv by the time interval of 5.20 s, we get the average acceleration av = (-2.50)i + (197)j + (6.70)k m/s^2.

(b) The magnitude of the average acceleration av can be calculated using the formula
|av| = √(avx^2 + avy^2 + avz^2),
where avx, avy, and avz are the components of av in the x, y, and z directions, respectively.

Substituting the values, we get |av| = √((-2.50)^2 + (197)^2 + (6.70)^2) = 198 m/s^2.

(c) The angle between the average acceleration av and the positive direction of the x-axis can be determined using the formula
θ = arctan(avy / avx).

Substituting the values, we get θ = arctan(197 / (-2.50)) ≈ 95.4 degrees.

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The speed of an electromagnetic wave is 299,792,458 meters per second (m/s) or the speed of light.

The direction of the vector product of E (electric field) and B (magnetic field) indicates the direction of energy transfer in an electromagnetic wave. This direction is perpendicular to both the E and B fields. The wave propagates in this direction as well. The direction of the vector product is referred to as the Poynting vector.

The Poynting vector, S, provides information about the direction and intensity of the electromagnetic energy flux or radiation pressure density. Its SI unit is watt per square meter (W/m²). It can be mathematically expressed as:S = E × BIn an electromagnetic wave, the E and B fields oscillate in mutually perpendicular planes. The direction of energy transfer is also perpendicular to both the E and B fields. An electromagnetic wave propagates perpendicular to both E and B fields and the direction of energy transfer. It has both electric and magnetic properties and carries energy. Therefore, an electromagnetic wave can be defined as a wave of energy produced by the acceleration of an electric charge and propagated through a vacuum or a medium.

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The cosine of the angle between the vectors ⟨1, 1, 1⟩ and ⟨6, -10, 11⟩ is 7 / (√3)(√257). we can use the dot product formula.

To find the cosine of the angle between two vectors, we can use the dot product formula.

The dot product of two vectors A and B is given by:

A · B = |A| |B| cos(θ)

Where A · B represents the dot product, |A| and |B| are the magnitudes of the vectors A and B respectively, and θ is the angle between the two vectors.

Given the vectors A = ⟨1, 1, 1⟩ and B = ⟨6, -10, 11⟩, we can calculate their dot product as follows:

A · B = (1)(6) + (1)(-10) + (1)(11) = 6 - 10 + 11 = 7

Now, we need to calculate the magnitudes of vectors A and B:

|A| = √(1^2 + 1^2 + 1^2) = √3

|B| = √(6^2 + (-10)^2 + 11^2) = √(36 + 100 + 121) = √257

Now, we can substitute the values into the formula:

A · B = |A| |B| cos(θ)

7 = (√3) (√257) cos(θ)

Dividing both sides by (√3)(√257), we get:

cos(θ) = 7 / (√3)(√257)

Therefore, the cosine of the angle between the vectors ⟨1, 1, 1⟩ and ⟨6, -10, 11⟩ is 7 / (√3)(√257).

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Category II electric meters are safe for working on low voltage circuits that have a current of less than or equal to 10A. The low voltage circuits with currents less than or equal to 10A are the types of circuits that Category II electric meters are safe for working on.

Category II electric meters are considered safe for low-voltage circuits with currents up to 10 amps. The 10-ampere maximum rating ensures that the electric meter's internal components are secure and the electric meter is not damaged by higher currents.

Since low-voltage circuits are commonly utilized for electronic devices, measuring and testing these circuits frequently need a category II electric meter.

Therefore, category II electric meters are safe for use in low-voltage circuits with currents of less than or equal to 10A.

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The numerical value of the mean voltage in the circuit is 57.27.

Suppose the voltage in an electrical circuit varies with time according to the formula v(t) = 90 sin(t) for t in the interval [0,].

The numerical value of the mean voltage in the circuit is 0.

The voltage is given by v(t) = 90 sin(t).To find the mean voltage, we need to find the average value of the voltage over the interval [0,].

The formula for the mean value of the voltage over an interval is:

Mean value of v(t) = (1/b-a) ∫aᵇv(t)dt

Where a and b are the limits of the interval.

In our case, a = 0 and b = π.

The integral is: ∫₀ᴨ 90sin(t) dt = -90 cos(t) between the limits 0 and π.

∴ Mean value of v(t) = (1/π-0) ∫₀ᴨ 90sin(t)dt

= (1/π) x [-90 cos(t)]₀ᴨ

= (1/π) x (-90 cos(π) - (-90 cos(0)))

= (1/π) x (90 + 90)

= 180/π

= 57.27 approx

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Option c) A and C statements are TRUE about a body moving in circular motion.

a) For a body moving in circular motion at a constant speed, the direction of the velocity vector is the same as the direction of the acceleration. This is true because in circular motion, the velocity vector is always tangential to the circular path, and the acceleration vector is directed towards the center of the circle, perpendicular to the velocity vector.

b) Increasing the mass of an object moving in a circular path will not directly affect the net force. The net force is determined by the centripetal force required to keep the object in circular motion, which is determined by the object's mass, speed, and radius of the circular path. Increasing the mass alone does not change the net force.

c) If an object moves in a circle at a constant speed, its velocity vector will be constant in magnitude but changing in direction. This is because the object is constantly changing its direction while maintaining the same speed. Velocity is a vector quantity that includes both magnitude (speed) and direction, so if the direction is changing, the velocity vector is also changing.

Therefore, the correct statements are A and C.

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The displacement of a car moving with a constant velocity of 9.5 m/s in a time interval between 3 seconds to 5 seconds is 19 meters.

It given by the formula: Δx = vΔt where Δx = displacement v = velocity Δt = time interval Substituting the given values, we get:Δx = 9.5 m/s × (5 s - 3 s)Δx = 9.5 m/s × 2 sΔx = 19 m, the displacement of the car during the given interval is 19 meters.

The given formula is derived from the definition of velocity which is the change in displacement per unit time. Since the velocity of the car is constant, we can assume that its acceleration is zero. Therefore, the car is not changing its velocity, which means that the displacement during that interval is equal to the product of velocity and time.In this case, we are given the initial and final times, and we need to find the displacement during that time interval.

The difference between the two times is 2 seconds. Multiplying the velocity with the time interval, we get the displacement of the car. The unit of displacement is meter, which is the same as the unit of distance.

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The heat transfer that occurs from 1.1 kg of water at 40°C to 1.1 kg of water at 20°C is 92,270 J.

To calculate the heat transfer that occurs when two substances reach thermal equilibrium, we can use the equation Q = mcΔT, where Q is the heat transfer, m is the mass, c is the specific heat, and ΔT is the change in temperature.

In this case, we have two equal masses of water, each weighing 1.1 kg. The specific heat of water, c, is given as 4186 J/(kg°C).

First, we need to calculate the change in temperature, ΔT, which is the difference between the final equilibrium temperature and the initial temperature. Since the masses are equal, the equilibrium temperature will be the average of the initial temperatures, which is (40°C + 20°C) / 2 = 30°C.

Next, we can calculate the heat transfer for each mass of water using the equation Q = mcΔT. For the water at 40°C, the heat transfer is Q₁ = (1.1 kg) * (4186 J/(kg°C)) * (30°C - 40°C) = -45,530 J (negative because heat is transferred out of the water). Similarly, for the water at 20°C, the heat transfer is Q₂ = (1.1 kg) * (4186 J/(kg°C)) * (30°C - 20°C) = 137,800 J.

The total heat transfer is the sum of the individual heat transfers: Q_total = Q₁ + Q₂ = -45,530 J + 137,800 J = 92,270 J.

Therefore, the heat transfer that occurs from 1.1 kg of water at 40°C to 1.1 kg of water at 20°C is 92,270 J.

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Complete Question:

(a) In reaching equilibrium, how much heat transfer occurs from 1.1 kg of water at 40€ when it is placed in contact with 1.1 kg of 20€ water? Specific heat of water c=4186 J/(kg) Hint: If the masses of water are equal, what is the equilirium temperature of the water mixture?

The car reaches its maximum speed of 106 m/s in 18.93 seconds and travels approximately 3366.26 meters. The motorbike reaches its maximum speed of 58.8 m/s in 7 seconds and travels 2058 meters. The car never catches up with the motorbike.

(a) To find the time it takes for the car and the motorbike to reach their maximum speeds, we can use the formula:

Time = (Final Speed - Initial Speed) / Acceleration

For the car:

Initial Speed = 0 m/s (rest)

Final Speed = 106 m/s

Acceleration = 5.6 m/s²

Time = (106 m/s - 0 m/s) / 5.6 m/s² = 18.93 seconds

For the motorbike:

Initial Speed = 0 m/s (rest)

Final Speed = 58.8 m/s

Acceleration = 8.4 m/s²

Time = (58.8 m/s - 0 m/s) / 8.4 m/s² = 7 seconds

(b) To find the distance traveled by the car and the motorbike when they reach their respective maximum speeds, we can use the formula:

Distance = (Initial Speed × Time) + (0.5 × Acceleration × Time²)

For the car:

Initial Speed = 0 m/s (rest)

Time = 18.93 seconds

Acceleration = 5.6 m/s²

Distance = (0 m/s × 18.93 seconds) + (0.5 × 5.6 m/s² × (18.93 seconds)²)

Distance = 0 + 0.5 × 5.6 m/s² × 357.2049 seconds²

Distance ≈ 3366.26 meters

For the motorbike:

Initial Speed = 0 m/s (rest)

Time = 7 seconds

Acceleration = 8.4 m/s²

Distance = (0 m/s × 7 seconds) + (0.5 × 8.4 m/s² × (7 seconds)²)

Distance = 0 + 0.5 × 8.4 m/s² × 49 seconds²

Distance = 2058 meters

(c) To find how long it takes the car to catch up with the motorbike, we need to determine the time at which their positions are equal. Since the car continues to accelerate while catching up, we can use the equation:

Distance = (Initial Speed × Time) + (0.5 × Acceleration × Time²)

Let's assume the time it takes for the car to catch the motorbike is t.

For the car:

Initial Speed = 0 m/s (rest)

Acceleration = 5.6 m/s²

For the motorbike:

Initial Speed = 0 m/s (rest)

Acceleration = 8.4 m/s²

Setting the distances equal to each other:

(0 m/s × t) + (0.5 × 5.6 m/s² × t²) = (0 m/s × t) + (0.5 × 8.4 m/s² × t²) + (58.8 m/s × t)

Simplifying the equation:

(0.5 × 5.6 m/s² × t²) = (0.5 × 8.4 m/s² × t²) + (58.8 m/s × t)

Since the term (0.5 × 5.6 m/s² × t²) equals (0.5 × 8.4 m/s² × t²), they cancel out, and we are left with:

0 = 58.8 m/s × t

This implies that t = 0, which is the unphysical solution since it means the car catches up with the motorbike instantaneously. Therefore, there is no valid solution for the car catching up with the motorbike.

In conclusion, the car and motorbike reach their maximum.

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The control limits for 3-sigma x chart are 718.5 mL and 731.5 mL.

An x-chart is a graph that shows a collection of data points on a line that corresponds to the sample mean. It's created by calculating the mean of the data and plotting it on a chart in the middle. The upper and lower control limits, or UCL and LCL, are also represented on the graph. The control limits show when a process is out of control or exceeding its predicted performance limits. The x-chart is used to monitor variables data, such as the sample mean, to detect changes in a process. The average range R is a measure of process variability. The average range R is a measure of process variability. It is calculated by taking the average of the ranges from several samples.

The X-bar chart is a type of Shewhart control chart used in industrial statistics to monitor the arithmetic means of successive samples of the same size, n. This control chart is used for characteristics like weight, temperature, thickness, and so on that can be measured on a continuous scale.

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High-mass stars, like the Sun, undergo stellar evolution in a different manner compared to lower-mass stars. Here are the primary differences in the stages of stellar evolution between a high-mass star and a star like the Sun:

Sun-like Star:

Nebula: A cloud of gas and dust collapses under its gravity, forming a protostar.

Main Sequence: The protostar reaches equilibrium, and nuclear fusion begins in its core, converting hydrogen into helium. This phase lasts for about 10 billion years.

Red Giant: As hydrogen fuel depletes, the star expands and becomes a red giant, burning helium in its core while outer layers expand.

Planetary Nebula: The red giant sheds its outer layers, creating an expanding shell of gas and exposing the core.

White Dwarf: The remaining core, composed of a dense, hot, degenerate gas, becomes a white dwarf, gradually cooling over billions of years.

High-Mass Star:

Nebula: Similar to the Sun-like star, a nebula collapses to form a protostar.

Main Sequence: The protostar becomes a high-mass main sequence star, undergoing nuclear fusion at a higher rate due to its higher mass.

Red Supergiant: The high-mass star exhausts its hydrogen quickly and expands to become a red supergiant, fusing heavier elements in its core.

Supernova: Once fusion ceases, the core collapses, resulting in a catastrophic explosion called a supernova, releasing an enormous amount of energy and creating heavy elements.

Neutron Star or Black Hole: The core of the high-mass star collapses further, forming either a neutron star or a black hole, depending on its mass.

In summary, the primary differences in stellar evolution between a high-mass star and a star like the Sun lie in their mass-dependent stages. High-mass stars burn through their fuel more rapidly, leading to shorter lifetimes and more energetic events such as supernovae. The remnants of high-mass stars can form neutron stars or black holes, while lower-mass stars like the Sun end their lives as white dwarfs. These differences highlight the profound influence of stellar mass on the evolutionary path of stars.

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The distance of closest approach is the minimum distance between the moving alpha particle and the fixed gold nucleus. At this distance, the kinetic energy of the alpha particle is converted into potential energy of electrostatic repulsion, which causes the alpha particle to reverse direction. For the alpha particle to get to this distance of closest approach, the initial speed must be calculated. We can apply conservation of energy, which states that the total energy of a system is constant, and is equal to the sum of the kinetic and potential energies.The potential energy is given byCoulomb's law : $U = \frac{kq_1q_2}{r}$where k is Coulomb's constant, $q_1$ and $q_2$ are the charges of the two particles, and r is the separation distance between the particles. At the distance of closest approach, the potential energy is maximum, and the kinetic energy is zero. Thus, we can equate the potential energy at the distance of closest approach to the initial kinetic energy of the alpha particle. That is,$U = \frac{kq_1q_2}{r} = \frac{2(79)e^2}{4\pi\epsilon_0(2.0\times10^{-10})}$ $= 9.14 \times 10^{-13} J$The initial kinetic energy of the alpha particle is given by$K = \frac{1}{2}mv^2$where m is the mass of the alpha particle and v is the initial speed. We can equate K to U. That is,$\frac{1}{2}mv^2 = \frac{kq_1q_2}{r}$Substituting the values,$\frac{1}{2}(6.64\times10^{-27})v^2 = 9.14\times10^{-13}$Solving for v,$v^2 = \frac{2(9.14\times10^{-13})}{6.64\times10^{-27}}$$v = 2.21\times10^7 m/s$Thus, the initial speed of the alpha particle is $2.21\times10^7 m/s$.

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25 million micrometeorites hit the surface of the moon daily. The Apollo astronauts' footprint will stay on the surface of the moon if it takes around 20 micrometeorites to damage it.

So, to calculate the duration, we'll need to find the number of footprints that have been damaged. We don't know how many footprints there are, so let's estimate that. Assume the average person walks at a rate of 1 step per second. Assume that each step is one foot in length. Assume the average person walks for 2 hours. Then, each person walks for 7200 seconds. The number of footprints per individual is 7200 x 1 = 7200. If we presume 12 people in total, the total number of footprints is 7200 x 12 = 86400.

Therefore, assuming that the footprints are uniformly distributed on the surface of the moon and that 25 million micrometeorites hit the moon's surface daily, the footprints are destroyed at a rate of 25,000,000/20 = 1,250,000 footprints per day.

The duration for the Apollo astronaut's footprints on the moon to remain intact:86400/1,250,000 = 0.06912 days, or roughly 1 hour and 40 minutes.

To calculate how long an Apollo astronaut's footprint would stay on the surface of the Moon, given that it takes around 20 micrometeorites to destroy a footprint, and given that 25 million micrometeorites hit the Moon's surface every day, we'll need to do some calculations. We'll begin by assuming that the footprints were uniformly distributed on the surface of the moon. We'll also assume that each person took 1 step per second, that each step is one foot in length, and that the average person walked for 2 hours. That means each person walked for 7200 seconds, or took 7200 steps. If we assume that there were 12 people on the Apollo mission, then the total number of footprints left by the astronauts would be 12 x 7200 = 86400.

Now, we need to figure out how quickly these footprints are being destroyed. Given that it takes around 20 micrometeorites to destroy a footprint, and given that 25 million micrometeorites hit the Moon's surface every day, we can calculate that the footprints are being destroyed at a rate of 25,000,000/20 = 1,250,000 footprints per day.

So, to find out how long it would take for the footprints to be destroyed, we divide the total number of footprints by the rate at which they are being destroyed:86400/1,250,000 = 0.06912 days, or roughly 1 hour and 40 minutes. Therefore, the length of time for the footprint to remain intact is approximately 1 hour and 40 minutes.

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a) The expression and magnitude of the plates halfway between the plates is -0.594 × 10⁶ V/m. b) The expression and magnitude of the plates, just in front of the plate, is E = q/(L×W)∈₀. c) the charge density is

-0.052×10⁻⁶ C/m².

Given information,

Distance between the plates, d = 0.1 m

Area, L×W = 0.19 m

Q = -1μC

a) The expression for the electric field,

E = q/(L×W)∈₀

E = -1×10⁻⁶/(0.19)8.85× 10⁻¹²

E = -0.594 × 10⁶ V/m

Hence, the electric field is -0.594 × 10⁶ V/m.

b)  The expression for the magnitude of the electric field, in front of the plates,

E = q/(L×W)∈₀

Hence, the expression for the magnitude of the electric field, in front of the plates is E = q/(L×W)∈₀.

c)  The charge density σ,

σ = Q/A

σ =   -1×10⁻⁶/0.19

σ = -0.052×10⁻⁶ C/m²

Hence, the charge density is -0.052×10⁻⁶ C/m².

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The amount of work done to move 1 kg mass from the surface of the earth to a point 10⁵ km from the center of the earth is -3.748 × 10^9 J.

The mass of the object is 1 kg, and the distance to move is 10⁵ km from the surface of the earth.

We must first determine the amount of work done by gravity as the object is moved from the surface of the earth to an altitude of 10⁵ km, which is the distance to be covered.

The formula for work done by gravity is given by;

Work done by gravity = -GmM/rwhere G = 6.674 × 10^-11 N.m^2/kg^2 is the gravitational constant, M = 5.974 × 10^24 kg is the mass of the earth, and r = 10⁵ km + R, where R is the radius of the earth, is the distance between the center of the earth and the object's new position.

Therefore,r = 10^5 km + 6.37 × 10^3 km = 1.06 × 10^8 m

The work done is given by the formula above.

Substituting the values,

Work done by gravity = -6.674 × 10^-11 × 1 × 5.974 × 10^24 / 1.06 × 10^8= -3.748 × 10^9 J

Therefore, the amount of work done to move 1 kg mass from the surface of the earth to a point 10⁵ km from the center of the earth is -3.748 × 10^9 J.

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