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Answer:

the average force exerted by the rain on the roof is 1.065 N

Explanation:

Given;

speed of the rainfall, v = 15 m/s

the mass of the rate of the rainfall, m' (m/t) = 0.071 kg

Let the average force exerted by the rain on the roof  = F

The average force exerted by the rain on the roof  is calculated by Newton's second law of motion;

[tex]F = ma = m \frac{v}{t} = \frac{m}{t} \times v \\\\Recall, \ m' = \frac{m}{t} \\\\F = m'\times v\\\\F = 0.071 \times 15\\\\F = 1.065 \ N[/tex]

Therefore, the average force exerted by the rain on the roof is 1.065 N

(a) Attached to the response as Figure 1.

(b) 35.0Ω

(c) Across 5.0Ω = 1.3V

   Across 10.0Ω = 2.6Ω

   Across 20.0Ω = 5.2Ω

(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.

(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e

R = R₁ + R₂ + R₃               -------------------(i)

Where;

R₁ = 5.0 Ω

R₂ = 10.0 Ω

R₃ = 20.0 Ω

Substitute these values into equation (i) as follows;

∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω

∴ R = 35.0 Ω

Therefore, the equivalent resistance is ∴ R = 35.0Ω

(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;

i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.

ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;

V = I R              ------------------(ii)

Where;

V = Voltage supplied to the circuit = 9.0V

I = Current through the circuit

R = Resistance of the equivalent resistor = 35.0Ω

Substitute these values into equation (ii)

9.0 = I x 35.0

I = [tex]\frac{9.0}{35.0}[/tex]

I = 0.26A

This is also the current flowing through each of the resistors separately.

iii. Calculate the voltage drop across

1. 5.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 5.0Ω resistor

I = current through the 5.0Ω resistor = 0.26A

R = resistance of the 5.0Ω resistor = 5.0Ω

=> V = 0.26 x 5.0

=> V = 1.3V

2. 10.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 10.0Ω resistor

I = current through the 10.0Ω resistor = 0.26A

R = resistance of the 10.0Ω resistor = 10.0Ω

=> V = 0.26 x 10.0

=> V = 2.6V

3. 20.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 20.0Ω resistor

I = current through the 20.0Ω resistor = 0.26A

R = resistance of the 20.0Ω resistor = 10.0Ω

=> V = 0.26 x 20.0

=> V = 5.2V

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

(c) vf = 7.17 m/s

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

[tex]2gh = v_f^2 - v_i^2[/tex]

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]

h = 0.82 m

Now, for the time in air during upward motion we use first equation of motion:

[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]

(c)

Now we will consider the downward motion and use the third equation of motion:

[tex]2gh = v_f^2-v_i^2[/tex]

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]

vf = 7.17 m/s

Now, for the time in air during downward motion we use the first equation of motion:

[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

t = 1.14 s

Answer:

D. Non- polar solvant

Explanation:

l think that's it

Answer:

I think the answer is D polar solvent

Answer:

Tenemos dos problemas a resolver acá:

Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.

Acá podemos usar la conservación de la energía.

E = U + K

U = energía potencial = m*g*H

m = masa

g = aceleración gravitatoria = 9.8m/s^2

H = altura

K = energía cinética = (m/2)*V^2

donde V es la velocidad.

Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:

Ei = U = m*(9.8m/s^2)*2.5m

Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:

Ef = (m/2)*V^2

Y como la energía se conserva, la energía final es igual a la inicial, entonces:

m*(9.8m/s^2)*2.5m = (m/2)*V^2

Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.

V = √(2*(9.8m/s^2)*2.5m) = 7m/s

Ahora respondamos la segunda parte.

Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:

A(t) = -9.8m/s^2

Para obtener su velocidad integramos:

V(t) = (-9.8m/s^2)*t + V0

donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s

V0 = (3/4)*7m/s = (21/4) m/s

Así, la ecuación de la velocidad es:

V(t) = (-9.8m/s^2)*t + (21/4) m/s

Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:

V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s

         t =  (21/4) m/s/9.8m/s^2 = 0.54 s

Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.

Para ello integramos de vuelta:

P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0

donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:

P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t  

La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s

P(0.54s) =  (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m

La altura máxima es 1.81 metros.

Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.

The question is incomplete, the complete question is;

In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?

A) The atom moves to a state of lower energy

B) The atom is ionized

C) One of the electrons leaves the atom

D) The atom can be excited to a higher energy state

Answer:

The atom can be excited to a higher energy state

Explanation:

According to the Bohr model of the atom, electrons in an atom can be excited from a lower to a higher energy level when energy is absorbed by the atom.

If electrons having an energy of 10.5ev are incident on a hydrogen atom, this energy is transferred to the atom by collision. Since the energy transferred is less than the ionization energy of hydrogen atom in its ground state(13.6ev), the atom is not ionized.

Rather, the atom is excited from ground state to a higher energy level.

Explanation:

a) [tex]I=\displaystyle \sum_{i}m_ir_i^2[/tex]

where [tex]r_i[/tex] is the distance of the mass [tex]m_i[/tex] from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,

[tex]I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2[/tex]

b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,

[tex]\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2[/tex]

Answer:

3.034 m

Explanation:

From the law of conservation of energy, the energy at the top of the incline equals the energy at the bottom of the incline since at the top of the incline, the horizontal surface is frictionless and along the incline there is no friction.

The work done in moving the crate a distance, d = 2.7 m with a force of F = 121 N to the top of the incline is W = Fd = 121 N × 2.7 m = 326.7 J.

From work-kinetic energy principles, this work W = kinetic energy of the crate at the top of the incline, K₁.

Now, the total mechanical energy at the top of the incline, E equals the total mechanical energy at the bottom of the incline E' since there is no friction along the incline.

So, E = E'

U₁ + K₁ = U₂ + K₂ where U₁ = potential energy of crate at top of incline = mgh where m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s², h = height of incline = 1.8 m, K₁ = kinetic energy of crate at top of incline = 326.7 J, U₂ = potential energy of crate at bottom of incline = 0 J(since it is at an elevation h = 0) and K₂ = kinetic energy of crate at bottom of incline

So, substituting the values of the variables into the equation, we have

U₁ + K₁ = U₂ + K₂

mgh + K₁ = U₂ + K₂

28.9 kg × 9.8 m/s² × 1.8 m + 326.7 J = 0 J + K₂

509.796 J + 326.7 J = K₂

K₂ = 836.496 J

K₂ ≅ 836.5 J

Now since the vertical wall is a distance d2 away and the long ideal spring has a length dy = 3.4 m, let x be the compression of the spring. So, the distance moved by the crate is thus D = d2 - dy - x.

Now, the change in kinetic energy of the crate ΔK equals the work done by friction and that done by the spring W.

So ΔK = -W (from work-kinetic energy principles)

Let W' = work done by friction = μmgD where  μ = coefficient of kinetic friction between surface and crate = 0.41, m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s² and D = distance moved by crate = D = d2 - dy - x = 5.2 m - 3.4 m - x = 1.8 - x

So, W' = μmgD

W' = 0.41 × 28.9 kg × 9.8 m/s² (1.8 - x)

W' = 116.12(1.8 - x)

W' = 2090.16 - 116.12x

The work done by the spring W" = 1/2k(x₀² - x²) where k = spring constant = 154 N/m, x₀ = initial spring length = dy = 3.4 m and x = final spring compression.

So,  W" = 1/2k(x₀² - x²)

W" = 1/2 × 154 N/m[(3.4 m)² - x²]

W" = 77 N/m[11.56 m² - x²]

W" = 890.12  - 77x²

So, W = W' + W"

W = 2090.16 - 116.12x + 890.12  - 77x²

W = 2980.28 - 116.12x - 77x²

Since the crate stops, final kinetic energy K₃ = 0. So, ΔK = K₃ - K₂ = 0 - 836.5 J = -836.5 J

Also, ΔK = -W

-836.5 = -(2980.28 - 116.12x - 77x²)

836.5 = 2980.28 - 116.12x - 77x²

77x² + 116.12 -2980.28 + 836.5 = 0

77x² + 116.12x -2143.78 = 0

dividing through by 77, we have

x² + 1.508x -27.841 = 0

Using the quadratic formula to find x, we have

[tex]x = \frac{-1.508 +/-\sqrt{1.508^{2} - 4 X 1 X (-27.841)} }{2 X 1.508} \\x = \frac{-1.508 +/-\sqrt{2.274064 + 111.364} }{3.016} \\x = \frac{-1.508 +/-\sqrt{113.638064} }{3.016} \\x = \frac{-1.508 +/- 10.66}{3.016} \\x = \frac{-1.508 - 10.66}{3.016} or x = \frac{-1.508 + 10.66}{3.016} \\x = \frac{-12.168}{3.016} or x = \frac{9.152}{3.016} \\x = -4.03 or 3.034[/tex]

x = -4.03 or 3.034

Since the compression of the spring is positive, we choose x = 3.034

So, the crate compresses the spring 3.034 m

Answer:

The correct solution is "11.51 mA".

Explanation:

Given:

Time average power,

[tex]P_{avg}=0.05 \ W/m^2[/tex]

n = 377

As we now,

⇒ [tex]P_{avg}=\frac{E_0^2}{n}[/tex]

or,

⇒ [tex]E_0^2=0.05\times 377[/tex]

⇒       [tex]=4.341 \ V[/tex]

hence,

⇒ [tex]H_0=\frac{E_0}{n}[/tex]

By putting the values, we get

         [tex]=\frac{4.341}{377}[/tex]

         [tex]=11.51 \ mA[/tex]

Answer:

8.5 feet.

Explanation:

A sketch of the position of the three people gives a right angled triangle. The hypotenuse of the triangle gives the resultant displacement, while the two other sides are 6 feet each.

The resultant displacement, R, is the overall displacement covered by the doctor. This can be determined by;

R = [tex]\sqrt{d_{1} ^{2} + d_{2} ^{2} }[/tex]

where: [tex]d_{1}[/tex] = 6 feet and [tex]d_{2}[/tex] = 6 feet.

Thus,

R = [tex]\sqrt{6^{2} + 6^{2} }[/tex]

  = [tex]\sqrt{72}[/tex]

R = 8.49

The resultant displacement covered by the doctor is 8.5 feet.

Answer:

The interpersonal environment is considered to be a subset of the organizational environment – defined as the employee’s perception of the practices, policies, and processes of an organization

Explanation:

Answer:

wait

Explanation:

Answer:

English please!!!!!!!!!!!!!!!!!!!!!!!

Answer:

M' = μ₀n₁n₂πr₂²

Explanation:

Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.

So, M = N₂Ф₂₁/i₁

substituting the values of the variables into the equation, we have

M = N₂Ф₂₁/i₁

M = N₂B₁A₂/i₁

M = n₂lμ₀n₁i₁πr₂²/i₁

M = lμ₀n₁n₂πr₂²

So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²

M' = μ₀n₁n₂πr₂²

Answer:

O The environment did work on an object

Explanation:

Answer:

8 N

Explanation:

Applying,

(F'+F) = ma............... Equation 1

Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.

From the question,

Given: F = 12 N, m = 4 kg, a = 5 m/s²

Substitute these values into equation 1

(F'+12) = 4×5

(F'+12) = 20

F' = 20-12

F' = 8 N.

Hence Ahmed's force is 8 N

Answer:

LOL PUEDO HABLAR CON

Explanation:

Answer:

Whats the anwser/????

Explanation:

Answer:

The power is 465.44 W.

Explanation:

mass, m = 70 kg

number of steps,  n = 1576

height of each step, h = 0.241 m

time taken, t = 9.33 min= 9.33 x 60 s

The power is given by the rate of doing work.

W = n m g h

W = 1576 x 70 x 9.8 x 0.241

W = 260553.776 J

The power is given by

[tex]P = \frac{W}{t}\\\\P = \frac{260553.776}{9.33\times 60}\\\\P = 465.44 W[/tex]  

Answer: When it comes to the movement of air, friction is greater near the ground surface.

Explanation:

A resistance in motion observed by an object while on another object is called friction.

For example, a vehicle moving on road will have friction between its tires and the road.

Friction is more near the ground surface rather than away from the ground surface.

Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.

Accuracy of measurement is the degree of how close the measurement is, to the actual measurement.

For a measurement of 2.00 m, 1.99 m is more accurate because it is closer to 2.00 m than 1.95 m.

The set of data is not given. So, I will give a general explanation.

For a measurement to be the accurate, the measurement must be close to the actual measurement, and in some cases, the measurement should be approximated to the actual measurement.

Given that:

[tex]Length =2.00m[/tex]

An example of an accurate measurement is:

[tex]Length =1.95m[/tex]

Why? The reasons are

Another example is:

[tex]Length = 1.99m[/tex]

1.99 m is also an accurate measurement because of the reasons above.

However, 1.99 m is more accurate because it is closer to 2.00 m than 1.95 m

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Answer:

B

Explanation:

Answer:

Sultan Kösen

here is a pic

Answer:

a) B = 3.11 km.  θ= 54.7º E of S

b) B = 3.11 km  θ= 54.7º W of S

Explanation:

a)

        [tex]C=\sqrt{(2.2km)^{2} + B^{2} } = 3.81 Km (1)[/tex]

       [tex]B=\sqrt{(3.81km)^{2} -(2.2km)^{2} } = 3.11 Km (2)[/tex]

        [tex]\theta = arc tg (\frac{B}{A} )= arc tg ( \frac{3.11}{2.2} )= arctg (1.414) = 54.7 deg (3)[/tex]

b)

      [tex]\theta = arc tg (\frac{-B}{A} )= arc tg ( \frac{-3.11}{2.2} )= arctg (-1.414) = -54.7 deg (4)[/tex]

Answer:

a. The disk

b. Because it has the smallest rotational inertia

Explanation:

a. Which object do you expect to reach the bottom of the inclined plan first?

I would expect the disk to reach the bottom first.

b. Why?

This is because the disk has the smallest rotational inertia.

The rotational inertial of the hollow sphere, disk and ring are 2/3MR², 1/2MR² and MR² respectively.

Since the three objects are rolling from the same height, they have the same mechanical energy.

But, since the disk has the smallest rotational inertia, it would have the smallest rotational kinetic energy and largest translational kinetic energy.  The disk's smaller rotational kinetic energy will cause  to rotate less but translate more than the other objects and thus reach the bottom first.

The object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.

Moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.

There are three objects, hollow sphere, disk and ring.

       [tex]I=\dfrac{2}{3}mr^2[/tex]

        [tex]I=mr^2[/tex]

Here, (m) is the mass and (r) is the radius of the object.

These three objects are going to race by releasing from rest at the top of an inclined plane to the bottom of the plane.

As moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.

Thus the less the value of inertia will result in less the time required to reach at the bottom of the inclined plane.

Hence, the object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.

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Answer:

O Distant objects are blurry. describes farsightedness.

Explanation:

Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.

A neutron changes into a proton during a radioactive decay process called beta decay (positron), which is option D.

A beta decay in physics is a nuclear reaction in which a beta particle (electron or positron) is emitted.

A positron is an electron with a positive charge.

During a beta decay, a neutron in the nucleus of the radioactive material suddenly changes into a proton, causing an increase in the atomic number of an element.

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#SPJ1

Answer:

I HOPE THIS IS CORRECT

Explanation:

Power of water =2 kw=2000w

Mass of water =200kg

difference in temperature ΔT=70−10=60oC

Concept

energy required to heat the water = energy given by water in time t=pt

energy required to increase tempeature of water by 60oC,Q=msΔT

S= specific heat =4200J/kgoC

              pt=msΔT

   2000×t=200×4200×60

      t=25200  

or   t=25.2×103sec.

The quantity of heat generated is 720,000 Joules or 720 kiloJoules.

Given the following data:

To find the quantity of heat generated when the electrical coil is used:

Note: the quantity of heat generated is energy.

First of all, we would convert the time in hour to seconds.

Conversion:

1 hour = 3600 seconds

Mathematically, the energy of a device is given by the formula;

[tex]Energy = power[/tex] × [tex]time[/tex]

Substituting the values into the formula, we have;

[tex]Energy = 200[/tex] × [tex]3600[/tex]

Energy = 720,000 Joules or 720 kiloJoules.

Therefore, the quantity of heat generated is 720,000 Joules or 720 kiloJoules.

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Answer:

B = 2.9 mT

Explanation:

Given that,

The distance between the wires, r = 2 cm

Distance halfway between them d = 1 cm = 0.01 m

The current in the wires, I = 145 A

We need to find the strength of the magnetic field they produce, halfway between them. The formula for the magnetic field in the wire is given by :

[tex]B=\dfrac{\mu_o I}{2\pi d}\\\\B=\dfrac{4\pi \times 10^{-7}\times 145}{2\pi \times 0.01}\\\\B=2.9\ mT[/tex]

So, the required magnetic field is 2.9 mT.

a. The force exerted by the ground on each of the front wheels is 4918.16 Newton.

b. The force exerted by the ground on each of the back wheels is 2627.84 Newton.

Given the following data:

a. To determine the force exerted by the ground on each of the front wheels:

First of all, we would take moment about the rear wheels.

[tex]F(3.13) - 1540(9.8) \times (3.13 - 1.09) = 0\\\\3.13F - 15092 \times 2.04 =0\\\\3.13F -30787.68=0\\\\F=\frac{30787.68}{3.13}[/tex]

Force, F = 9836.32 Newton

For each front wheel:

[tex]Force = \frac{9836.32}{2}[/tex]

Force = 4918.16 Newton.

b. To determine the force exerted by the ground on each of the back wheels:

We would determine the sum of the vertical forces acting on the wheels.

[tex]9836.32 + B - 1540(9.8) = 0\\\\9836.32 + B - 15092 = 0\\\\B=15092-9836.32[/tex]

B = 5255.68 Newton.

For each back wheel:

[tex]Force = \frac{5255.68}{2}[/tex]

Force = 2627.84 Newton.

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