If a force of 50 N stretches a spring 0.40 m, what is the spring constant?

Answer:

wool and fibers

Explanation:

Answer:

Rahul's weight

Explanation:

In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.

Using formulas, Rahul's weight is equal to

W=mg

where m is Rahul mass and g is the gravitational acceleration (g=9.81 m/s^2).

Answer:

a) By [tex]v^2 = u^2 + 2as[/tex] => a= 70291.70.

(b)By [tex]v = u + at[/tex] => t= 1.58 ms.

(c)By [tex]v^2 = u^2 - 2gh[/tex] => H = 46045.92 m.

Explanation:

a) By [tex]v^2 = u^2 + 2as[/tex]

[tex](950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70[/tex]

(b)By [tex]v = u + at[/tex]

[tex]950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms[/tex]

(c)By [tex]v^2 = u^2 - 2gh[/tex]

[tex]0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m[/tex]

Answer:

a)  [tex]X_1=7.36rev[/tex]

b)  [tex]X_2=46.22radians[/tex]

c) [tex]X_3=2649.6^o[/tex]

Explanation:

From the question we are told that:

Diameter [tex]d=9.00[/tex]

Distance [tex]x=208[/tex]

Generally the equation for circumference of a circle is mathematically given by

[tex]C=2 \pi r\\\\C=2*\pi*4.5[/tex]

[tex]C=28.3[/tex]

Therefore

Angular distance that the pie plate has moved through in revolutions is

[tex]X_1=\frac{x}{C}[/tex]

[tex]X_1=\frac{208}{28.3}[/tex]

[tex]X_1=7.36rev[/tex]

Generally Angular distance that the pie plate has moved through in radians is

[tex]X_2= 7.36rev* 2 \pi[/tex]

[tex]X_2=46.22radians[/tex]

Generally Angular distance that the pie plate has moved through in degrees is

[tex]X_3=7.36rev* 360[/tex]

[tex]X_3=2649.6^o[/tex]

Answer:

a)3pF  has the greatest charge

b) 3pF to have the greatest voltage.

Explanation:

From the question we are told that:

1pF is parallel to [tex]2pF =1pF//2pF[/tex]

And 3pF is in with series 1pF is parallel to[tex]2pF =3pF+(1pF//2pF)[/tex]

Generally the equation for Resultant capacitor is mathematically given by

[tex]C=3pF+(1pF//2pF)[/tex]

[tex]C=\frac{3}{2}pF[/tex]

Ohm's law

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points

a)

Since

The same charge flowing through [tex]1pF//2pF[/tex] flow through 3pF

Therefore

3pF  has the greatest charge

b)

Voltage drop in series according to ohms law

Therefore the parallel share same voltage

Given

3pF to have the greatest voltage.

Answer:

An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume. ... In electric circuits the charge carriers are often electrons moving through a wire.

Answer:

The flow of moving electrons

Answer:

c. More than 2 s

Explanation:

First, we will find the length of the pendulum:

[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\2\ s = 2\pi \sqrt{\frac{l}{9.81\ m/s^2}}\\\\4\ s^2 = 4\pi^2 (\frac{l}{9.81\ m/s^2})\\\\l = \frac{(4\ s^2)(9.81\ m/s^2)}{4\pi^2} \\\\l = 0.99\ m[/tex]

Now, the value of g becomes 1.625 m/s² on the surface of the moon. So the time period will be:

[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\T = 2\pi \sqrt{\frac{0.99\ m}{1.625\ m/s^2}}\\\\[/tex]

T = 4.9 s

Therefore, the correct option is:

c. More than 2 s

Answer:

Gravitational potential energy is mass of the object times the gravitational constant times the height of the object:

U = mgh (I will use 10 for the gravitational constant but you can use 9.8 or 9.81 or something even more accurate)

U = 280

The gravitational potential of the object is 280 joules

Answer:

Explanation:

Look at the color scheme. That will help you a lot.

The metals are Na Mg and Al. They are colored Blue.

The Non metals are colored yellow.

Seven of the eight entries are taken up by yellow or blue. There is only 1 element left over and that is Si. So it must the metalloid. It has properties of the both the metals and the non metals.

Answer: silicon

Explanation:

Answer:

1. Carpenter

2. True

Explanation:

While a mason was working concrete into the formwork, the formwork collapses. The best person to rectify this problem is CARPENTER.

This is because it is the job of the Carpenter to design and build formwork, most especially wooden formwork. Formwork is like casing built to receive concrete and reinforcement during construction. Hence, when formwork collapses either due to stress, tension, or improper construction, it is the job of Carpenter to reconstruct the formwork or rectify the problem.

It is TRUE that when a device made in a workplace had defects. To address this issue the workshop manager should communicate directly with the workshop​. However, this communication will be an instruction on what to do next, and it usually directs those responsible to take action where necessary. For example, a workshop manager communicates to a carpenter about the need to rectify a chair or table that has a defect.

Answer:

B

Explanation:

The liquid evaporates in the solid is left in the dish..

Answer:

    v₂ = 63.62 m / s

Explanation:

For this exercise in fluid mechanics we will use Bernoulli's equation

         P₁ + ρ g v₁² +  ρ g y₁ = P₂ +  ρ g v₂² +  ρ g y₂

where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.

We will assume that the distance between the two parts is small, so y₁ = y₂

        P₁-P₂ =  ρ g (v₂² - v₁²)

pressure is defined by

        P = F / A

we substitute

        ΔF / A =  ρ g (v₂² - v₁²)

         v₂² = [tex]\frac{\Delta F}{A \ \rho \ g} + v_1^2[/tex]

suppose that the area of ​​the wing is A = 1 m²

we substitute

         v₂² = [tex]\frac{1000}{1 \ 1.29 \ 9.8} + 63^2[/tex]

         v₂² = 79.10 + 3969

         v₂ = √4048.1

         v₂ = 63.62 m / s

Work done by  man will be A. 640 J

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

according to work energy theorem

Work done = final Kinetic energy - initial kinetic energy

                   = KE (final) - KE (initial )

                   = 1/2 m ([tex]v^{2}[/tex])  - 1/2 m ([tex]u^{2}[/tex])

                  = 1/2 m ([tex]v^{2}[/tex] - [tex]u^{2}[/tex])

                  = 1/2 * 500 * ( [tex]2^{2}[/tex] - [tex]1.2^{2}[/tex])

                  = 250 * 2.56 = 640 J

correct answer is A. 640 J

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Answer:

(a) V = 3.75 x 10^5 V

(b) q = 5.2 x 10^-6 C

Explanation:

Diameter, d = 25 cm

radius, r = 12.5 cm = 0.125 m

Electric field, E = 3 x 10^6 V/m

(a) The maximum potential is given by

[tex]V = E \times r \\\\V = 3\times 10^6\times 0.125\\\\V = 3.75\times10^5 V[/tex]

(b) The charge is given by

[tex]V = \frac{k q}{r}\\\\3.75\times10^5=\frac{9\times10^9\times q}{0.125}\\\\q = 5.2\times 10^{-6} C[/tex]

Answer:

Force is the strength or weight of things that depends on movement. The types of forces are conteact force, spring force, applied force, air resistance force, normal force, frictional force, tension force, and non-contact force.

Answer:

Coefficient of friction =  Magnitude depends on the interacting materials

Friction = A force that acts parallel to the surface.

kinetic friction = A force that is constant regardless of the applied force

Normal = A force that acts perpendicular to the surface

Static friction = A force that increases as applied force increases up to some maximum value

Explanation:

Let's define all the forces, and then let's solve the problem.

Normal force:

When an object rests on some place (like a book on the table) the force that causes the book to not fall through the table is called the normal force, which is usually equal to the weight of the object and acts perpendicular to the surface where the object is resting.

Friction force.

When an object moves (or tries to move) parallel to a surface, such that the object is in contact with that surface, there appears a force that opposes to the movement (the force is parallel to the surface, and in the opposite direction to the movement).

And this force can be written as:

F = -N*μ

Where μ is the coefficient of friction and N is the normal force.

If the object is not moving yet (but there is applied a force that would move the object) the coefficient is called the coefficient of static friction which increases in a given range, until it can't keep increasing and the object starts to move, while if the object is moving, the coefficient is called the coefficient of kinetic friction and it is constant, where usually the first one is larger than the second, and these coefficients depend on both materials, the surface one and the object one.

Then we have two friction forces, one called the kinetic friction and the other called the static friction.

Then:

Coefficient of friction =  Magnitude depends on the interacting materials

Friction = A force that acts parallel to the surface.

kinetic friction = A force that is constant regardless of the applied force

Normal = A force that acts perpendicular to the surface

Static friction = A force that increases as applied force increases up to some maximum value

Answer:

As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. The black-body radiation graph is also compared with the classical model of Rayleigh and Jeans.

So as you see the wavelengths are in the x axis so all wavelengths are covered.

Black-body radiation provides insight into the thermodynamic equilibrium state of cavity radiation. If each Fourier mode of the equilibrium radiation in an otherwise empty cavity with perfectly reflective walls is considered as a degree of freedom capable of exchanging energy, then, according to the equipartition theorem of classical physics, there would be an equal amount of energy in each mode. Since there are an infinite number of modes this implies infinite heat capacity (infinite energy at any non-zero temperature), as well as an unphysical spectrum of emitted radiation that grows without bound with increasing frequency, a problem known as the ultraviolet catastrophe. Instead, in quantum theory the occupation numbers of the modes are quantized, cutting off the spectrum at high frequency in agreement with experimental observation and resolving the catastrophe. The study of the laws of black bodies and the failure of classical physics to describe them helped establish the foundations of quantum mechanics.

The above explains why the classical assumptions lead to a wrong spectrum.

Explanation:

i don't know if It helps you..parang Ang layo naman Ng sagot ko sa tanong mo

The wave model of light cannot explain the energy emissions from a blackbody radiator, but the particle model of light can be because the electromagnetic wave theory does not explain the black body radiation and the particle model of light can explain it.

Black body radiation is defined as the surface which absorbs all the energy and radiant light falling on it because it absorbs all light of color.

An example of black body radiation is a cavity with a small hole in it.

The Electromagnetic Wave is also called EM Waves which is defined as the waves that are created as a vibration between a magnetic field and electric field.

Light is defined as the electromagnetic radiation which propogates as waves and allows us to make the object visible.

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Explanation:

Reaction quotient is defined as the ratio of the concentration of the products and reactants of a reaction at any point of time with respect to some unit. It is represented by the symbol Q.

The ratio of the concentration of products and reactants of a reaction in equilibrium with respect to some unit is said to be equilibrium constant expression. It is represented by the symbol K.

The relationship between Gibbs free energy change and reaction quotient of the reaction is:

[tex]\Delta G=\Delta G^o+RT ln Q[/tex]           ......(1)

where,

[tex]\Delta G[/tex] = Gibbs free energy change

[tex]\Delta G^o[/tex] = Standard Gibbs free energy change

R = Gas constant

T = Temperature

At equilibrium, the free energy change of the reaction becomes 0 and standard Gibbs free energy change can be related to the equilibrium constant by the equation:

[tex]\Delta G^o=-RT ln Q[/tex]           ...(2)

Answer:Increasing

Explanation:

Given

Car is driven on the straight road with  instantaneous acceleration in the direction of car's motion.

If instanateneous acceleration is constant then speed of car is increasing at a constant pace. As there are no turns on the road, therefore speed of car is increasing.

The speed of the car is "decreasing". A further description is provided in the below paragraph.

Thus the above answer is correct.

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Answer:

L = 11.014 cm

Explanation:

Halfway between the two lenses is 19/2 = 9.5 cm.

Thus, this means virtually with respect to lens, the final image is at -9.5 cm

Thus, from here, we will work this out backwards.

Let's first solve for the initial position of the object for the second lens;

(1/S2) + (1/s'2) = (1/f2)

Where s'2 is the real image.

F2 is focal length

Thus;

(1/s'2) = (1/f2) - (1/s2)

(1/s'2) = (1/15) - (1/-9.5)

(1/s'2) = 0.1719

s'2 = 5.82 cm

The object for the second lens is located at 5.82 cm in front of the second lens.

Now, The object for the second lens and the image for the first lens will be the same.

This means the distance of the image from the first lens is at; 19 - 5.82 = 13.18 cm.

Now let's solve for the object distance of the first lens which will be denoted by L.

1/L = (1/f1) + (1/s'1)

Where f1 = 6 cm

1/L = (1/6) - (1/13.18)

1/L = 0.090794

L = 1/0.090794

L = 11.014 cm

Answer:

9.965 nF

Explanation:

The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m

So, C = εA/d

C = ε2πrL/d

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

C = ε2πrL/d

C =  24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m

C =  9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m

C =  996463 × 10⁻¹⁴ F

C = 9.96463 × 10⁻⁹ F

C = 9.96463 nF

C ≅ 9.965 nF

Answer:

the answer is c which is a+2 charge

Explanation:

Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.

The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.

In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.

This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.

Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.

The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.

Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.

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Answer:

a) the cheetah's top speed is 64.4 miles/hr

b) time taken to reach top speed is 3.3 seconds

Explanation:

Given the data in the question;

Cheetahs can accelerate to a speed of 21.8 m/s in 2.50 s.

They can continue to accelerate to reach a top speed of 28.8 m/s.

a) Express the cheetah's top speed in mi/h. (Express your answer to three significant figures.)

The cheetah's top speed = 28.8 m/s = ( 28.8 × 2.237 ) miles/hr

= 64.4256 ≈ 64.4 miles/hr

Therefore, the cheetah's top speed is 64.4 miles/hr

b) Starting from a crouched position, how long does it take a cheetah to reach its top speed.

given that

v₁ = 21.8 m/s   and    t₁ = 2.50 s

let a represent the acceleration of the cheetah

From the First Equation of Motion;:

v = u + at

we substitute

21.8 = 0 + ( a × 2.50 )

21.8 = a × 2.50

a = 21.8 / 2.50

a = 8.72 m/s²

Now, let the time taken by cheetah to reach top speed ( 28.8 m/s ) be t

so from the first equation of motion;

v = u + at

we substitute

28.8 = 0 + ( 8.72 × t )

t = 28.8 / 8.72

t = 3.3 seconds

Therefore, time taken to reach top speed is 3.3 seconds

Answer:

Explanation:

Wind is caused by the uneven heating of the atmosphere by the sun, variations in the earth's surface, and rotation of the earth. ... Wind turbines convert the energy in wind to electricity by rotating propeller-like blades around a rotor. The rotor turns the drive shaft, which turns an electric generator

Answer:

x = 132.7 m

y = 51.34 m

Explanation:

Given :

Initial speed, u = 49.5 m/s²

Angle of projection, θ = 40°

Time, t = 3.50 seconds

The distance, x = horizontal component ;

Distance = speed * time

Distance = uCosθ * 3.50

Distance = 49.5 * Cos40° * 3.50

Distance = 49.5 * Cos40° * 3.50

Horizontal distance = 132.7 m

Vertical distance, y :

Sy = ut + 1/2gt²

Sy = Vertical distance ; g = 9.8 m/s²

Sy = 49.5 * sin40 * 3.5 - (0.5 * 9.8 * 3.5²)

Sy = 111.36295 - 60.025

Sy = 51.33795 m

x = 132.7 m

y = 51.34 m

Answer:

The angle of minimum deviation is 28.72°

Explanation:

u welcome

Answer:

ΔV = 2 10¹ V

Explanation:

The calculation of the uncertainty or error in an expression is given by

         ΔV = [tex]\frac{dV}{di}[/tex]  |Δi| + [tex]\frac{dV}{dR}[/tex]  |ΔR |

         V = i R

let's make the derivatives

        [tex]\frac{dV}{di}[/tex] = R

        [tex]\frac{dV}{dR}[/tex] = i

we substitute

         ΔV = R | Δi | + i | ΔR |

in the exercise give the values

         i = (5.9 ± 0.4) A

         R = (42.7 ± 0.6) Ω

we calculate

          ΔV = 42.7  0.4 + 5.9  0.6

          ΔV = 20.6 V

          ΔV = 2 10¹ V

the voltage is

         V = i R

         V = 5.9  42.7

          V = 251.9 V

the result is

         V = (25 ± 2) 10¹ V

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

f is the force

m is the mass

From the question we have

[tex]a = \frac{4200}{1500} = \frac{42}{15} \\ = 2.8[/tex]

We have the final answer as

Hope this helps you

Answer:

Approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] (assuming that the boiling point of water in this experiment is [tex]100\; \rm ^\circ C\![/tex].)

Explanation:

Latent heat of condensation/evaporation of water: [tex]2260\; \rm J \cdot g^{-1}[/tex].

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to [tex]\rm J \cdot g^{-1}[/tex].

Specific heat of water: [tex]4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}[/tex].

Specific heat of copper: [tex]0.39\; \rm J \cdot g^{-1}\cdot K^{-1}[/tex].

The temperature of this calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains increased from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]. Calculate the amount of energy that would be absorbed:

[tex]\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}[/tex].

[tex]\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}[/tex].

Hence, it would take an extra [tex]585\; \rm J + 31500\; \rm J = 32085\; \rm J[/tex] of energy to increase the temperature of the calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].

Assume that it would take [tex]x[/tex] grams of steam at [tex]100\; \rm ^\circ C[/tex] ensure that the equilibrium temperature of the system is [tex]50\; \rm ^\circ C[/tex].

In other words, [tex]x\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] would need to release [tex]32085\; \rm J[/tex] as it condenses (releases latent heat) and cools down to [tex]50\; \rm ^\circ C[/tex].

Latent heat of condensation from [tex]x\; \rm g[/tex] of steam: [tex]2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J[/tex].

Energy released when that [tex]x\; {\rm g}[/tex] of water from the steam cools down from [tex]100\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]:

[tex]\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}[/tex].

These two parts of energy should add up to [tex]32085\; \rm J[/tex]. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].

[tex](2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J[/tex].

Solve for [tex]x[/tex]:

[tex]x \approx 13[/tex].

Hence, it would take approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] for the equilibrium temperature of the system to be [tex]50\; \rm ^\circ C[/tex].

Answer: A 10,000-Hz sound is 10 times more intense as compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness.

Explanation:

The formula used is as follows.

[tex]\beta = 10 dB log (\frac{I}{I_{o}})\\60 = 10 dB log (\frac{I}{I_{o}})[/tex]

[tex]I_{o} = 10^{-12}[/tex] normal threshold

The difference is sound level is as follows.

60 - 60 = 0

Hence,

[tex]0 = 10 dB [log (\frac{I_{f}}{I_{o}}) - log (\frac{I_{i}}{I_{o}})]\\log (\frac{1000}{I_{o}}) = log (\frac{10000 x}{I_{o}})\\log (10^{15}) = log (10^{16}x)\\15 = 16 + log x\\log x = 1\\x = 10[/tex]

This means that 10,000 Hz sound is 10 times more intense.

Thus, we can conclude that a 10,000-Hz sound is 10 times more intense as compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness.