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3 years ago

7

Two blocks a and b ($m_a>m_b$) are pushed for a certain distance along a frictionless surface. how does the magnitude of the

work that a does on b compare to the magnitude of the work that b does on a?

1 answer:

Yuki888 [10]3 years ago

3 0

Answer:

the magnitude of the work done by the two blocks is the same.

Explanation:

The work done by block a on block b is given by:

$W_a = F_a d$

where Fa is the force exerted by block a on block b, and d is the distance they cover.

The work done by block b on block a is given by:

$W_b = F_b d$

where Fb is the force exerted by block b on block a, and d is still the distance they cover.

For Newton's third law, the force exerted by block a on block b is equal to the force exerted by block b on block a, therefore

$F_a = F_b$

and so

$W_a=W_b$

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What prevents the pressure from increasing as a cloud contracts due to its gravity?

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Thermal energy is converted to radiative energy via molecular collisions and released as photons.

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If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di

xeze [42]

Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

So , displacement is :

D = 30 - 25 m = 5 m .

Therefore , distance and displacement covered is 55 m and 5 m respectively .

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3 years ago

A dragster starts with zero velocity and completes a 404.5 m (0.2528 mile) run in 4.922 s. If the car had a constant acceleratio

KIM [24]

Answer:

a. a=33.34ms⁻², V=164.4m/s

Explanation:

Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.

Below are the data given

Distance, s=404.5m,

time taken,t=4.922secs

Using the equation

S=ut+1/2at²

where u is the initial velocity and u=0

Making the acceleration the subject of the formula, we arrive at

a=2s/t²

a=(2*404.5)/4.922²

a=33.34ms⁻².

To determine the velocity, we use

V=u+at

V=0+33.34ms⁻² *4.922sec

V=164.4m/s

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3 years ago

Read 2 more answers

8. What is the frequency of green light waves that have a wavelength of 5.2 x 10-7 m.? The speed of light is 3.0 x 108 m/s

o-na [289]

Answer:

$f=5.76\times 10^{14}\ Hz$

Explanation:

We need to find the frequency of green light having wavelength o $5.2\times 10^{-7}\ m$ . It can be calculated as follows :

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{5.2\times 10^{-7}}\\\\f=5.76\times 10^{14}\ Hz$

So, the required frequency of green light is equal to $5.76\times 10^{14}\ Hz$ .

4 0

3 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

So

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

So

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

So

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

3 years ago