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3 years ago

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How much work did the movers do (horizontally) pushing a 41.0-kg crate 10.6 m across a rough floor without acceleration, if the

effective coefficient of friction was 0.60?

1 answer:

VLD [36.1K]3 years ago

5 0

Answer:

The required work done is $2555.448~J$

Explanation:

Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if ' $\mu_{k}$ ' be the coefficient of friction, then

$f = \mu_{k} \times N = \mu_{k} \times Mg$

where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.

Since the application of force by the movers does not create any acceleration to the block, we can write

$F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N$

So the work done (W) in moving the crate by a distance s = 10.6 m is

$W = F \times s = 241.08~N \times 10.6~m = 2555.448 J$

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A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force

Anettt [7]

Answer:

$VB - VA = - 33.4$

Explanation:

Generally the workdone in moving the proton is mathematically represented as

$W = KE_f - KE_i$

Where $KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy$

So

$KE_i = \frac{1}{2} m v_a^2$

Here $v_a$ is the velocity at A with value 50 m/s

So

$KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2$

$KE_i = 2.09 *10^{-24} \ J$

Also

$KE_f = \frac{1}{2} m v_b^2$

Here $v_a$ is the velocity at A with value $80 km/s = 80000 m/s$

=> $KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2$

=> $KE_f = 5.34 *10^{-18} \ J$

So

$W = 5.34 *10^{-18} - 2.09 *10^{-24}$

$W = 5.34 *10^{-18} m/s$

Now this workdone is also mathematically represented as

$W = q * V$

So

$q * V = 5.34 *10^{-18}$

Here $q = 1.60*10^{-19} C$

So

$V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}$

$V = 33.4 \ V$

Generally proton movement is in the direction of the electric field it means that $VA>VB$

So

$VB - VA = - 33.4$

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3 years ago

Suppose that the half-life of an element is 1000 years. How many half-lives will it take before one-eighth of the original sampl

34kurt

Answer:

3

Explanation:

The half-life of a radioactive isotope is the time it takes for the mass of the sample to halve.

This can be rewritten as follows:

$m(t) = m_0 (\frac{1}{2})^n$

where

m(t) is the mass of the sample at time t

m0 is the original mass of the sample

n is the number of half-lives that passed

We see that if we take n=3, the amount of original sample left is

$m(t) = m_0 (\frac{1}{2})^3 = m_0 (\frac{1}{8})$

So 3 (3 half-lives) is the correct answer.

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What will a spring scale read for the weight of a 75.0-kg woman in an elevator that moves upward with constant speed of 5.8 m/s

Sholpan [36]

<span>On the scale the only external forces are the man's weight acting downwards and the normal force which the scale exerts back to support his weight. So F = Ma = mg + Fs The normal force Fs (which is actually the reading on the scale) = Ma + Mg But a = 0 So Fs = Mg which is just his weight. Fs = 75 * 9.8 = 735N</span>

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Answer:

a new moon is quite near the Sun in the sky

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Answer:

A

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