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3 years ago

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A boy slides a book across the floor using a force of five and over a distance of 2M what is the kinetic energy of the book afte

r he slides at assume there is no friction

1 answer:

Alinara [238K]3 years ago

5 0

Answer: 10Nm or 10J

Explanation:

Given the following :

Force (f) = 5

Distance (d) = 2m

Calculate the kinetic energy assuming no friction

Work done = force × distance

Work done = 5N × 2m = 10Nm

Recall :

Work done = ΔK.E ( change in kinetic energy)

Therefore, kinetic energy of the book after sliding = ΔK. E, which is equal to work done.

Hence, K. E of book after sliding is 10Nm

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3 years ago

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In a particular metal, the mobility of the mobile electrons is 0.0033 (m/s)/(N/C). At a particular moment, the electric field ev

Lapatulllka [165]

Answer:

the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

Explanation:

Given;

mobility of the mobile electrons in the metal, μ = 0.0033 (m/s)/(N/C)

the electric field strength inside the cube of the metal, E = 0.033 N/C

The average drift speed of the mobile electrons in the metal is calculated as;

v = μE

v = 0.0033 (m/s)/(N/C) x 0.033 N/C

v = 1.089 x 10⁻⁴ m/s.

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3 years ago

Which warning sign suggests being bullied?

Vikentia [17]

Answer:

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Explanation:

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3 years ago

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa

Sonja [21]

Answer:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

$c_{pA} > c_{pB}$

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

$T_{iA}= T_{iB}$

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

$Q = m c_p \Delta T$

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

$Q_A = Q_B$

And if we replace the formula for sensible heat we got:

$m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B$

And if we replace for the change of the temperature we got:

$m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})$

And since $T_{iA}= T_{iB}= T$ we have this:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

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3 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

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3 years ago