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borishaifa [10]

3 years ago

8

A 30-kg child sits at the top of a 3-meter slide. After sliding down, the child is traveling 4 m/s. How much PE does he start wi

th? How much KE does he end with? How much energy is lost to friction?

1 answer:

Jobisdone [24]3 years ago

6 0

At the top:

   Potential Energy = (mass) x (gravity) x (height)

   = (30 kg) x (9.8 m/s²) x (3 meters)

   = 882 joules

At the bottom:

   Kinetic Energy = (1/2) x (mass) x (speed)²

   = (1/2) x (30 kg) x (3 m/s)²

   = (15 kg) x (9 m²/s²)

   = 135 joules .

He had 882 joules of potential energy at the top,
but only 135 joules of kinetic energy at the bottom.

Friction stole (882 - 135) = 747 joules of his energy while he slid down.
The seat of his jeans must be pretty warm.

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lyudmila [28]

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$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
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By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
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part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
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5 0

2 years ago

Calculate the Latent Heat of Vaporization. (Please see picture attached)

Hunter-Best [27]

Answer:

20 J/g

Explanation:

In this question, we are required to determine the latent heat of vaporization

To answer the question, we need to ask ourselves the questions:

What is latent heat of vaporization?

It is the amount of heat required to change a substance from its liquid state to gaseous state without change in temperature.

It is the amount of heat absorbed by a substance as it boils.

How do we calculate the latent heat of vaporization?

Latent heat is calculated by dividing the amount of heat absorbed by the mass of the substance.

In this case;

Mass of the substance = 20 g
Heat absorbed as the substance boils is 400 J (1000 J - 600 J)

Thus,

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ivolga24 [154]

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2 years ago

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

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2 years ago

falling objects drop with an average acceleration of 9.8m/sec/sec. if an object falls from a tall building how long will it take

sveta [45]

Answer:

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<em> $a = \frac{v - u}{t}$ </em>

<em> $9.8 = \frac{49 - 0}{t}$ </em>

<em> $9.8 = \frac{49}{t}$ </em>

<em> $9.8t = 49$ </em>

<em> $t = \frac{49}{9.8}$ </em>

<em> $t = 5$ </em>

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