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4 years ago

9

A copper telephone wire has essentially no sag between poles 36.0 m apart on a winter day when the temperature is −20.0°C. How m

uch longer is the wire on a summer day when the temperature is 34.0°C?

1 answer:

RideAnS [48]4 years ago

4 0

Answer:

The extension is $\Delta L = 0.033 \ m$

Explanation:

From the question we are told that

The length of the wire on a winter day is $L_w = 36.0 \ m$

The temperature on the winter day is $T_w = -20.0 ^o C$

The temperature on a summer day is $T_s = 34.0 ^0 C$

The the extension of the wire on a summer day is mathematically represented as

$\Delta L = \alpha L_w [T_s - T_w]$

Where

$\alpha$ is the coefficient of linear expansion of copper with a values $\alpha = 17 *10^{-6}$

substituting value

$\Delta L = 17 *10^{-6} * 36.0 [34 - [-20]]$

$\Delta L = 0.033 \ m$

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3 years ago

A 24 kg child slides down a 3.3-m-high playground slide. She starts from rest, and her speed at the bottom is 3.0 m/s.a. What en

Gelneren [198K]

Answer:

(a) Potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

(b) $U=668.16\ J$

Explanation:

Given:

mass of the child, $m=24\ kg$

height of the slide, $h=3.3\ m$

initial velocity of the child at the slide, $v_i=0 m.s^{-1}$

final velocity of the child at the bottom of slide, $v_f=3\ m.s^{-1}$

(a)

∴The initial potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

Initial potential energy:

$PE=m.g.h$

$PE=24\times 9.8\times 3.3$

$PE=776.16\ J$

Kinetic energy at the bottom of the slide:

$KE=\frac{1}{2} m.v^2$

$KE= 0.5\times 24\times 3^2$

$KE= 108\ J$

(b)

Now, the difference in the potential and kinetic energy is the total change in the thermal energy of the slide and the seat of her pants.

This can be given as:

$U=PE-KE$

$U=776.16-108$

$U=668.16\ J$

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3 years ago

You and your neighbour are shovelling snow. If your neighbour pushes a shovel four times as far as you do but exerts only a half

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Answer:

The neighbour

Double your work

Explanation:

Work is a product of force and distance, expressed as W=Fd where W is work in Nm, F is applied force and d is the distance in m

Since your neighbour pushes 4 times then you just push 1m when he pushes 4m, that is an example. Also, since the neighbour only exerts half the force ie 0.5N then you exert full force of 1N

Work from your neighbour will be 4d*0.5F=2dF

Work from you will be dF

Clearly, 2dF=2(dF) hence the neigjbour exerts double work

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4 years ago

(based on 15-103 in the text) At an initial instant, a 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a sp

lisov135 [29]

Answer:

a

$v_r =8.65 \ ft/s$

b

$W_{1-2} = 3.24 \ ft \cdot lb$

Explanation:

From the question we are told that

The mass of the ball is $m = 4 \ lb$

The radius is $r= 3 \ ft$

The speed is $v_B_1 = 4.8 \ ft /s$

The speed of the attached cord is $v_c =2.2 \ ft$

The position that is been considered is $r_1 = 2 \ ft$ h

Generally according to the law of angular momentum conservation

$L_a = L_b$

Here $L_a$ is the initial momentum of the ball which is mathematically represented as

$L_a = m* v_B_1 * r$

while

$L_b$ is the momentum of the ball at r = 2 ft which is mathematically represented as

$L_a = m* v_B_2 * r_1$

So

$m* v_B_1 * r = m* v_B_2 * r_1$

=> $4.8 * 3 = v_B_2 * 2$

=> $v_B_2 = 7.2 \ ft/s$

Generally the resultant velocity of the ball is

$v_r = \sqrt{v_B_2^2 + v_B_1^2 }$

=> $v_r = \sqrt{7.2^2 + 4.8^2 }$

=> $v_r =8.65 \ ft/s$

Generally according to equation for principle of work and energy we have that

$K_1 + \sum W_{1-2} = K_2$

Here $K_1$ is the initial kinetic energy of the ball which is mathematically represented as

$K_1 = \frac{1}{2} * m* v_B_1^2$

While $\sum W_{1-2}$ is the sum of the total workdone by the ball

and $K_2$ is the final kinetic energy of the ball which is mathematically represented as $K_2 = \frac{1}{2} * m* v_r^2$

So

$\sum W_{1-2} = \frac{1}{2} * m (v_r^2 - v_B_1^2)$

Here m is the mass which is mathematically represented as

$m = \frac{W}{g}$ here W is the weight in lb and g is the acceleration due to gravity which is $g = 32 \ ft/s^2$

So

$\sum W_{1-2} = \frac{1}{2} * \frac{4}{32} * (8.65^2 - 4.8^2)$

=> $W_{1-2} = 3.24 \ ft \cdot lb$

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3 years ago

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Soloha48 [4]

answer would be A becuz that's the only reasonable choice according to real life situations

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4 years ago