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3 years ago

What is the resistance of a 3.5 m copper wire (Rho= 1.7x10-8 Ohm·m) that 1 point

Physics

1 answer:

VikaD [51]3 years ago

4 0

Answer:

(D)

Explanation:

Given :

l=3.5 m

$A=5.26*10^{-6}$ $m^{2}$

$p=1.7*10^{-8} ohm.m$

Resistance can be calculated as :

$R=p\frac{l}{A} \\R=1.7*10^{-8} \frac{3.5}{5.26*10^{-6} }$

$R=\frac{5.95*10^{-2} }{5.26} \\R=1.13*10^{-2}$

Resistance of the wire will be 1.1× $10^{-2}$ ohms

Option D is correct

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First of all we state the formula
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3 years ago

A heat engine does 200 j of work per cycle while exhausting 600 j of heat to the cold reservoir. what is the engine's thermal ef

AveGali [126]

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3 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

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$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

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Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$

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3 years ago

In three sentences, please describe SIMPLE HARMONIC motion, and give two examples. Thank you! :-)

11Alexandr11 [23.1K]

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3 years ago