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marysya [2.9K]
2 years ago
7

A new mechanic foolishly connects an ammeter with 0.1 Ω resistance directly across a 12-V car battery with internal resistance o

f 0.01 Ω. What’s the power dissipation in the meter? (No wonder it gets destroyed!)
Physics
1 answer:
NemiM [27]2 years ago
8 0

Answer:

The power dissipated by the meter is 1188W

Explanation:

Here we have a circuit constituted with a power source and two resistors in series, we can calculate the power dissipated by the meter using the following formula:

P=I^2R_m

We first need to fin the current going through the circuit:

I=\frac{V}{R}\\where:\\V=voltage\\R=resistance

R=R_s+R_m

because they are connected in series. So:

I=\frac{12V}{(0.01+0.1)}=109A

P=(109)^2*0.1=1188W

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By vector addition.
In fact, velocity is a vector, with a magnitude intensity, a direction and a verse, so we can't simply do an algebraic sum of the two (or more velocities). 
First we need to decompose each velocity on both x- and y-axis (if we are on a 2D-plane), then we should do the algebraic sum of all the components on the x- axis and of all the components on the y-axis, to find the resultants on x- and y-axis. And finally, the magnitude of the resultant will be given by
R= \sqrt{(R_x)^2+(R_y)^2}
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3 years ago
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In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

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So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

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V_{Ay}=(21m/s)sin(-14^{o})

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we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

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we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

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