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bija089 [108]
1 year ago
5

Open the resistance in a wire simulation, adjust only the length of the resistor, what happens to resistor?

Physics
1 answer:
Eduardwww [97]1 year ago
4 0

If we increase the length of the resistor, resistance will also increase and vice-versa.

It is given that in a wire simulation, we have to adjust only the length of the resistor.

It required to find the effect on resistor when adjust only the length of the resistor.

<h2>What happens to resistor when adjust only the length of the resistor?</h2>

As we know that the length is directly proportional to the resistance of the wire that means is we increase the size or length of the wire then it also increase the resistance present in the wire. On the other hand if we decrease the length of the wire, it will decrease the resistance. This can be represented as:

Resistance = resistivity × length/area

This is also because there is more collision of the flowing electrons in the wire to the metal ions present there.

Thus if we increase the length of the resistor, resistance will also increase and vice-versa.

Learn more about the resistance here:

brainly.com/question/11431009

#SPJ4

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If a car is traveling forward at 15 m/s, how fast will it be going in 1.2 seconds if the acceleration is
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Answer:

3

Explanation:

v = v⁰ (its original speed) + a (negative acceleration) X t² (time)

v = 15 - 10 x 1.2 = 15 - 12 = 3 (it's slowing down)

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1 year ago
A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t
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Answers:

a) 65.075 kgm/s

b) 10.526 s

c) 61.82 N

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

I=\Delta p=p_{2}-p_{1} (1)

Where:

I is the impulse

\Delta p is the change in momentum

p_{2}=mV_{2} is the final momentum of the ball with mass m=0.685 kg and final velocity (to the right) V_{2}=57 m/s

p_{1}=mV_{1} is the initial momentum of the ball with initial velocity (to the left) V_{1}=-38 m/s

So:

I=\Delta p=mV_{2}-mV_{1} (2)

I=\Delta p=m(V_{2}-V_{1}) (3)

I=\Delta p=0.685 kg (57 m/s-(-38 m/s)) (4)

I=\Delta p=65.075 kg m/s (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 1.0 cm=0.01 m:

V_{2}=V_{1}+at (6)

V_{2}^{2}=V_{1}^{2}+2ad (7)

Where:

a is the acceleration

d=0.01 m is the length the ball was compressed

t is the time

Finding a from (7):

a=\frac{V_{2}^{2}-V_{1}^{2}}{2d} (8)

a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)} (9)

a=90.25 m/s^{2} (10)

Substituting (10) in (6):

57 m/s=-38 m/s+(90.25 m/s^{2})t (11)

Finding t:

t=1.052 s (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force F is proportional to the variation of momentum  \Delta p in time  \Delta t:

F=\frac{\Delta p}{\Delta t} (13)

F=\frac{65.075 kgm/s}{1.052 s} (14)

Finally:

F=61.82 N

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Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that ap
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Answer:

The object can have zero velocity and, simultaneously, nonzero acceleration.

The object can have zero acceleration and, simultaneously, nonzero velocity.

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Explanation:

An object in simple harmonic motion has a total mechanical energy (sum of elastic potential energy and kinetic energy) that is constant:

E=U+K=1/2kx^2 + 1/2}mv^2

where,

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We can note that the force on the spring is given by Hook's law:

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When the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

In all the other intermediate situations, both velocity and acceleration are nonzero.

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The object can have nonzero velocity and nonzero acceleration simultaneously.

The object can have zero velocity and, simultaneously, nonzero acceleration.

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