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3 years ago

14

How will a current change if the resistance of a circuit remains constant while the voltage across the circuit decreases to half

its original value?
1. The current will drop to half of its original value.
2. It is impossible to predict.
3. The current will triple.
4. The current will remain the same.
5. The current will double.

1 answer:

beks73 [17]3 years ago

4 0

Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

$V=RI$

where

V is the voltage across the circuit

R is the resistance of the circuit

I is the current

We can rewrite it as

$I=\frac{V}{R}$

In this problem, we have:

- the resistance of the circuit remains the same: R' = R

- the voltage is decreased to half of its original value: $V'=\frac{V}{2}$

So, the new current will be

$I'=\frac{V'}{R'}=\frac{V/2}{R}=\frac{1}{2}\frac{V}{R}=\frac{I}{2}$

so, the current will drop to half of its original value.

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3000 N is exerted for 4.0 seconds on a 9500 kg object.<br><br> What is the change in momentum?

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Force is the change in momentum over a specific time. The change of momentum is therefore the force multiplied by the time that the force acts, so 3000x4.0=12000 N s=12000 kg m/s

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After robbing a bank, a criminal tries to escape from the police by driving at a constant speed of 55 m/s (about 125 mph). A pol

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Answer:

18.03 s

Explanation:

We have two different types of motions, the criminal moves with uniform motion while the police do it with uniformly accelerated motion. Therefore we will use the equations of these cases. We know that by the time the police reach the criminal they will have traveled the same distance.

$x=vt\\x=x_{0}+v_{0}t+\frac{a}{2}t^2$

The distance between the police and the criminal when the first one starts the persecution is 0, its initial speed is also zero. So:

$x=(55m/s)t\\x=\frac{6.1m/s^2}{2}t^2=(3.05m/s^2)t^2$

Equalizing these two equations and solving for t:

$(55m/s)t=(3.05m/s^2)t^2\\(3.05m/s^2)t^2-(55m/s)t=0\\t((3.05m/s^2)t-55m/s)=0\\t=0 \\(3.05m/s^2)t-55m/s=0\\t=\frac{55m/s}{3.05m/s^2}=18.03 s$

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3 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

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Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

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And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

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We know that the kinetic energy for a mass m moving at speed v is:

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so, for the pellet will be

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<h3>All together</h3>

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Answer:

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Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They determine that a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s. What is the acceleration value of a cart with a mass of 2M when acted upon by a net force of 2F?

from newtons second law of motion ,

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we can say that

f=m(v-u)/t

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t=time

v=final velocity

u=initial velocity

since a=(v-u)/t

f=m*a

force applied is F

m =mass of the object involved

a is the acceleration of the object involved

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3 years ago

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Answer:

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