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3 years ago

Can someone help me solve this its a simple series circuit

Physics

1 answer:

daser333 [38]3 years ago

8 0

Answer:

See the explanation below.

Explanation:

Indeed this is a serial circuit, in order to calculate the different questions, we must calculate the current of the circuit. The total resistance of the circuit can be calculated by means of an algebraic summation.

Then using ohm's law,we can calculate the current.

Rt = 2 + 4 + 6 = 12 [ohm]

The current

I = V / Rt

I = 12 / 12

I = 1 [amp]

Therefore the voltage in each resistance will be:

V1 = 2 * 1 = 2 [V]

V2 = 4 * 1 = 4 [V]

V3 = 6 * 1 = 6 [V]

The total voltage will be:

V = V1 + V2 + V3 = 12 [V]

The power can be calculated by multiplying the voltage by the total current.

P = V * I

P = 12 * 1

P = 12 [W]

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(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

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Answer:

1.43 s

Explanation:

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The vertical distance covered by an object in free fall is given by

$S=ut + \frac{1}{2}at^2$

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