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3 years ago

5

HELP I NEED THIS ANSWERED AS FAST AS POSSIBLE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

WITCHER [35]3 years ago

8 0

Displacement is d

Vf² = Vi² + 2 g d

(-20²) = (+10²) + 2 (-9.8) d

-19.6 d = 300

d = -15.3 m

negative means lower

time is t

d = Vi t + 1/2 g t²

-15.3 = 10 t + (-4.9) t²

4.9 t² - 10 t -15.3 = 0

t = 3.06 s

Hope this helps -John

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Read 2 more answers

A flat coil of wire consisting of 15 turns, each with an area of 40 cm 2, is positioned perpendicularly to a uniform magnetic fi

zheka24 [161]

Answer:

0.54 A

Explanation:

Parameters given:

Number of turns, N = 15

Area of coil, A = 40 cm² = 0.004 m²

Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T

Time interval, Δt = 2 secs

Resistance of the coil, R = 0.2 ohms

To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:

|V| = |(-N * ΔB * A) /Δt)

|V| = | (-15 * 3.6 * 0.004) / 2 |

|V| = 0.108 V

According to Ohm's law:

|V| = |I| * R

|I| = |V| / R

|I| = 0.108 / 0.2

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The magnitude of the current in the coil of wire is 0.54 A

6 0

3 years ago

(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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2 years ago

KATRIN_1 [288]

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