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NeX [460]
3 years ago
6

A roller coaster car rapidly picks up speed as it rolls down a slope. As it starts down the slope, its speed is 4m/s. But 3 seco

nds later at the bottom of the slope its speed is 22m/s. What is its average acceleration?
Physics
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

6 m/s^2

Explanation:

The average acceleration of the car is given by:

a=\frac{v-u}{t}

where

v is the final speed

u is the initial speed

t is the time elapsed

For the car in this problem,

v = 22 m/s

u = 4 m/s

t = 3 s

Therefore, the acceleration of the car is

a=\frac{22 m/s-4 m/s}{3 s}=6 m/s^2

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1-Calculate Req
ArbitrLikvidat [17]

Answer:

1. 21.66 Ohms

2.  3.38 A

3. 6.7 V

Explanation:

1. Req = 6+2 = 8 Ohms (2 and 6 are in a series circuit)

  Req = 1/8 +1/4 = 3/8 = 8/3 = 2.66 Ohms (8 and 4 are parallel, so we will add them using this equation)

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2. I = V/R = 9/2.66 = 3.38 A (In a series circuit, the current is the same across the resistors, so we will add them and divided them by 9 volts)

3. V = IR = 3.38 x 2 = 6.7 V (In a series circuit, the voltage is different, so each resistor will have a different voltage.)

I hope this helps.  I am not an expert in physics but its ok :)

<u><em>Note: If the answer benefited u, mark me as the brainliest answer if u can, thx.</em></u>

3 0
2 years ago
A car accelerates from rest at a constant rate of 2m/s2 for 5s. what is the speed of a car at the end of that time?
Fiesta28 [93]
Good afternoon.


We have:

\mathsf{V_0 = 0}\\ \mathsf{a = 2 \ m/s^2}\\ \mathsf{t = 5 \ s}

The function of velocity:

\mathsf{V = V_0+at}\\ \\ \mathsf{V = 0 + 2t}\\ \\ \mathsf{V = 2t}


For t = 5 s:

\mathsf{V = 2\cdot 5}\\ \\ \boxed{\mathsf{V = 10 \ m/s}}
4 0
3 years ago
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