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Arisa [49]
3 years ago
12

A car sitting at a red light begins to accelerate at 2.0 m/s2 when the light turns green. It continues with this acceleration un

til it reaches a speed of 20 m/s. It then travels at this speed for another few minutes. How far does the car travel in the first 46 s after the light changes to green?
Physics
1 answer:
KonstantinChe [14]3 years ago
8 0

Answer:

d = 820 m

Explanation:

here we know that car starts from rest and continue to accelerate till it will reach to maximum speed of 20 m/s

so we will have

v_f - v_i = at

20 - 0 = 2t

t = 10 s

so car will accelerate till t = 10 then it will move with uniform speed

so the distance moved by the car till it accelerates is given as

d = \frac{1}{2}at^2

d = \frac{1}{2}(2)(10^2)

d_1 = 100 m

now it will move with uniform speed for next 36 s

so we have

d_2 = 20(36) = 720 m

so total distance moved by the car is given as

d = d_1 + d_2

d = 100 + 720

d = 820 m

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static

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static friction pushes in the direction you are walking.

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3 years ago
During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 33.5 rad/s. Find the angular displ
zavuch27 [327]

Answer: angular displacement in rad = 3038.45 rad

angular displacement in rev = 483.589 rev

Explanation: mathematically

Angular velocity = angular displacement / time taken.

Angular velocity = 33.5 rad/s, time taken = 90.7s

33.5 = angular displacement /90.7

Angular displacement = 33.5 * 90.7 = 3038.45 rad

But 1 rev =2π

Hence 3038.45 rad to rev is

3038.45/2π = 483.599 rev

7 0
4 years ago
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Answer:

16km

Explanation:

First change the minutes into hours then multiply by the distance.

(8÷60)×120=16km

5 0
3 years ago
Before partacting, if correct marked as brainest
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C. Basic swimming capability.

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3 years ago
A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Serggg [28]

Answer:

The spring force constant is  k=243\ \frac{N}{m} .

Explanation:

We are told the mass of the ball is m=0.0555\ kg, the height above the spring where the ball is dropped is h=0.536\ m,  the length the ball compresses the spring is d=0.04897\ m and the acceleration of gravity is 9.8\ \frac{m}{s^{2}} .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy E_{mi} is equal to the final mechanical energy E_{mf} :

                                                    E_{mf}=E_{mi}

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

                                                   \frac{1}{2}kd^{2}=mgh

If we manipulate the equation we have that:

                                                    k=\frac{2mgh}{d^{2} }

                                         k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}

                                              k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}

                                                     k=243\ \frac{N}{m}

                                                   

                             

5 0
4 years ago
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