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3 years ago

9

In an experiment, students measure the position x of a cart as a function of time t for a cart that starts at rest and moves wit

h a constant acceleration. The following data are collected for the cart.
t(s) x(m)
0 0
1 4
2 16
3 36
4 64

The acceleration of the cart is most nearly:_____

1 answer:

Minchanka [31]3 years ago

5 0

Given :

Initial velocity , u = 0 m/s² .

To Find :

The acceleration of the cart.

Solution :

Since, acceleration is constant.

Using equation of motion :

$x = ut + \dfrac{at^2}{2}\\\\x = \dfrac{at^2}{2}$

Putting, t = 1 s and x = 4 m in above equation, we get :

$4 = \dfrac{a(1)^2}{2}\\\\a = 8 \ m/s^2$

Therefore, the acceleration of the cart is 8 m/s².

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What is the period of a wave traveling 5 m/s if its wavelength is 20 m/s

Ilia_Sergeevich [38]

Speed of wave is given as

$v = 5 m/s$

Wavelength of the wave is given as

$\lambda = 20 m$

now from the formula of wave time period we can say

$speed = \frac{wavelength}{time period}$

$5 = \frac{20}{T}$

$T = \frac{20}{5}$

$T = 4 s$

so it will have time period of T = 4 s

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3 years ago

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa

tester [92]

Answer:

Part a)

$t = 2.83 s$

Part b)

Ball thrown downwards = $v_f = 23.8 m/s$

Ball thrown upwards = $v_f = 23.8 m/s$

Part c)

$d = 22.24 m$

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

$\Delta y = 0 = v_y t + \frac{1}{2}at^2$

$0 = 13.9 t - \frac{1}{2}(9.81) t^2$

$t = 2.83 s$

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

$v_f^2 - v_i^2 = 2 a \Delta y$

$v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)$

$v_f^2 = 567.9$

$v_f = 23.8 m/s$

Part c)

Relative speed of two balls is given as

$v_{12} = v_1 - v_2$

$v_{12} = (13.9) - (-13.9) = 27.8 m/s$

now the distance between two balls in 0.8 s is given as

$d = v_{12} t$

$d = 27.8 \times 0.8$

$d = 22.24 m$

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3 years ago

You are driving at 50 miles per hour. If you decrease the time it takes you to travel 1 mile by 8 seconds, what is your new spee

Dmitriy789 [7]

Answer:

The new speed is 56.25 miles/hour.

Explanation:

Since speed = distance/time;

time = distance/speed.

While driving at 50 miles/hour, time taken for one to complete 1 mile is (1/50) hour

(1/50) hour = (1/50) × 3600s = 72 seconds.

So, if this time to complete 1 mile (72 seconds) is reduced by 8 seconds,

New time to complete 1 mile will be = 72 - 8 = 64 seconds = (64/3600) hour = 0.0178 hour

New speed would be = (1 mile/64 seconds) = (1 mile/0.0178 hour) = 56.25 miles/hour.

Hope this Helps!!!

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3 years ago

What is the momentum of a 1200 kg car traveling with a speed of 27 m/s (60 mph)?

serious [3.7K]

Answer:

This is your answer

Explanation:

Actually I took this from go ogle

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2 years ago

A box falls out of a stationary helicopter hovering 135 m above the ground. How long will it take to hit the ground?

Marrrta [24]

We have the equation of motion $s = ut + \frac{1}{2} at^2$ , where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.

Here displacement = 135 m, Initial velocity = 0 m/s, acceleration = 9.81 $m/s^2$

Substituting

$135 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2 =135\\ \\ t =5.25 seconds$

A box falls out of a stationary helicopter hovering 135 m above the ground will take 5.25 seconds to reach the ground.

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3 years ago