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2 years ago

6

A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc

eleration of the cart

1 answer:

amid [387]2 years ago

8 0

A= f/m

a= 19/2

a= 9.5m/s^2

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Read 2 more answers

A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar

sesenic [268]

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

$v^2-u^2 = 2gd$

where

u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

h = 24 m is the distance covered

Solving for h,

$v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s$

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

$y_1 = h - \frac{1}{2}gt^2$

While the vertical position of the ball thrown upward is

$y_2 = ut - \frac{1}{2}gt^2$

The two balls meet when

$y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s$

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

$y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m$

So the distance below the top of the cliff is

$d=24.0 - 18.0 = 6.0 m$

4 0

3 years ago