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A cart is moving to the right with a constant speed of 20 m s . A box of mass 80 kg moves with the cart without slipping. The co

nignag [31]

Answer:

Explanation:

Given

mass of box $m=80 kg$

coefficient of kinetic friction $\mu _k=0.15$

coefficient of Static friction $\mu _s=0.30$

cart is moving with constant velocity therefore Net Force is zero

Since there is no net acceleration therefore friction force will be zero

mathematically

$f_r=ma$

where $f_r=frictional\ Force$

$a=acceleration$

$a=0$

$f_r=0$

4 0

2 years ago

The gravitational pull of the sun on Earth keeps Earth orbiting around the sun. Which statement is correct about the force that

Furkat [3]

Answer:

Earth pulls the sun towards itself with a force equal to the ratio of the mass of the sun to the mass of Earth

5 0

2 years ago

A 10-kg object is pushed across a rough surface. It begins at rest and accelerates to a speed of 4 m/s in a distance of 5m. Whic

brilliants [131]

Answer:

Explanation:

Unbalanced Force according to newton's second law is the one which causes the object to move or break its state of rest on application of force.

Here the object accelerate to a speed of 4 m/s and moves a distance of 5 m on application of force .

Thus we can say that the applied force is unbalanced in nature.

5 0

2 years ago

Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o

klio [65]

Answer:

The acceleration of Abbie is half of the Zak's.

Explanation:

The centripetal acceleration of an object on a circular path is given by :

$a=r\omega^2$

Two children are riding on a merry-go-round that is rotating with a constant angular speed. Let $r_1$ is distance of Abbie from the merry-go-round and $r_2$ is distance of Zak's from the merry-go-round. Acceleration of Abbie is :

$a_1=r_1\omega^2$ ...... (1)

$r_1=1\ m$

Acceleration of Zak's is :

$a_2=r_2\omega^2$ .......(2)

$r_2=2\ m$

Dividing equation (1) and (2) we get :

$\dfrac{a_1}{a_2}=\dfrac{r_1}{r_2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{2}\\\\a_1=\dfrac{a_2}{2}$

So, the acceleration of Abbie is half of the Zak's.

7 0

2 years ago

A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a

Readme [11.4K]

Answer:

Part a)

a = -9.81 m/s/s

Part b)

v = 0

Part c)

v = 9.81 m/s

Part d)

$H = 4.905 m$

Explanation:

Part a)

During the motion of ball it will have only gravitational force on the ball

so here the acceleration of the ball is only due to gravity

so it is given as

$a = g = 9.81 m/s^2$

Part b)

As we know that ball is moving against the gravity

so here the velocity of ball will keep on decreasing as the ball moves upwards

so at the highest point of the motion of the ball the speed of ball reduce to zero

$v_f = 0$

Part c)

We know that the total time taken by the ball to come back to the initial position is T = 2 s

so in this time displacement of the ball will be zero

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

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

$v_y = 9.81 m/s$

Part d)

at the maximum height position we know that the final speed will be zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

here we have

$0 - (9.81^2) = 2(-9.81)H$

$H = 4.905 m$

4 0

3 years ago