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3 years ago

If a book is knocked off a desk that is .75 m tall at a rate of 2.0 m/s, how far away from the desk does it fall

Physics

1 answer:

ANEK [815]3 years ago

7 0

Answer: 0.8 m

Explanation:

In the vertical direction, the speed is zero, u = 0.

Distance covered in the vertical direction, s = 0.75 m.

The book would fall with acceleration due to gravity in the vertical direction, a = g = 9.8 m/s²

From the equation of motion,

s = u t + 0.5 a t²

Substituting the above values, we will find out the time taken for the book to hit the ground.

⇒0.75 m=0+0.5×9.8 m/s²×t²

⇒t = √0.153 = 0.39 s ≈ 0.40 s

Now, the horizontal distance covered,

d = v×t ⇒d= 2.0 m/s × 0.40 s =0.8 m

Hence, the book falls 0.8 m away from the desk.

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A basketball player wishes to make a stunning last-second basket. With .8 seconds on the clock, she wants the ball to enter the

zheka24 [161]

Answer:

(1) $46.86^\circ$

(2) Diagram has been attached in the solution.

Explanation:

This question is from projectile motion.

From the given question, we will discuss the motion of the basket ball only in the vertical direction from which we will be able to find out the angle of the initial velocity with the horizontal with which it should be shoot to enter the hoop.

Part (1):

Let us assume:

$y_i$ = initial position of the basket ball = 2.1 m
$y_f$ = final position of the basket ball = 3.05 m
$a_y$ = acceleration of the ball along the vertical = $-9.8\ m/s^2$
$t$ = time taken to reach the goal = 0.8 s
$\theta$ = angle of the initial velocity with the horizontal
$u$ = initial speed of the ball = 7 m/s
$u_y$ = initial vertical velocity of the ball = u\sin \theta

Using the equation of motion for constant acceleration, we have

$y_f-y_i=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 3.05-2.1=u\sin \theta (0.8) +\dfrac{1}{2}\times (-9.8)(0.8)^2\\\Rightarrow 0.95=7\times \sin \theta (0.8) -3.136\\\Rightarrow 0.95=5.6\sin \theta -3.136\\\Rightarrow 5.6\sin \theta= 0.95+3.136\\\Rightarrow 5.6\sin \theta= 4.086\\\Rightarrow \sin \theta= \dfrac{4.086}{5.6}\\\Rightarrow \sin \theta=0.729\\\Rightarrow \theta=\sin^{-1}(0.729)\\\Rightarrow \theta=46.86^\circ$

Hence, the angle of the shoot of the basket ball with the horizontal is $46.86^\circ$ such that it reaches the hoop on time.

Part (2):

For this part, a diagram has been attached.

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3 years ago

1. A 1.30 kg ball strikes a wall with a velocity of -10.5 m/s. The ball bounces off with a velocity of 6.50 m/s. If the ball is

bekas [8.4K]

Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is

Ft = F (0.0210 s)

and this impulse is equal to the change in the ball's momentum,

m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)

Solve for F :

F (0.0210 s) = (1.30 kg) (17.0 m/s)

F = (1.30 kg) (17.0 m/s) / (0.0210 s)

F ≈ 1050 N

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3 years ago

1. Take the speed at which the body moves as 10 m/s 2. Calculate the distance travelled by the body every second 3. Take the tim

exis [7]

Answer:

a)this graph is also a line b) in both cases we have a uniform movement

Explanation:

In this exercise we have a uniform movement

v = d / t

d = v t

in the table we give some values to make the graph

t (s) d (m)

1 10

2 20

3 30

In the attached we can see the graph that is a straight line

we have another vehicle at v = 50 me / S

t (s) d (m)

1 50

2 100

3 150

this graph is also a line

b) in both cases we have a uniform movement

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4 years ago