Ask question

Ask question

All categories

3 years ago

7

A body staring for rest acquries a velocity of 10ms in 5 seconds calculate (a) the accleration(b) the distance covered by the bo

dy in5 second

1 answer:

larisa86 [58]3 years ago

3 0

Answer:

Acceleration of the body=2m/s square

Distance covered= 25m

Explanation:

Acceleration is found using formula a=v-u divided by t.

Distance covered is found using formula= ut+1 by 2 × at square.

Where initial velocity(u)=0 (body starts from rest)

You might be interested in

2. A ball is dropped from rest. The acceleration due to gravity is 10m/s? and the time it

d1i1m1o1n [39]

Answer:

STEP BY STEP

V = S/T

V= 10/5

V = 2 m/s

3 0

3 years ago

Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th

Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

$\theta=30^{\circ}$

So, the ball's horizontal velocity is

$v_x = u cos \theta = (15)(cos 30)=13.0 m/s$

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

$y=u_yt+\frac{1}{2}at^2$

where

$u_y = u sin \theta = (15)(sin 30) = 7.5 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

$2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0$

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

$d=v_x t =(13.0)(1.19)=15.5 m$

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

$v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s$

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

$d=98 m$

while the horizontal velocity of the ball is

$v_x = u cos \theta = (34)(cos 30)=29.4 m/s$

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

$t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s$

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

$y= h + u_y t + \frac{1}{2}at^2$

where

h = 1.2 m is the initial height

$u_y = u sin \theta = (34)(sin 30)=17.0 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting t = 3.33 s,

$y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m$

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

$u_y' = 7 m/s$

So its position of the glove at time t' is

$y'= h' + u_y' t' + \frac{1}{2}at'^2$

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

$3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0$

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

$t'' = t -t'$

So we have two solutions:

$t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s$

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

It is epistemically possible that a completed physics might include which of the following propositions? O Souls and other spiri

ohaa [14]

Answer:

option D.

Explanation:

The correct answer is option D.

At present the knowledge of physics does not includes the study related to souls , other spiritual substance.

Physics does not give the answers of question like God created universe , matter is unreal, etc.

So, there might will be possibility that in future complete physics will give the answer of all these questions i.e. correct answer is All of the above.

3 0

3 years ago

Which list includes translucent objects only

Lunna [17]

Objects that let in light and blurry images are translucent.

Translucent is a term that refers to an adjective. This characteristic is the property of an object to allow the passage of light, without allowing visibility with high clarity through it.

This term is often confused with transparency. However, they differ because a transparent object lets light through easily and allows you to see clearly through it.

According to the above, objects that allow light to pass through but do not allow clear vision are translucent.

Learn more in: brainly.com/question/10626808

5 0

3 years ago

In an inelastic collision a 2.5 kg ball moving at 7.5 m/s is caught by a 70kg man while the man is standing on ice. What is the

MrRa [10]

The velocity of the ball and the man is 0.259 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, in an isolated system, the total momentum before and after the collision must be conserved. Therefore, for the ball-man system, we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 2.5 kg$ is the mass of the ball

$u_1 = 7.5 m/s$ is the initial velocity of the ball

$m_2 = 70 kg$ is the mass of the man

$u_2 = 0$ is the initial velocity of the man

$v$ is the final velocity of the man and the ball after the collision

Re-arranging the equation and substituting the values, we find the final velocity:

$v=\frac{m_1 u_1}{m_1+m_2}=\frac{(2.5)(7.5)}{2.5+70}=0.259 m/s$

So, the man and the ball slides on the ice at 0.259 m/s.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

3 0

3 years ago