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Leya [2.2K]
3 years ago
15

John bounces a ball off the ground, and it leaves the ground with a velocity of 10 m/s. How long will it take the ball to reach

maximum height?

Physics
2 answers:
Maksim231197 [3]3 years ago
8 0

Explanation:

This is true because at maximum height, the velocity is 0

wariber [46]3 years ago
3 0

Gravity makes an upward-moving object move 9.8 m/s slower every second.

So if it leaves the ground at 10 m/s, it'll run out of upward speed in

(10 m/s) / (9.8 m/s²)  =  <em>1.02 seconds </em>

At that time, since it has no more upward speed, it'll start falling. So at that instant, it was exactly at its maximum height.

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kirza4 [7]

The parietal layers of the membranes line the walls of the body cavity.


hope this helps

3 0
3 years ago
7.
jeka94

الملك - الإتاوات - المحاكم - المجلس الملكي - القانون الروماني

5 0
3 years ago
I help much help on 18!!!
ra1l [238]
I think it's something like electrons don't attract, cuz you know the saying "Opposites attract." Cause electrons are negative... Ahaha... sorry, I don't know the answer.
4 0
3 years ago
I got part c right but idk why the other parts are wrong HELP!
dedylja [7]

a) The impulse is 76.5 Ns

b) The average force is 546.4 N

c) The final speed is 31.5 m/s

Explanation:

a)

The impulse exerted on an object is defined as

J=\int F\Delta t

where

F is the magnitude of the force exerted on the object

\Delta t is the time interval during which the force is applied

If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.

Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are

B=0.14-0 = 0.14s\\b=8-5=3s

and whose height is

h=900 N

Therefore, the area (and the impulse) is

J=\frac{(B+b)h}{2}=\frac{(0.14+0.03)(900)}{2}=76.5 Ns

b)

In this problem, the force applied is not constant. However, we can rewrite the impulse also as

J=F_{avg} \Delta t

where

F_{avg} is the average force exerted during the whole time \Delta t

In this problem we have

J = 76.5 Ns is the impulse (calculated in part a)

\Delta t = 0.14 s is the time interval

Solving for the average force, we find

\Delta t = \frac{J}{F_{avg}}=\frac{76.5}{0.14}=546.4 N

c)

According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:

J=\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have

J = 76.5 Ns

m = 3.0 kg is the mass

u = 6.0 m/s is the initial velocity

Solving for v, we find the final velocity (and speed):

v=u+\frac{J}{m}=6.0+\frac{76.5}{3}=31.5 m/s

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

6 0
3 years ago
A flexible shaft consists of a 3 mm diameter steel wire in a flexible hollow tube which imposes a frictional torque of 0.04 N m
Rom4ik [11]

Answer:

2.5m

Explanation:

Torque is defined as the rotational effect of a force on a body.

The torque T for the maximum shear stress is given as 0.1 Nm

Frictional torque is the torque caused by a frictional force

The frictional torque F is given as 0.04 Nm/m

The maximum length of the shaft is thus given as

L = T / F

  = 0.1/0.04

L= 2.5 m

6 0
3 years ago
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