What is Restoring force for large waves is A. Wind, B. Capillary Action, C. Diffusion D.Friction, or E. Gravity

Find the pressure exerted by a 3000 N crate that has an area of 2m squared

ipn [44]

Pressure = Force/ Area = 3000/2 = 1500 pascal.

8 0

2 years ago

A basketball player drops a 0.4-kg basketball vertically so that it is traveling at 5.7 m/s when it reaches the floor. The ball

Alik [6]

Answer:

(a) p = 3.4 kg-m/s (b) 37.78 N.

Explanation:

Mass of a basketball, m = 0.4 kg

Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)

It rebounds upward at a speed of 2.8 m/s (as it rebounds so positive)

(a) Change in momentum = final momentum - initial momentum

p = m(v-u)

p = 0.4 (2.8-(-5.7))

p = 3.4 kg-m/s

(b) Impulse = change in momentum

Ft = 3.4

We have, t = 0.09 s

$F=\dfrac{3.4}{0.09}\\\\F=37.78\ N$

Hence, this is the required solution.

4 0

2 years ago

Will give correct answer brainliest 5 kg m/s 8kg m/s 80 kg m/s 200 kg m/s

o-na [289]

Answer: Here this will help you..

Explanation:

1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second

5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second

10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second

20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second

30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second

40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second

50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second

75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second

100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second

8 0

2 years ago

A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?

Yanka [14]

Answer:

$\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}$

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

$\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}$

Substituting values of m & v in the equation:

$\sf \implies KE = \frac{1}{2} \times 380.2 \times {11}^{2}$

$\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 190.1 \times 121$

$\sf \implies KE = 190.1 \times 121$

$\sf \implies KE = 23002.1 \: J$

$\therefore$

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0

3 years ago

A plane starting from rest accelerates at 3m/s2 for 25s. Calculate the increase in velocity after:

horsena [70]

Initial velocity=u=0m/s
Acceleration=a=3m/s^2

Time not needed now

Case-1:-

Time=1s
Final velocity=v

$\\ \rm\hookrightarrow v=u+at$

$\\ \rm\hookrightarrow v=0+3(1)$

$\\ \rm\hookrightarrow v=3m/s$

Case-2:-

Time=3s

$\\ \rm\hookrightarrow v=0+3(3)$

$\\ \rm\hookrightarrow v=9m/s$

Case-3:-

Time=25s

$\\ \rm\hookrightarrow v=0+3(25)$

$\\ \rm\hookrightarrow v=75m/s$

5 0

2 years ago