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3 years ago

State the equation of momentum impulse theorem.

Physics

1 answer:

Paladinen [302]3 years ago

3 0

Answer:

J = Δp

Explanation:

The impulse-momentum theorem says that the impulse J is equal to the change in momentum p.

J = Δp

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4.0 kg objects mive a distance of 7.8 m under the action of a cinstant gorce of 5.6 N. how much work is dine on object?

alexdok [17]

Answer:

43.68 J

Explanation:

Distance moved= 7.8 m

Force = 5.6 N

Work Done= Distance moved * Force

= 7.8 *5.6

=43.68 Joules

6 0

3 years ago

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.

Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m$

The maximum height above the ground that the ball reaches is 7.09683 m

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s$

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

$v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s$

The velocity just before it hits the ground is 11.8 m/s

6 0

3 years ago

Object C has a mass of 3,600 kilograms. Object D has a mass of 900 kilograms. Both objects were placed on different planets so t

FinnZ [79.3K]

The correct answer would be 4

7 0

3 years ago

The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.

sattari [20]

Answer:

$d=691.71km$

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

$t=\frac{d}{v_t}-\frac{d}{v_l}$

Here d is the distance at which the earthquake take place and $v_t, v_l$ is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

$t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km$

8 0

3 years ago

Will mark as brainliest if correct!!!!!!!!!!!

Aneli [31]

Answer:

The answer is A.

Explanation:

The diagram shows the light ray bending away from the normal. Light rays bend away from the normal when their speed increases. This means that in the diagram, the light ray moves from a medium in which light has a lower speed to a medium in light has a higher speed. The only choice where the speed of light increases from A to B is answer A. So that has to be the answer.

7 0

3 years ago