What is the momentum of a 5.6 kg ball going 22m/s ?

Physics

1 answer:

maks197457 [2]3 years ago

6 0

<h2>momentum(p) = mass(m)×acceleration (a)</h2>

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reasons. 5. Why is the unit of temperature called a fundamental unit? Give reasons. ring derived unit.

Furkat [3]

Explanation:

It doesn't depends upon other.

It have it's own identity.

It's a lot easier to measure temperature than to measure the motion of component particles.

8 0

3 years ago

A child walks due east on the deck of a ship at 2 miles per hour. the ship is moving north at a speed of 18 miles per hour. find

garik1379 [7]

Refer to the figure shown below.

The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.

The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph

The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.

Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.

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3 years ago

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

FrozenT [24]

Answer:

$\omega=0.31\frac{rad}{s}$

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

$a_c=\frac{v^2}{r}(1)$

Here, r is the radius and v is the tangential speed, which is given by:

$v=\omega r(2)$

Here $\omega$ is the angular velocity, we replace (2) in (1):

$a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r$

Recall that $r=\frac{d}{2}=\frac{200m}{2}=100m$ .

Solving for $\omega$ :

$\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}$

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3 years ago

A car, starting from the origin, travels

sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.