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evablogger [386]
4 years ago
6

What force is required to accelerate a body with the mass of 15 kilograms at a rate of 8m/s2?

Physics
1 answer:
Tom [10]4 years ago
8 0
Force = mass * acceleration
F = 15 *8 = 120 Newton
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Which of the following is not in the form of waves?
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C. Seismic energy
This is energy that is released in earthquakes.
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1. The Moon's mass is 7.34 x 1022 kg, and it is 3.8 x 105 km away from Earth. Earth's
sineoko [7]

Answer:

2.03 x 10²⁴N

Explanation:

Given parameters:

Mass of moon = 7.34 x 10²²kg

Mass of the earth  = 5.97 x 10²⁴kg

Distance  = 3.8 x 10⁵km

Unknown:

Gravitational force of attraction  = ?

Solution:

To find the gravitational force of attraction between the masses, we use the expression below;

   F = \frac{Gm_{1} m_{2}  }{r^{2} }

G is the universal gravitation constant

m is the mass

1 and 2 represents moon and earth

r is  the distance

  F = \frac{6.67 x 10^{-11}  x 7.34 x 10^{22} x 5.97 x 10^{24}  }{(3.8 x 10^{5})^{2}  }

 F = \frac{2.92 x 10^{35} }{1.44 x 10^{11} }  = 2.03 x 10²⁴N

8 0
3 years ago
A 70kg man runs at a pace of 4 m/s and a 50g meteor travels at 2 km/s. Which has the most kinetic energy?.
kirill115 [55]

Answer:

the 70kg man

Explanation:

because he has more weight and is moving faster

4 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
makkiz [27]

Answer:

They move outwards.

They don't intersect each other at any point.

They show the electric field.

Explanation:

8 0
3 years ago
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a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
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