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2 years ago

A force of 8 N accelerates by 4 m/s^2. What would be the amount of force needed to give a final acceleration of 5.3 m/s^2

Physics

1 answer:

Elis [28]2 years ago

4 0

Answer:

10.6 N

Explanation:

F = ma

8 = m * 4

m = 2 kg

F = 2 * 5.3

F = 10.6 N

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A 30-gram bullet is fired and a 50-gram bullet is dropped simultaneously from the same height. Which will hit the ground first?

Marina86 [1]

Answer: the 30gram will hit the ground first

Explanation:the 30gram bullet will hit the ground first because it is fired

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3 years ago

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The speed of a sound wave traveling through air is 400 meters per second. The wavelength of the wave is 25 meters What is the

sweet [91]

Answer:

16Hz

Explanation:

Given parameters:

Speed of sound = 400m/s

Wavelength = 25m

Unknown:

Frequency of the wave = ?

Solution:

To solve this problem;

V = F ∧

V is the velocity

F is the frequency

∧ is the wavelength

400 = F x 25

F = 16Hz

7 0

2 years ago

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3. A model rocket takes 0.05 seconds to speed up from rest to its maximum velocity of 80 m/s.

nikklg [1K]

Answer:

$1600 \frac{m}{s^2}$

Explanation:

Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

$a = \frac{change-in-velocity}{time} = \frac{Vf-Vi}{t}$

Where Vf is the final velocity of the object, (in our case 80 m/s)

Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)

and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.

Therefore we have:

$a = \frac{80 m/s - 0 m/s}{0.05 sec} = 1600 \frac{m}{s^2}$

Notice that the units of acceleration in the SI system are $\frac{m}{s^2}$ (meters divided square seconds)

7 0

2 years ago

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(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

4 0

2 years ago

The Earth and the Moon are attracted to each other by universal gravitation. The Earth is much more massive than is the Moon. Do

OverLord2011 [107]

Answer:

Earth attract the Moon with a force that is greater.

Explanation:

According to the law of gravitation, the gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically, F1 = Gm1m2/r²... 1

Let m1 be the mass of the earth and m2 be that of the moon

If the Earth is much more massive than is the Moon, the new force of attraction between them will become;

F2= G(2m1)m2/r²

F2 = 2Gm1m2/r² ... (2)

Dividing eqn 1 by 2 we have;

F1/F2 = (Gm1m2/r²)÷(2Gm1m2/r²)

F1/F2 = Gm1m2/r²×r²/2Gm1m2

F1/F2 = 1/2

F2=2F1

This shows that that the earth will attract the moon by a force 2times the initial force of the masses(i.e a much greater force)

6 0

3 years ago