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dangina [55]
3 years ago
9

T/F climate can change quickly

Physics
1 answer:
Vitek1552 [10]3 years ago
4 0

bhdfbvhfdbvhhfhvfhdf vfdhybvghbhdf

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Two satellites are in circular orbits around a planet that has radius 9.00×106m. One satellite has mass 68.0 kg, orbital radius
34kurt

Answer: 6782 m/s

Explanation:

Given

Radius of the planet, r = 9*10^6 m

Mass of satellite 1, m1 = 68 kg

Radius of satellite 1, r1 = 6*10^7 m

Orbital speed of satellite 1, vs1 = 4800 m/s

Mass of satellite 2, m2 = 84 kg

Radius of satellite 2, r2 = 3*10^7 m

Orbital speed of satellite 2, vs2 = ?

We know that magnitude of gravitational force, F = (G.m.m•) / r²

Where,

m = mass of satellite

m• = mass of planet

r = radius of orbit

If we consider Newton's second law that states that, F = ma, thus

F(g) = ma(rad)

Where, a(rad) = v²/r

F(g) = mv²/r

Substituting in the initial equation

mv²/r = (G.m.m•) / r²

v² = (G.m•) / r

v = √[G.m•/r]

To find vs2, we first need to find mass of the planet, m• we know that G is a gravitational constant, so we plug in the values

vs1 = √[G.m•/r1]

4800 = √[(6.67*10^-11 * m•) / 6*10^7]

4800² = (6.67*10^-11 * m•) / 6*10^7

2.3*10^7 * 6*10^7 = 6.67*10^-11 * m•

1.38*10^15 = 6.67*10^-11 * m•

m• = 1.38*10^15 / 6.67*10^-11

m• =2.07*10^25 kg

Having found that, we use the value to find our vs2

vs2 = √[(G.m•) / r2]

vs2 = √[(6.67*10^-11 * 2.07*10^25) / 3*10^7]

vs2 = √(1.38*10^15 / 3*10^7)

vs2 = √4.6*10^7

vs2 = 6782.33 m/s

Therefore, the orbital speed of the second satellite is 6782 m/s

4 0
4 years ago
Read 2 more answers
2) A man squeezes a pin between his thumb and finger, as shown in Fig. 6.1.
Salsk061 [2.6K]
<h3>pressure = force / area</h3>

<h3>force = 84 N</h3><h3>pressure = 6 × 10 - 5 = 55 m2</h3>

<h3>pressure = 84 / 55</h3>

<h3>pressure = 1.53 pascals</h3>

hope that helps and please tell me if i am wrong :)

8 0
3 years ago
The distance from earth to a NASA satellite traveling through the solar system is 6.0 × 109m. A command is sent to the satellite
alexandr1967 [171]

Answer:

the signal it takes 40 s to get the answer

Explanation:

As the satellite is in space the speed of radiation (waves9 is the speed of light in a vacuum, which is constant 3. 108 m / s, so we can use the uniform motion ratios to find the time.

    v = d / t

    t = d / v

    t = 6.0 109 / 3. 108

    t = 2 101 s

    t = 20 s

This is the time it takes to get the signal from the earth to the satellite and it takes the same time to return, so since they send the signal it takes 40 s to get the answer

5 0
4 years ago
If a device uses 280 watts of power a day, about how many kilowatt-hours will it use in 30 days?
kvasek [131]
Given: 280 watts power a day
            
Unknown: KwH=?in 30 days
Solution: 280 watt-hours per day/1000x30 days
               .28 x 30=n
Answer:8.4 kwh
4 0
3 years ago
Read 2 more answers
Consider a rifle, which has a mass of 2.44 kg and a bullet which has a mass of 150 grams and is loaded in the firing chamber.Whe
svetlana [45]

Answer:

a. 0 kgm/s

b. 0 kgm/s

c. 66 kgm/s

d. -66 kgm/s

e. 0 kgm/s

f. -27.05 m/s

g. 173.68 N

h. 12.58 m/s

i. 0.772 m

j. 14487 J

Explanation:

150 g = 0.15 kg

a. Before the bullet is fired, both rifle and the bullet has no motion. Therefore, their velocity are 0, and so are their momentum.

b. The combination is also 0 here since the bullet and the rifle have no velocity before firing.

c. After the bullet is fired, the momentum is:

0.15*440 = 66 kgm/s

d. As there's no external force, momentum should be conserved. That means the total momentum of both the bullet and the rifle is 0 after firing. That means the momentum of the rifle is equal and opposite of the bullet, which is -66kgm/s

e. 0 according to law of momentum conservation.

f. Velocity of the rifle is its momentum divided by mass

v = -66 / 2.44 = -27.05 m/s

g. The average force would be the momentum divided by the time

f = -66 / 0.38 = 173.68 N

h. According the momentum conservation the bullet-block system would have the same momentum as before, which is 66kgm/s. As the total mass of the bullet-block is 5.1 + 0.15 = 5.25. The velocity of the combination right after the impact is

66 / 5.25 = 12.58 m/s

i. The normal force and also friction force due to sliding is

F_f = N\mu = Mg\mu = 5.25*9.81*0.83 = 42.75N

According to law of energy conservation, the initial kinetic energy will soon be transformed to work done by this friction force along a distance d:

W = K_e

dF_f = 0.5Mv_0^2

d = \frac{Mv_0^2}{2F_f} = \frac{5.25*12.58^2}{2*42.75} = 0.772 m

j.Kinetic energy of the bullet before the impact:

K_b = 0.5*m_bv_b^2 = 0.5*0.15*440^2 = 14520 J

Kinetic energy of the block-bullet system after the impact:

K_e = 0.5Mv_0^2 = 0.5 * 5.25*12.58^2 = 33 J

So 14520 - 33 = 14487 J was lost during the lodging process.

3 0
3 years ago
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