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levacccp [35]
3 years ago
8

The sputnik one satellite orbiting earth mass equals 5.98×10 to the 24th power kilograms in a circle of radius 6.96×10 to the si

x power meters what was its orbital velocity
Physics
1 answer:
nika2105 [10]3 years ago
6 0

Answer:

7572 m/s

Explanation:

The force between two masses separated by a distance r is given as:

F=G\frac{m_1m_2}{r^2}

Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is 6.674*10^-11 Nm^2/kg^2

The mass of the earth (m_1) is far greater than the mass of the sputnik (m_2). Therefore m_1m_2=m_1. The mass of the sputnik is neglected, therefore:

F=G\frac{m_1}{r^2}=\frac{(6.674*10^{-11})(5.98*10^{24})}{(6.96*10^6)^2} = 8.2389N

But F is actually centripetal acceleration, a = v²/r

8.2389 = v^2 / 6.96*10^6\\v=7572m/s

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In a closed energy system, energy is?​
Alisiya [41]

Answer:

Explanation:

A closed system can exchange energy but not matter, with its surroundings. An isolated system cannot exchange any heat, work, or matter with the surroundings, while an open system can exchange energy and matter.

Hope this helped you!

3 0
3 years ago
A silver bar 0.125 meter long is subjected to a temperature change from 200 C to 100 C . What will be the length of the bar afte
dimulka [17.4K]
\Delta L= \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
So your answer is B.
8 0
3 years ago
Read 2 more answers
Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary
kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

I=mr^2

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

I_A=1 \times 0.5^2\ kg.m^2

I_A=0.25\ kg.m^2

T= I α

0.5= 0.25 α

\alpha = 2\ rad/s^2

For Wheel B

m= 1 kg

d= 2 m,r=1 m

I_B=1 \times 1^2\ kg.m^2

I_B=1 \ kg.m^2

Given that angular acceleration is same for both the wheel

\alpha = 2\ rad/s^2

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0
3 years ago
If arts gravitational pool just got three times stronger what would happen to our weight
IRINA_888 [86]
We would just get three times heavier.

weight = mass x gravity (9.8 m/s/s)
4 0
3 years ago
A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency
I am Lyosha [343]

A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine

Explanation:

A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine

8 0
3 years ago
Read 2 more answers
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