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Harman [31]
3 years ago
8

What time does a first quaeter moon rise?​

Physics
1 answer:
butalik [34]3 years ago
7 0

Answer:

Noon

Explanation:

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When a 100-N force acts horizontally on an 8.0-kg chair, the chair moves at a constant speed across the level floor. Which state
TiliK225 [7]

Answer:

The applied force is greater than the frictional force.

Explanation:

the chair moves at <u>a constant speed</u><u> </u><u>therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>answer</u><u> </u><u>is</u><u> </u><u>not</u><u> </u><u>A</u><u> </u><u>or</u><u> </u><u>C</u><u>.</u>

if there is no friction then the chair <u>would accelerate and it would not be at a constant speed</u><u>.</u>

hence, the only possible answer is B.

7 0
3 years ago
Consider an ideal gas-turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each s
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Answer:

Please see attached

Explanation:

Please see attached

4 0
4 years ago
Read 2 more answers
A driver drives a car up a hill. The driver keeps pressing the gas so that the engine increases the force it applies to the car.
iris [78.8K]

Answer:

because it's on an upward climb therefore newton's law of motion is in place and you're car is pulled by gravity

6 0
3 years ago
How do electromagnetic waves move through space?
lesantik [10]

Answer:through a small space through atoms

Explanation:

4 0
4 years ago
A 782-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Vadim26 [7]

Answer:

a) v = 5.59x10³ m/s

b) T = 4 h

c) F = 1.92x10³ N

Explanation:

a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:

F_{c} = F_{G}

\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}

v = \sqrt{\frac{gr^{2}}{r+h}          

Where:

g is the gravity = 9.81 m/s²        

r: is the Earth's radius = 6371 km

h: is the satellite's height = r = 6371 km      

v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s                                      

b) The period of its revolution is:

T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h

c) The gravitational force acting on it is given by:

F = \frac{GMm}{(r + h)^{2}}

Where:

M is the Earth's mass =  5.97x10²⁴ kg    

m is the satellite's mass = 782 kg

G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²

F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N

I hope it helps you!

3 0
3 years ago
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