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4 years ago

Which one of the following substances is a liquid fuel used in rocket engines?

Physics

2 answers:

melomori [17]4 years ago

8 0

There are none on the list you included with your question.

Mumz [18]4 years ago

5 0

Answer:

Ammonia

Explanation:

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The fixed hydraulic cylinder C imparts a constant upward velocity v = 2.2 m/s to the collar B, which slips freely on rod OA. Det

Olenka [21]

Answer:

so angular velocity is 7.13128 sec−1

Explanation:

velocity v = 2.2 m/s

displacement s = 220 mm = 0.220 m

distance d = 510 mm = 0.510 m

to find out

angular velocity

solution

we know that

angular velocity will be velocity ( v) / (displacement² + distance²) .....1

now put all these value in equation 1 and we get angular velocity i.e.

angular velocity = velocity ( v) / (displacement² + distance²)

angular velocity = 2.2 / (0.22² + 0.51²)

angular velocity = 2.2 / 0.3085

angular velocity = 7.13128

so angular velocity is 7.13128 sec−1

6 0

4 years ago

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the

Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

P sun = I sun-earth A sun-earth

where A = 2.863 10²³×m², and I  is 1360 W/m²

P sun = 2.863 10²³ × 1360

P sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

I sun-mars = 3.85×10²⁶W / 6.53 × 10²³m²

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 590 W/m²

6 0

4 years ago

According to Coulomb's law, what happens to the attraction of two oppositely charged objects as their distance of separation inc

NARA [144]

Answer:

Option B. Decreases

Explanation:

Coulomb's law states that:

F = Kq₁q₂ / r²

Where:

F => is the force of attraction between two charges

K => is the electrical constant.

q₁ and q₂ => are the two charges

r => is the distance apart.

From the formula:

F = Kq₁q₂ / r²

The force of attraction (F) is inversely proportional to the square of their separating distance (r).

This implies that as the distance between them increase, the force of attraction between the two charges will decrease and as the distance between two charges decrease, the force of attraction between them will increase.

Considering the question given above and the illustration given above, the force of attraction will decrease as their distance of separation increases.

Option B gives the right answer to the question.

7 0

3 years ago

What are the four layers of the earth

kumpel [21]

Answer:

earth's crust

meatle

outercore

innercore

8 0

3 years ago