Ask question

Ask question

All categories

3 years ago

14

How high does a rocket have to go above the Earth’s surface so that its weight is reduced to 58.8 % of its weight at the Earth’s

surface? The radius of the Earth is 6380 km and the universal gravitational constant is 6.67 × 10−11 N · m 2 /kg2 . Answer in units of km

1 answer:

julsineya [31]3 years ago

4 0

Explanation:

Below is an attachment containing the solution.

You might be interested in

Samuel is running in a parks nature trail.at first checkpoint he ran 925 m in 10 minutes.at second checkpoint he slowed down and

geniusboy [140]

Answer:

hey did you complete the whole test?

Explanation:

7 0

3 years ago

Will Give Brainliest and 25 Points

Vikentia [17]

This is the Doppler effect.

1. As the sound leaves the horn the sound waves are at first close to each other and as they move outwards they become further apart. The closer the sound waves are the louder the noise.

As the car gets the closer the sound waves get closer, so the horn becomes louder.

2. As the horn moves away, the sound waves become less frequent, causing the pitch to get lower.

5 0

3 years ago

The kind of heat transfer that travels through space in electromagnetic waves is?

Masteriza [31]

Answer:

radiation!

Explanation:

7 0

2 years ago

Read 2 more answers

A car drives 215 km east and then 45 km north. What is the magnitude of the car’s displacement? Round your answer to the nearest

34kurt

Displacement = (straight-line distance between the start point and end point) .

Since the road east is perpendicular to the road north,
the car drove two legs of a right triangle, and the magnitude
of its final displacement is the hypotenuse of the triangle.

Length of the hypotenuse = √ (215² + 45²)

= √ (46,225 + 2,025)

= √ 48,250

= 219.7 miles .

8 0

3 years ago

Read 2 more answers

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$

$kx_1 = mg$

The extension of the spring due to the weight of the object on Moon is a value of $x_2$ , then

$kx_2 = mg_m$

Recall that gravity on the moon is a sixth of Earth's gravity.

$kx_2 = m\frac{g}{6}$

$kx_2 = \frac{1}{6} mg$

$kx_2 = \frac{1}{6} kx_1$

$x_2 = \frac{1}{6} x_1$

We have that the displacement at the earth was $x_1 = 0.3m$ , then

$x_2 = \frac{1}{6} 0.3$

$x_2 = 0.05m$

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0

3 years ago