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2 years ago

What would be the weight of a 59.1-kg astronaut on a planet with the same density as Earth and having twice Earth's radius?

1 answer:

5 0

You could always directly use F = G M m / r², but that takes forever.

It is best to know that, on Earth's surface, the astronaut would weigh 
F = m g = 579.2 N 
where m = 59.1 kg and g = 9.8 m/s². 

You then notice from F = G M m / r² that doubling the mass of the earth would double his weight and doubling the radius would give 1/4 his original weight. So... 
F = 1/4 * 2 * 579.2 N = 290 N

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A 5 kg block is resting on a ramp inclined at 35 degrees above the horizontal. What is the magnitude of the normal force acting

Mamont248 [21]

I'm pretty sure the answer is b 28n hope helps :)

4 0

2 years ago

Sonia was experimenting with electric charges. She tied two inflated balloons together, held them next to each other, and rubbed

lions [1.4K]

Answer: They will repel each other.

Explanation:

Two inflated balloons when rubbed with woolen cloth will lead to repeal each other because of the similar charges on both the balloons.

Rubbing both the balloons together by the woolen cloth will introduce negative charge in the balloons.

As, we know that the same charges repeal each other both of the balloons with be apart from each other.

This is due to the static electricity, the negatively charged particles jump to the positive one. When balloons are rubbed they become negatively charged.

3 0

3 years ago

Read 2 more answers

Two of the three tuning forks have known frequencies. When a 610 Hz fork and the unknown fork are struck together, four beats pe

RUDIKE [14]

-- In combination with 610 Hz, the beat frequency is 4 Hz.

So the unknown frequency is either (610+4) = 614 Hz
or else (610-4) = 606 Hz.

In combination with 605 Hz, the beat frequency will be
either (614-605) = 9 Hz or else (606-605) = 1 Hz.

-- In actuality, when combined with the 605 Hz, the beat
frequency is too high to count accurately. That must be
the 9 Hz rather than the 1 Hz.

So the unknown is (605+9) = 614 Hz.

6 0

3 years ago

On a topographic map, contour lines create a group of concentric, closed loops. Which of the following features could this indic

posledela

Answer:

b. Hill top

Explanation:

On a topographic map, the closed circles are meant to represent a hill. So if the contour lines are creating a group of concentric closed loops then it must be an indication of a hill.

3 0

2 years ago

A 5 kg ball of clay is moving with a speed of 25 m/s directly toward a 10 kg ball of clay which is at rest. The two clay balls c

Dafna11 [192]

Answer:

8.3m/s

Explanation:

Given parameters:

mass of clay ball = 5kg

Speed of clay ball = 25m/s

mass of clay ball at rest = 10kg

speed of clay ball at rest = 0m/s

Unknown:

Velocity after collision = ?

Solution:

Since the balls stick together, this is an inelastic collision:

m1v1 + m2v2 = v(m1 + m2)

5(25) + 10(0) = v (5 + 10)

125 = 15v

v = 8.3m/s

7 0

2 years ago