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3 years ago

If an astronaut travels to different planets, which of the following planets will the astronaut’s weight be the same as on Earth

Physics

2 answers:

AysviL [449]3 years ago

5 0

Answer:

D. none

Explanation:

Nataly_w [17]3 years ago

3 0

<h2>Astronaut travels to different planets - Option 4 </h2>

If an astronaut travels to different planets, none of the planets will the astronaut’s weight be the same as on Earth. On jupiter, weight will be more than the weight on earth. For instance if an astronaut has 100kg on earth then he will have 252 kg on jupiter.

On Mars, weight will be less than the weight on the earth. For instance, if an astronaut has 68 kg on earth then he will has 26 kg on mars. On Mercury, weight of an astronaut will be less than the weight on earth. Example if he has 68 kg on earth then he will have 25.7kg on mercury.

Hence, none of these planets the weight of astronaut will be same as on earth.

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Which is more useful for storing thermal energy, the block of lead or the water in the calorimeter?

leonid [27]

I believe it is the block of lead

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3 years ago

How do sea surface temperatures affect evaporation rate?

kotegsom [21]

<span>Answer: The temperature doesn't affect the evaporation rate, but affects on how much of water a parcel of air can contain when saturated which is known by the absolute humidity. Hurricanes are usually happening when the temperature of the sea water west of the Cape Verde islands is over 27 degrees Celsius. If ahead of the path of a hurricane, the sea water temperature drops then it will be less moisture in the air and perhaps the hurricane will fade out. But it is not as simple. How strong a tropical storm is is relative to the difference of temperture between ground level and the top of the troposphere. The greater the difference, the faster the air will rise and the deeper the pressure will be, forcing surrounding air to rush in, thus forming a hurricane force wind. Then there is the fact that the wet adiabatic lapse rate is about half that of dry air. It means that rising moist air cools down slower and therefore rises higher. Hence water is the true fuel of bad weather. But it can't be isolated from the fact that the difference of temperature must be great too. What we often forget is that the tropopause (the border to the stratosphere) is much higher over the equator and therefore, much colder than e.g. the poles.</span>

8 0

3 years ago

A ball is droped from a height of 16m how much time will pass before the ball hits the ground

sergey [27]

Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

$h= \frac{1}{2}g{t}^{2}$

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

Also, let take

$g = 9.8m{s}^{-2}$

By substitution we obtain,

$16= \frac{1}{2}\times 9.8{t}^{2}$

$\implies32=9.8{t}^{2}$

Diving through by 9.8

$\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}$

$\implies{t}^{2} =3.265$

square root both sides, we obtain

$\implies t= \sqrt{3.265}$

$t=1.8s$

4 0

3 years ago

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$