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Galina-37 [17]
3 years ago
6

The downward acceleration of a falling body on Earth is 9.81m/s2. On the moon the same quantity is 1.62m/s2. An astronaut in a s

pace suit has a mass of 145kg. (a) What is the astronaut’s weight on Earth? (b) On the moon? (c) What is the astronaut’s mass on the moon?
Physics
1 answer:
Bas_tet [7]3 years ago
5 0
The astronaut's mass doesn't change.  It's the same wherever he goes,
because it doesn't depend on what else is around him.

His weight depends on what else is near him, so it changes, depending
on where he is.

         Weight  =  (mass) x (gravity)

On Earth,  Weight = (145 kg) x (9.81 m/s²)  =  1,422.5 newtons.
                                                               (about 320 pounds)

On the moon, Weight = (145 kg) x (1.62 m/s²)  =  234.9 newtons.
                                                                   (about 53 pounds)
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An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force
sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = 10ms^{-2}, F = ma = 15*10 = 150N

2) m = 3kg, v = 4ms^{-1}, r = 4m, F = \frac{mv^{2} }{r}

\frac{3*4^{2} }{4} = 12N

3) a = 10ms^{-2}, r = 10m, v=?

F = \frac{mv^{2} }{r} and F = ma

equating the two equations and cancelling a, we have:

\frac{v^{2} }{r} = a

making v the subject of formula, we have:

v = \sqrt{ar}

= \sqrt{100}

= 10ms^{-1}

4) r = 10m, v = 5ms^{-1}, a = ?

F = \frac{mv^{2} }{r}

F = ma

equating the above equations and making a subject of formula, we get:

a = \frac{v^{2} }{r}

a = 25/10 = 2.5ms^{-2}

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is tripled

F = (3v)^{2}

F = 9v^{2}

We can see that the force will be 9X greater than it was.

9) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is doubled

F = (2v)^{2}

F = 4v^{2}

We can see that the force will be 4X greater than it was.

10) F = \frac{mv^{2} }{r}

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0
3 years ago
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