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2 years ago

A 60 kilogram astronaut weighs 96 newtons on the surface of the moon. calculate the acceleration due to gravity on the moon.

Physics

2 answers:

Charra [1.4K]2 years ago

8 0

Answer:

Acceleration due to gravity on moon, $1.6\ m/s^2$

Explanation:

Mass of the astronaut, m = 60 kg

Weight of astronaut on the surafce of moon, W = 96 N

The weight of astronaut is given by :

W = m g

$g=\dfrac{W}{m}$

$g=\dfrac{96\ N}{60\ kg}$

$g=1.6\ m/s^2$

So, the acceleration due to gravity on the surface of moon is $1.6\ m/s^2$

Also, the acceleration due to gravity on the surface of moon is (1/6)th of the acceleration due to gravity on the surface of earth.

Hence, this is the required solution.

34kurt2 years ago

6 0

Given: Mass m = 60 Kg

Weight W = 96 N

Required: Acceleration due to gravity, g = ?

Formula: W = mg

g = W/g

g = 96 Kg.m/s²/60 Kg (note: this is the derive unit for Newton "N")

g = 1.6 m/s²

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Convert 23 miles into feet.<br> **There are 5280 feet in 1 mile** *

zhenek [66]

Answer:

121440 feet is 23 miles.

5280 x 23= 121440

To check your answer just divide 121440 by 5280, which gives you 23.

Hopefully this helps you!

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2 years ago

A person weighing 0.9 kN rides in an elevator that has a downward acceleration of 1.9 m/s^2. The acceleration of gravity is 9.8

Keith_Richards [23]

Answer:

The magnitude of the force is 0.7255kN

Explanation:

The elevator floor acts on the person with a force that is due to the gravitational acceleration less the downward acceleration of the elevator:

(force of floor F) = (mass of person m) x [ (grav. acceleration g) - (elevator acceleration a) ]

in other words, considering the elevator floor as a reference frame in the Earth's gravitational field, the person's weight decreases due to the downward acceleration, as follows:

$F = m\cdot(g-a)$

We are given the person's weight at rest, 0.9kN, from which the mass can be determined as:

$900 N = m\cdot g \implies m = \frac{900N}{9.8 \frac{m}{s^2}}$

$F = \frac{900N}{9.8 \frac{m}{s^2}}\cdot(9.8-1.9)\frac{m}{s^2}\approx 725.5N=0.7255kN$

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2 years ago

Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?

noname [10]

Here we can use coulomb's law to find the force between two charges

As per coulombs law

]tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we have

$k = 9 * 10^9$

$q_1 = 2pC$

$q_2 = 4pC$

$r = 6cm = 0.06 m$

now by using the above equation we have

$F = \frac{9*10^9 * 2*10^{-12} * 4*10^{-12}}{0.06^2}$

$F = 2 * 10^{-11} N$

so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.

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2 years ago

If an astronaut throws an object in space, the object's speed will _____________. A) decrease. B) increase. C) remain constant.

GenaCL600 [577]

The object's speed will remain constant after the it leaves his hand.

So will HIS speed in the opposite direction.

7 0

2 years ago

Read 2 more answers

If I were to transmit a radio wave in our three dimensional world could a fourth dimensional “being” be able to receive it?

Morgarella [4.7K]

Depends. Are you talking about a mathematical 4th dimension (in which there is infinite dimensions) or some sort of etheral dimension (in which there is no scientific evidence for)

If you mean the first then yes. But it depends how these beings exist. From our understanding we only can theorize shapes in 4-d and if we assume that there is only one universe these "beings" arleady exist and thus any message in 3-d would be sent to them like a shadow ("flat").
If they exist in a alternate "plane" then you would need some method to transverse this plan and if u did, then we would easily be able to communicate, but we would at first sound like a wild animal. They either would ignore us, not understand or perceive us, or they would attempt to send back a signal (essential they are ET's)

IF you mean the second then thats some mystic stuff and its pretty creepy (although a fun read for me :P)

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2 years ago