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timama [110]
3 years ago
5

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]3 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.

34kurt3 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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Explanation:

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F=\dfrac{m(v-u)}{t}

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Which of the following hypotheses is written correctly?*
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A student of Physics applies a constant force of 25Newtons while lifting a heavy Physics book up into the air a distance of 2.0m
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In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.46 m. The mug
Tresset [83]

Answer:

(a). The horizontal velocity is 1.46 m/s.

(b). The direction of the mug's velocity just before it hit the floor is 74.7° below the horizontal.

Explanation:

Given that,

Height of the counter = 1.46 m

Distance = 0.80 m

We need to calculate the time

Using equation of motion

s_{y}=ut+\dfrac{1}{2}gt^2

s_{y}=\dfrac{1}{2}gt^2

Put the value into the formula

1.46=\dfrac{1}{2}\times9.8\times t^2

t^2=\dfrac{1.46\times 2}{9.8}

t=\sqrt{\dfrac{1.46\times2}{9.8}}

t=0.545\ sec

Here, horizontal velocity is constant

(a). We need to calculate the velocity

Using formula of velocity

v_{x}=\dfrac{d}{t}

Put the value into the formula

v_{x}=\dfrac{0.80}{0.545}

v_{x}=1.46\ m/s

(b). We need to calculate the final velocity

Using equation of motion

v_{f}^2=u^2+2as

Put the value into the formula

v_{f}^2=0+2\times9.8\times1.46

v_{f}=\sqrt{28.616}

v_{f}=5.34\ m/s

The velocity is 5.34 m/s downward.

We need to calculate the direction

Using formula of direction

\tan\theta=\dfrac{v_{y}}{v_{x}}

\theta=\tan^{-1}(\dfrac{5.34}{1.46})

\theta=74.7^{\circ}

Hence, (a). The horizontal velocity is 1.46 m/s.

(b). The direction of the mug's velocity just before it hit the floor is 74.7° below the horizontal.

5 0
4 years ago
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