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Sonbull [250]
3 years ago
11

A 600 N astronaut travels to an asteroid where the gravitational force is one-hundredth that of Earth. What is the astronaut's w

eight on the asteroid?
A
6 N

B
6 kg

C
600 kg

D
60,000 N
Physics
1 answer:
Ganezh [65]3 years ago
4 0

Answer:

A) 6N

Explanation:

the weight of the astronaut on earth can be calculated with the formula

Weight = mass * gravity of earth

Since the gravitational force is one-hundredth of Earth, the formula should be

Weight = mass * (gravity of earth / 100)

Weight = (mass * gravity of earth) / 100

Since you already know the weight, you only need to divide by 100

Weight = (mass * gravity of earth) / 100

Weight = 600N / 100

Weight = 6N

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1 year ago
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a blackbody is radiating with a characteristic wavelength of 9 microns what is the blackbody temperature answer in kelvin
Daniel [21]

This question involves the concepts of Wein's displacement law and characteristic wavelength.

The blackbody temperature will be "3.22 x 10⁵ k".

<h3>WEIN'S DISPLACEMENT LAW</h3>

According to Wein's displacement law,

\lambda_{max} T = c\\\\T=\frac{c}{\lambda_{max}}

where,

  • \lambda_{max} = characteristic wavelength = 9 μm = 9 x 10⁻⁹ m
  • T = temperature = ?
  • c = Wein's displacment constant = 2.897 x 10⁻³ m.k

Therefore,

T=\frac{2.897\ x\ 10^{-3}\ m.k}{9\ x\ 10^{-9}\ m}

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Learn more about characteristic wavelength here:

brainly.com/question/14650107

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What force is needed to give a 4.5-kg bowling ball an acceleration of 9 m/s2?
Mrac [35]
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It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

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0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

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Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

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3 years ago
What is an emergent ray​
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Answer:

Explanation:

the light ray leaving a medium in contrast to the entering or incident ray.

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