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antiseptic1488 [7]
3 years ago
8

Which of the following is an accurate comparison of the weight of an astronaut on the moon and the Earth? The weight of the astr

onaut is the same on both. The weight of the astronaut on the moon is greater than the weight of the astronaut on the Earth. The weight of the astronaut on the moon is less than the weight of the astronaut on the Earth. The weight of the astronaut on the moon is half the weight of the astronaut on the Earth.
Physics
2 answers:
LiRa [457]3 years ago
7 0

The mass of the astronaut is the same on both, but weight is actually a force and it depends on the acceleration due to gravity.  On the moon, the acceleration due to gravity is 1/6 of the Earth’s so the astronaut’s weight will be 1/6 lighter on the moon.

Lorico [155]3 years ago
7 0

Answer:

The weight of the astronaut on the moon is less than the weight of the astronaut on the Earth.            

Explanation:

Matter contained in a body is known as mass. It remains constant even if one goes to another celestial body. Weight is the force due to gravity acting on a mass. Since, gravitational force varies at each celestial body, the weight also changes.

The acceleration due to gravity on the Moon is \frac{1}{6}^{th} the acceleration due to gravity on the Earth. Therefore, the astronaut would weigh less on moon than on the Earth. His weight would be \frac{1}{6}^{th} of that on Earth.

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A 5 kg object moving to the right at 4 m/s collides inelastically with a 5 kg object
skad [1K]

Answer:

The linear momentum of a particle with mass m moving with velocity v is defined as

p = mv (7.1)

Linear momentum is a vector . When giving the linear momentum of a particle you must

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dt = ma = Fnet

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5 0
3 years ago
The velocity of a motor car moving along a road increases from 10m/s to 50{1} ms in 8s. Find its avarege acceleration. RESULT
lesantik [10]

Answer:

hope it helps you................

7 0
3 years ago
How many protons, neutrons, and electrons does a neutral atom of this element have? (round atomic mass to nearest whole number)
adelina 88 [10]

Answer:

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8 0
3 years ago
Read 2 more answers
Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441
Tasya [4]
The period of a simple pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
g= \frac{4 \pi^2}{T^2}L (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz
And its period is the reciprocal of its frequency:
T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s

So now we can use eq.(1) to find the gravitational acceleration of the planet:
g= \frac{4 \pi^2}{T^2}L =  \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2
3 0
3 years ago
Alarge plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa m1/2. If the plate is expos
Korvikt [17]

Answer:

minimum length of a surface crack is 18.3 mm

Explanation:

Given data

plane strain fracture toughness K = 82.4 MPa m1/2

stress σ = 345 MPa

Y = 1

to find out

the minimum length of a surface crack

solution

we will calculate length by this formula

length = 1/π ( K / σ Y)²

put all value

length = 1/π ( K / σ Y)²

length = 1/π ( 82.4 10^{3/2} / 345× 1)²

length = 18.3 mm

minimum length of a surface crack is 18.3 mm

4 0
3 years ago
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