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
2 answers:
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 the acceleration due to gravity on the Earth. Therefore, the astronaut would weigh less on moon than on the Earth. His weight would be
of that on Earth.
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Answer:
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The period of a simple pendulum is given by:
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
(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
And its period is the reciprocal of its frequency:
So now we can use eq.(1) to find the gravitational acceleration of the planet:
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
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
And its period is the reciprocal of its frequency:
So now we can use eq.(1) to find the gravitational acceleration of the planet:
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 / 345× 1)²
length = 18.3 mm
minimum length of a surface crack is 18.3 mm