Answer:
The speed of space station floor is 49.49 m/s.
Explanation:
Given that,
Mass of astronaut = 56 kg
Radius = 250 m
We need to calculate the speed of space station floor
Using centripetal force and newton's second law




Where, v = speed of space station floor
r = radius
g = acceleration due to gravity
Put the value into the formula


Hence, The speed of space station floor is 49.49 m/s.
Shadows are formed when an opaque object or an object that doesn't allow light to pass through is in the way or infront of etc. a source of light.
The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
Read more about vertical weight here:
brainly.com/question/15244771
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Answer:
Explanation:
First of all we shall find the velocity at equilibrium point of mass 1.2 kg .
It will be ω A , where ω is angular frequency and A is amplitude .
ω = √ ( k / m )
= √ (170 / 1.2 )
= 11.90 rad /s
amplitude A = .045 m
velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s
= .5355 m /s
At middle point , no force acts so we can apply law of conservation of momentum
m₁ v₁ = ( m₁ + m₂ ) v
1.2 x .5355 = ( 1.2 + .48 ) x v
v = .3825 m /s
= 38.25 cm /s
Let new amplitude be A₁ .
1/2 m v² = 1/2 k A₁²
( 1.2 + .48 ) x v² = 170 x A₁²
( 1.2 + .48 ) x .3825² = 170 x A₁²
A₁ = .0379 m
New amplitude is .0379 m