The question is incomplete.
The distance between the Moon and Earth influences: 1) the attractive gravitational force between them, 2) the tides, 3) the eclipses, 4) the period of each full turn of the moon around the Earth.
Assuming the question refers to the gravitational attraction, we must use the fact that, as per, Newton's Universal Gravitaional Law, the attractive force between the two bodies is inversely related to the square distance that separates them.
Then, if the Moon were twice as far, the gravitational pull would be one fourth (1/4) of actual pull.
A) 140 degrees
First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is
T = 32 s
So the angular velocity is
Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:
and substituting t = 75 seconds, we find
In degrees, it is
So, the new position is 140 degrees from the initial position at the top.
B) 2.7 m/s
The tangential speed, v, of a point at the egde of the wheel is given by
where we have
r = d/2 = (27 m)/2=13.5 m is the radius of the wheel
Substituting into the equation, we find
Answer:W = 1.23×10^-6BTU
Explanation: Work = Surface tension × (A1 - A2)
W= Surface tension × 3.142 ×(D1^2 - D2^2)
Where A1= Initial surface area
A2= final surface area
Given:
D1=0.5 inches , D2= 3 inches
D1= 0.5 × (1ft/12inches)
D1= 0.0417 ft
D2= 3 ×(1ft/12inches)
D2= 0.25ft
Surface tension = 0.005lb ft^-1
W = [(0.25)^2 - (0.0417)^2]
W = 954 ×10^6lbf ft × ( 1BTU/778lbf ft)
W = 1.23×10^-6BTU
Answer:
What is a Pure Substance?
Explanation:
It is something which cannot be divided into parts by physical means, as it's all made up of the same thing. Pure substances are either elements or compounds. Elements can NOT be separated into other types of matter (physically or chemically).
Answer:
The time period of the motion is, T = 0.03 s
The frequency of the rotation is, f = 30 Hz
Explanation:
Given data,
The rotational speed of an object, ω = 30 rpm
ω = 188.5 rad/s
The time period of motion is,
T = 2π / ω
Substituting the given values in the above equation
= 2π / 188.5
T = 0.03 s
The time period of the motion is, T = 0.03 s
The frequency of rotation,
f = 1 /T
= 1 / 0.03
= 30 Hz
Hence, the frequency of the rotation is, f = 30 Hz
