In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.
Static frictional force = ƒs = (Cs) • (Fɴ)
2.26 = (Cs) • m • g
2.26 = (Cs) • (1.85) • (9.8)
Cs = 0.125
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
1.49 = (Cκ) • m • g
1.49 = (Cκ) • (1.85) • (9.8)
Cκ = 0.0822
Answer:
<em>The range is 35.35 m</em>
Explanation:
<u>Projectile Motion</u>
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and
the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:



The range is 35.35 m
Answer:
1047 miles
Explanation:
The radius of the Earth is
(miles)
So its circumference, which is the total length of the equator, is given by

Now we know that the Earth rotates once every 24 hours. So the distance through which the equator moves in one hour is equal to its total length divided by the number of hours, 24:

The elastic potential energy of the spring is 0.31 J
Explanation:
The elastic potential energy of a spring is given by

where
k is the spring constant
x is the compression/stretching of the spring
For the spring in this problem, we have:
k = 500 N/m (spring constant)
x = 0.035 m (compression)
Substituting, we find the elastic potential energy:

Learn more about potential energy:
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