Answer:
2. A force of 50 N is applied to the object for a distance of 2.0 m. Assume that object（the mass of the object is 3kg）
was at rest at the beginning, what speed did it achieved because of the work done on it? (Hint:
Calculate the works performed by the force first.)
I figured that is 8.2m/S，I am just not sure can anyone help me i much appreciate it.
The answer is 7 (its like the only question thats easy so far lol)
Answer: Search Results
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Answer: Surface waves can have characteristics of both longitudinal and transverse waves in the following way; The motion of the surface waves is up and down which is perpendicular to the direction of the wave. This is similar to the motion of transverse waves whereas the the motion of longitudinal.
Explanation:
the friction force provided by the brakes is 30000 N.
<h3>What is friction force?</h3>
Friction force is the force that opposes the motion between two bodies in contact.
To calculate the average friction force provided by the brakes, we apply the formula below.
Formula:
- K.E = F'd............. Equation 1
Where:
- K.E = Kinetic energy of the train
- F' = Friction force provided by the brakes
- d = distance
Make F' the subject of the equation
- F' = K.E/d............ Equation 2
From the question,
Given:
Substitute these values into equation 2
- F' = (8.1 ×10⁶)/270
- F' = 30000 N
Hence, the friction force provided by the brakes is 30000 N
Learn more about friction force here: brainly.com/question/13680415
Answer:
X = 6910319.7 m
Explanation:
let X be the distance where the acceleration of gravity is 85% of what it is on the surface and g1 be the acceleration of gravity at the surface and g2 be the acceleration of gravity at some distance X above the surface.
on the surface of the earth, the gravitational acceleration is given by:
g1 = GM/(r^2) = [(6.67408×10^-11)(5.972×10^24)]/[(6371×10^3)^2] = 9.82 m/s^2
at X meters above the earth's surface, g2 = 85/100(9.82) = 8.35m/s^2
then:
g2 = GM/(X^2)
X^2 = GM/g2
X = \sqrt{GM/g2}
= \sqrt{(6.67408×10^-11)(5.972×10^24)/ 8.35
= 6910319.7 m
Therefore, the acceleration of gravity becomes 85% of what it is on the surface at 6910319.7 m .
