A 8 kilogram cat is resting on top of a bookshelf that is 2 meters high. What is the cat’s gravitational potential energy relati
1 answer:
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Answer: 0.5 seconds or 2.625 seconds
Explanation:
At t = 0, The ball is 4 ft above the ground.
The height of the football varies with time in the following way:
s(t) = -16 t² + 50 t + 4
we need to find the time in which the height would of the football would be 25 ft:
⇒25 = -16 t² + 50 t + 4
we need to solve the quadratic equation:
⇒ 16 t² - 50 t + 21 = 0
⇒ t = 0.5 s or 2.625 s
Therefore, at t = 0.5 s or 2.625 s, the football would be 25 ft above the ground.
Answer:
The current in the wire under the influence of the force is 216.033 A
Solution:
According to the question:
Length of the wire, l = 0.676 m
Magnetic field of the Earth,
Forces experienced by the wire,
Also, we know that the force in a magnetic field is given by:
I = 216.033 A
To solve this problem, we use the Law of Universal Gravitation:
F = Gm1m2/d^2
where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:
F = k/d^2
k = Fd^2
The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1
(8E+2)(d1)^2 = F2(2d1)^2
(8E+2)(d1)^2 = 4F2(d1)^2
(8E+2)=4F2
F2 = 200 Newtons
F = Gm1m2/d^2
where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:
F = k/d^2
k = Fd^2
The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1
(8E+2)(d1)^2 = F2(2d1)^2
(8E+2)(d1)^2 = 4F2(d1)^2
(8E+2)=4F2
F2 = 200 Newtons