Answer:
The Gravitational Force between the 2 masses is approximately 1.209x10^32 Newton’s
Explanation:
Explanation:
Given that,
Length of the spring, l = 50 cm = 0.5 m
Mass connected to the end, m = 330 g = 0.33 kg
The mass is released and falls, stretching the spring by 30 cm before coming to rest at its lowest point. On applying Newton's second law, 10 cm below the release point, x = 15 cm
(a) When the mass is connected, the force of gravity is balanced by the force in spring.

(b) The amplitude of the oscillation will be 15 cm as it is half of the total distance travelled.
(c) The frequency of the oscillation is given by :

Hence, this is the required solution.
Complete Question:
In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.
a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.
b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.
c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.
d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?
Answer:
force, 1.318 ₓ 10⁻⁴
direction, 18.435°
Explanation:
The attached file gives a breakdown step by step solution to the questions
m = mass of the partner which the cheerleader lifts = 59.6 kg
h = height to which the partner is lifted by the cheerleader = 0.749 m
g = acceleration due to gravity = 9.8 m/s²
work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.
W = work done by the cheerleader in lifting the partner
PE = potential energy gained
so W = PE
potential energy is given as
PE = mgh
hence
W = mgh
inserting the values in the above formula
W = 59.6 x 9.8 x 0.749
W = 437.5 J
this is the work done in lifting the partner once.
the cheerleader does this 30 times , hence the total work done is given as
W' = 30 W
W' = 30 x 437.5
W' = 13125 J