Hello! Assuming that the only force acting on the mass is 30N...
Fnet = 30N
Fnet = ma (mass x acceleration)
ma = 30N
a = 30N / m
a = 30N / 7kg
a = 4.2857 m/s^2
a = 4 m/s^2
I hope this helps!
Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;

(a) the force constant of the spring

(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k

The time period for guilty party was between 1900-1988.
You cant skip nothing. All the lessons are important! Sorry about you free time :/
Answer:
<em>a) 3.56 x 10^22 N</em>
<em>b) 3.56 x 10^22 N</em>
<em></em>
Explanation:
Mass of the sun M = 2 x 10^30 kg
mass of the Earth m = 6 x 10^24 kg
Distance between the sun and the Earth R = 1.5 x 10^11 m
From Newton's law,
F = 
where F is the gravitational force between the sun and the Earth
G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2
m is the mass of the Earth
M is the mass of the sun
R is the distance between the sun and the Earth.
Substituting values, we have
F =
= <em>3.56 x 10^22 N</em>
<em></em>
A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>
b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>