It depends where you are.
-- If you weigh 120 pounds on the Moon,
then your mass is 329.1 kilograms.
-- If you weigh 120 pounds on Mars,
then your mass is 143.8 kilograms.
-- If you weigh 120 pounds on the Earth,
then your mass is 54.4 kilograms.
(a) The spring stiffness constant of the spring is 18,392 N/m.
(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.
<h3>Kinetic energy of the car</h3>
The kinetic energy of the car is calculated as follows;
K.E = ¹/₂mv²
K.E = ¹/₂ x 950 x 22²
K.E = 229,900 J
<h3>Stiffness constant of the spring</h3>
The stiffness constant of the spring is calculated as follows;
K.E = U = ¹/₂kx²
k = 2U/x²
k = (2 x 229,900)/(5)²
k = 18,392 N/m
<h3>Force exerted on the spring</h3>
F = kx
F = 18,392 x 5
F = 91,960 N
<h3>Time of impact</h3>
F = mv/t
t = mv/F
t = (950 x 22)/(91960)
t = 0.23 s
Learn more about spring constant here: brainly.com/question/1968517
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Hello. The answer to your question is ''<span>hypothesis''. I hope this helps! </span>
Answer:
In my opinion I think the answer is C you don't have to choose C
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
