Answer: 20,734.69 N/m
Explanation:
The elastic potential energy (ELPE) of the rubber band is given by
where
k is the spring constant
x = 0.035 m is the stretching of the rubber band
E = 12.7 J is the ELPE of the rubber band
Substituting the numbers and re-arranging the equation, we find
What is the magnitude of force required to accelerate a car of mass 1.7 × 10³ kg by 4.75 m/s²
Answer:
F = 8.075 N
Explanation:
Formula for force is;
F = ma
Where;
m is mass
a is acceleration
F = 1.7 × 10³ × 4.75
F = 8.075 N
28 times stronger than the force of gravity on the surface of the Earth.
Answer:
8.75
Explanation:
First, find the force of friction.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.9 m/s)² = F (1.4 m)
F = 11.7 N
Next, find the distance at the new velocity.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d
d = 8.75 m
For this case, the switch is located at point B of the diagram.
Remember that point D is the universal symbol for resistance.
In A what you have is a source of power and in C what you have is a cable.
Therefore, the answer for this case is B.
