Refer to the diagram shown below.
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)
Answer:
Energy Lost for group A's car = 0.687 J
Energy Lost for group B's car = 0.55 J
Explanation:
The exact question is as follows :
Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.
To find - How much energy is lost due to heat for group A's car ?
How much for Group B's car ?
Solution -
We know that,
GPE = 1 Joule (Potential Energy)
Now,
For Group A -
Energy Lost = GPE - KE
= 1 J - 0.313 J
= 0.687 J
So,
Energy Lost for group A's car = 0.687 J
Now,
For Group B -
Energy Lost = GPE - KE
= 1 J - 0.45 J
= 0.55 J
So,
Energy Lost for group B's car = 0.55 J
Answer:
S = V t where S is the horizontal distance traveled
1/2 g t^2 = H where H is the vertical distance traveled
t^2 = 2 H / g
V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations
tan theta = H / S
V^2 = S g / (2 tan theta)
Using S = L cos theta
V^2 = L g cos theta / (2 tan theta)
Giving V in terms of L and theta
Kinetic energy than parked
Answer:
Explanation:
Q1 = 35 nC = 35 x 10^-9 C
m = 3.5 micro gram = 3.5 x 10^-9 Kg
d = 35 cm = 0.35 m
(a) The electrostatic force between the two charges is balanced by the weight of another charge.
F = m g
(b) By substituting the values
Q2 = 13.34 x 10^-12 C
Q2 = 0.0134 nC
