Answer:
An Object is positively charged if it has more Positive Electrons in that object
Answer:
The ball has no momentum
Explanation:
The given parameters are;
The mass of the ball = 5 kg
The velocity of the ball = 0 (The ball is sitting on the floor without moving)
The momentum of the ball = The mass of the ball × Velocity of the ball
Therefore, the momentum of the ball = 5 kg × 0 m/s = 0
The momentum of the ball is zero, the ball has no momentum.
Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.
Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:
Pcos 15°-N=0
Psin15°-f= m*ac
from the first we obtain N, the normal force
N=750Kg*9.8* cos (15°)= 7.1 *10^3 N
Then to calculate the frictional force (f) we can use the second equation
f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r
f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N
Answer:
I = 8.75 kg m
Explanation:
This is a rotational movement exercise, let's start with kinetic energy
K = ½ I w²
They tell us that K = 330 J, let's find the angular velocity with kinematics
w² = w₀² + 2 α θ
as part of rest w₀ = 0
w = √ 2α θ
let's reduce the revolutions to the SI system
θ = 30.0 rev (2π rad / 1 rev) = 60π rad
let's calculate the angular velocity
w = √(2 0.200 60π)
w = 8.683 rad / s
we clear from the first equation
I = 2K / w²
let's calculate
I = 2 330 / 8,683²
I = 8.75 kg m
Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
