Answer:
B 1.3 m/s2
Explanation:
Newton's second law of motion states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force. This in form of an equation is F = m*a
Since you want to know the acceleration, you have to solve for a
F = m*a
a = F / m
and now you substitute the data
a = 3.2x10^3 N / 2.4x10^3 kg
a = 1.3 m/s^2
Answer:Average speed is greater than average velocity.
Explanation :
Given
student walks slowly along a straight line for a while ,then stops to rest a while, and finally runs quickly back to her initial position
Let x be the length of track and the whole process takes t time
For average speed 
Average speed
For average velocity 
Since displacement is zero as she returns to its initial position.
Average velocity=0
Therefore Average speed is greater than average velocity.
Answer:
w_f = m*V*cos(Q_n) / L*(m+M)
Explanation:
Given:
- mass of the putty ball m
- mass of the rod M
- Velocity of the ball V
- Length of the rod L
- Angle the ball makes before colliding with rod Q_n
Find:
What is the angular speed ωf of the system immediately after the collision,
Solution:
- We can either use conservation of angular momentum or conservation of Energy. We will use Conservation of angular momentum of a system:
L_before = L_after
- Initially the rod is at rest, and ball is moving with the velocity V at angle Q from normal to the rod. We know that the component normal to the rod causes angular momentum. Hence,
L_before = L_ball = m*L*V*cos(Q_n)
- After colliding the ball sicks to the rod and both move together with angular speed w_f
L_after = (m+M)*L*v_f
Where, v_f = L*w_f
L_after = (m+M)*L^2 * w_f
- Now equate the two expression as per conservation of angular momentum:
m*L*V*cos(Q_n) = (m+M)*L^2 * w_f
w_f = m*V*cos(Q_n) / L*(m+M)
Answer:
I) 57.5 m
Il) 50 m/s
Explanation:
Given that a boy stands at a certain distance from a large building and blows a whistle.
After 2.3s he hears the echo of the sound. The speed V of the sound will be:
V = 2X / T
Where X is the distance covered
V = 2X / 2.3
V = 0.87X ...... (1)
He moves 50m towards the building and blows his whistle again; this time the echo reaches him after 2.0s.
V = (2×50) / 2
V = 50 m/s
Substitutes the V into equation 1
50 = 0.87X
X = 50 / 0.87
57.5 m
Therefore, the Boys original distance from the building is 57.5m and the
Speed of sound in air is 50m/s
When a pendulum is at the midpoint of its oscillation, hanging straight down ...
-- that's the fastest it's going to swing, so its kinetic energy is maximum;
and
-- that's the lowest it's going to get, so its potential energy is minimum.
'c' is your choice.
