A pendulum at position A is released and swings through position B to position Con the other side.
B
1. Describe the total mechanical energy at each of the following positions. (3)
A.
B.
C
Answers
Explanation:
Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),
the conditions can be summarized bellow
point A
initial velocity= 0
final velocity=0
P.E= Max
K.E= 0
point B
initial velocity= maximum
final velocity=maximum
P.E=K.E
point C
initial velocity= min
final velocity=min
P.E= 0
K.E= max
Related Questions
1. Find the energy required to melt 255g of ice at 0°C into water at 0°C
Answers
Answer:
E = 85170 J (/ 85.2 kJ)
Explanation:
Take the latent heat of fusion of water be 334J / g.
From the equation E = ml,
E = energy required (unknown),
mass m = 255g,
latent heat of fusion l = 334J / g,
E = 255 x 334
E = 85170 J (/ 85.2 kJ)
Two teams are playing tug-of-war. Team A, on the left, is pulling on the rope with an effort of 5000 N. If the rope is moving at a constant velocity, how hard and in which direction is team B pulling?
A. 2500 N to the left
B. 5000 N to the right
C. 2500 N to the right
D. 5000 N to the left
Answers
Explanation:
If Team A is on the left, B is on the right
if the force is constant, it means that the effort applied is equal.
So Team B is pulling 5000N to the right.
A student throws a 120 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0.15 s
Answers
Answer:
The magnitude of the average force on the wall during the collision is 6 N.
Explanation:
Given;
mass of snowball, m = 120 g = 0.12 kg
velocity of the snowball, v = 7.5 m/s
duration of the collision between the snowball and the wall, t = 0.15 s
Magnitude of the average force can be calculated by applying Newton's second law of motion;
F = ma
where;
a is acceleration = v / t
a = 7.5 / 0.15
a = 50 m/s²
F = ma
F = 0.12 x 50
F = 6 N
Therefore, the magnitude of the average force on the wall during the collision is 6 N.
Choose the friction which opposes the relative motion between surfaces in motion a.Static friction b.Kinetic friction c.Sliding friction d.Both kinetic and sliding friction
Answers
Answer:
d. Both kinetic and sliding friction
Explanation:
Kinetic friction, commonly known as sliding friction, happens when a body with its surfaces in contact is in relative motion with another. It's the frictional force slowing it down, and finally stopping a moving body. One can describe sliding friction as the resistance any two objects create while sliding against each other. It is often documented as the force required to hold a surface moving along another surface. It is determined by two variables- one is material of the object and another is its weight.
when a 0.622kg basketbll hits the floor its velocit changes from 4.23m/s down to 3.85m/s up. if the averge force was 72.9N how much time was it in contact with the floor?
Answers
Answer:
Time, t = 3.2 ms
Explanation:
It is given that,
Mass of basketball, m = 0.622 kg
Initial velocity, u = 4.23 m/s
Final velocity, v = 3.85 m/s
Average force acting on the ball, F = 72.9 N
We need to find the time of contact of the ball with the floor. Let t is the time of contact. So,
[tex]F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-4.23)}{72.9}\\\\t=0.0032\ s\\\\\text{or}\\\\t=3.2\ ms[/tex]
So, the ball is in contact with the floor for 3.2 ms.