Answer:
(a) They must have same direction
(b) It is not necessary for them to have same magnitudes
Explanation:
(a)
Momentum is a vector quantity. It is the product of mass (scalar) and velocity (vector). Thus, if the direction of velocity is changed, then as a result the direction of momentum will also change or its magnitude or component in the same direction will change. Hence, for the two objects to have same momentum, the directions of their velocities must also be the same.
(b)
Since, the momentum is product of velocity and mass. It is possible that two bodies of different masses with different velocities might have same momentum, provided the direction of their velocities is same.
For example, take a body of mass 4 kg moving with speed 5 m/s. It will have a momentum of 20 N.s. Now, consider another body of mass 2 kg, moving with speed 10 m/s. It will also have a momentum of 20 N.s.
Thus, it is not necessary for two objects to have same magnitude of velocity to have same momentum.
Answer:
The correct option is a behavioural models of the the to-be system.
Explanation:
As the Use case analysis method generates the analysis classes list such that the classes are capable of performing the behavior needed to make the system function successfully. From these analysis classes list the responsibilities of each class are defined.
As the list is derived from the behavioural models of the system, thus option a is the correct option.
Answer;
A.It’s not moving.
Explanation;
Position-Time graphs display the motion of a object by showing the changes of velocity with respect to time.
The motion of a car on a position-time graph that is represented with a horizontal line indicates that the car has stopped moving.
A straight line with a positive slope indicates that the car is moving at a constant velocity, and thus the slope is constant. On the other hand, a curve with a changing slope, shows that the velocity is changing.
Answer:
vT = v0/3
Explanation:
The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R
Since F = F0 = F'
GMm/R² = m(v0)²/R
GM = (v0)²R (1)
Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R
Since F1 = F'
GMm/27R² = m(vT)²/3R
GM = 27(vT)²R/3
GM = 9(vT)²R (2)
Equating (1) and (2),
(v0)²R = 9(vT)²R
dividing through by R, we have
9(vT)² = (v0)²
dividing through by 9, we have
(vT)² = (v0)²/9
taking square-root of both sides,
vT = v0/3
There are mistakes in the question as the unit of speed and height is not mention here.The correct question is here
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0m/s when the hand is 1.80m above the ground.How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)
Answer:
t=3.37s
Explanation:
Given Data
As we have taken hand at origin and positive upward
So given data are
To find
time taken by the ball before it hits the ground
Solution
By using the common kinematic equation
Put the given values and find for t
So
Apply quadratic formula to solve for t
