Answer:
The instantaneous speed of the object after the first five seconds is 12.5 m/s.
(C) is correct option.
Explanation:
Given that,
An object starts at rest. Its acceleration over 30 seconds.
We need to calculate the instantaneous speed of the object after the first five seconds
We know that,
Area under the acceleration -time graph gives speed.
According to figure,




Hence, The instantaneous speed of the object after the first five seconds is 12.5 m/s.
Answer:
The answer to the question is
The two balls, although of different masses, could be made to have the same demolishing force by setting the velocity of the 100 kg ball to 1.5 times the velocity of the 150 kg ball.
That is if V₁ is the velocity of the 150 kg ball and V₂ is the velocity of the 100 kg ball then V₂ = 1.5×V₁ for the demolishing effect of the two balls to be equal.
Explanation:
To answer the we are required to explain the meaning of momentum and state its properties
Momentum is a physical property of an object in motion. It indicates the amount of motion inherent in the object. An object in motion is said to have momentum
The types of momentum possessed by an object can be classified into either
1, Linear momentum or
2. Angular momentum
An object moving with a velocity, v has linear momentum while a spinning object has an angular momentum
The momentum is given by the formula
P = m × V
Where m = mass and
V = velocity
Newtons second law of motion states that, the force acting on an object is equivalent to the rate of change of momentum produced and acting in the direction of the force
Properties of momentum
From the above statements it means that the two balls can be made equivalent by having the appropriate amount of speed. That iis the two balls can have the same momentum thus for equal momentum effect, we have
150 kg × V₁ = 100 kg × V₂
or V₂ = 1.5×V₁
Answer:
the acceleration during the collision is: - 5
Explanation:
Using the formula:

we get:

Answer:
T= 2p√m/k
Explanation:
This is because the period of oscillation of the mass of spring system is directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.
The period of a mass on a spring is given by the equation
T=2π√m/k.
Where T is the period,
M is mass
K is spring constant.
An increase in mass in a spring increases the period of oscillation and decrease in mass decrease period of oscillation.