Answer
It increases to B, then it decreases.
When the stone thrown upwards, it moves vertically upwards to a maximum height then is starts to come down. Due to gravitational force, the velocity of the stone projected upwards decreases but when coming down wards is increases.
The trajectory shown in the diagrams represents an example of an object projected upwards.
We can then says that the vertical component increases from A to B, then it decreases to C.
If one of the variables is changed, that tells nothing about what happens to the other one, or IF anything happens, or when, or how long it lasts. Because they are UN-RELATED. You just said so yourself.
None of the choices says this.
Answer:
60000N
Explanation:
acceleration is change in velocity
a =(v-u)/t where a is acceleration u is initial velocity and v is final velocity
a = (0-60)/5 = -60/5= - 12m/s^2
here minus sign shows that body is decelerating and force is force of friction Now f = ma here f is force of friction m is mass and a is acceleration
f= 5000×- 12= -60000N
MINUS SIGN HERE SHOWS FORCE OF FRICTION
Hence force of friction is 60000N
Answer:
a 
b 
Explanation:
Generally the force constant is mathematically represented as
substituting values given in the question
=> 
=> 
Generally the workdone in stretching the spring 3.5 m is mathematically represented as
=> 
=>
Generally the workdone in compressing the spring 2.5 m is mathematically represented as
=> 
=>
Answer:
0.62 m/s
Explanation:
Given that a model train, travelling at speed , approaching a buffer model train buffer spring
The train, of mass 2.5 kg, is stopped by compressing a spring in the buffer.
After the train has stopped, the energy stored in the spring is 0.48 J.
Calculate the initial speed v of the train
Solution
The total energy stored in the spring will be equal to the kinetic energy of the train.
That is,
1/2fe = 1/2mv^2
Substitutes the spring energy, and mass into the formula
0.48 = 1/2 × 2.5 × V^2
2.5V^2 = 0.96
V^2 = 0.96 / 2.5
V^2 = 0.384
V = sqrt ( 0.384 )
V = 0.62 m/s
Therefore, the initial velocity of the train is 0.62 metres per second.
