Explanation:
first answer is that the to and fro movement of a pendulum is known as amplitude
Answer : The motion of the object is, non-uniform
Explanation :
Uniform motion : It is defined as the movement of an object along a straight line with constant speed. It travels equal distances in equal time interval.
The average speed of an object is similar to the actual speed of an object. The distance-time graph shows a straight line.
Non-uniform motion : It is defined as the movement of an object along a straight line with variable speed. It travels unequal distances in equal time interval.
The average speed of an object is different to the actual speed of an object. The distance-time graph shows a curved line.
The given graph B is a non-uniform motion graph.
Hence, the motion of the object is, non-uniform
The rocket would have gone 125 m away in deep space.The speed when it has gone 300 m is 76.7 m/s.
Answer:
The distance the rocket traveled is 125 m. The speed when it has gone 300 m is 76.7 m/s.
Explanation:
According to the equations of motion, if an object is moving with constant acceleration, then the distance travelled by the object with that acceleration and within a specific time can be obtained as
Since the initial velocity is given as u = 0 at t = 0. Then at t = 10 s, the acceleration is given by 2.5 m/.
Thus,
s= 0 + (×2.5×)=125 m.
Thus, the rocket would have gone 125 m away in deep space.
Similarly, if it had gone 300 m then its speed can be determined using the third equation of motion.
Since, here v is unknown and u is known as 0 with a = 2.5 m/.
2× 9.8×300=v²
5880=v²
Thus, the speed v =
So, the speed when it has gone 300 m is 76.7 m/s.
Answer:
The acceleration of the car is 7.85 m/s²
Explanation:
Given;
vx(t)= (0.910m/s³)t², given time traveled by the car 't' = 5.0 s
⇒To determine the velocity for 5 seconds, we substitute in 5.0 s for t
vx(5)= (0.910m/s³)(5s)²
= (0.910m/s³)(25s²)
vx = 22.75 m/s
⇒To determine the acceleration of the car when vx=12.0m/s
Acceleration is change in velocity per unit time
when vx=12.0m/s, our new equation becomes; 12 = (0.910m/s³)t²
Solving for t: t² = 12/0.91
t² = 13.187
t = √13.187 = 3.63 s
Acceleration = Δv/Δt
Acceleration = 7.85 m/s²