The motorbike reaches 100 km/h in 3.5 seconds
Explanation:
The motion of the motorbike is a uniformly accelerated motion (= constant acceleration), therefore we can use the following suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
For the motorbike in this problem,
u = 0 (it starts from rest)
is the final velocity
is the acceleration
Solving for t, we find the time it takes for the bike to reach that velocity:

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Answer:
10kg
Explanation:
Let PE=potential energy
PE=196J
g(gravitational force)=9.8m/s^2
h(change in height)=2m
m=?
PE=m*g*(change in h)
196=m*9.8*2
m=10kg
Answer:
(a) 3.807 s
(b) 145.581 m
Explanation:
Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.
The distance traveled by car after Δt (seconds) at
speed is

The distance traveled by the motorcycle after Δt (seconds) at
speed and acceleration of a = 8 m/s2 is


We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:





(b)


Answer:
1. 
2. 
3. 
Explanation:
Given:
- mass of slinky,

- length of slinky,

- amplitude of wave pulse,

- time taken by the wave pulse to travel down the length,

- frequency of wave pulse,

1.



2.
<em>Now, we find the linear mass density of the slinky.</em>


We have the relation involving the tension force as:




3.
We have the relation for wavelength as:


