Answer:
9) This is a case of deceleration
10)-0.8 ms-2
b) acceleration is the change in velocity with time
11)
a) 100 ms-1
b) 100 seconds
12) 10ms-1
13) more information is needed to answer the question
14) - 0.4 ms^-2
15) 0.8 ms^-2
Explanation:
The deceleration is;
v-u/t
v= final velocity
u= initial velocity
t= time taken
20-60/50 =- 40/50= -0.8 ms-2
11)
Since it starts from rest, u=0 hence
v= u + at
v= 10 ×10
v= 100 ms-1
b)
v= u + at but u=0
1000 = 10 t
t= 1000/10
t= 100 seconds
12) since the sprinter must have started from rest, u= 0
v= u + at
v= 5 × 2
v= 10ms-1
14)
v- u/t
10 - 20/ 25
10/25
=- 0.4 ms^-2
15)
a=v-u/t
From rest, u=0
8 - 0/10
a= 8/10
a= 0.8 ms^-2
Answer:
They will come back at the same time.
Explanation:
The angular velocity equation of ω
where ω is the frequency of the movement, dependent on the angle. But since swings are simple pendulums and their angles of 8 and 4 degrees are small, they will come back to their starting points at the same time.
I hope this answer helps.
Answer:

Explanation:
We first identify the elements of this simple harmonic motion:
The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.
The angular frequency ω can be calculated with the formula:

Where k is the spring constant and m is the mass of the particle.
Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

Finally, the equation of the motion of the system is:
or

Answer:
248
Explanation:
L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H
= length of the slinky = 3 m
N = number of turns in the slinky
r = radius of slinky = 4 cm = 0.04 m
Area of slinky is given as
A = πr²
A = (3.14) (0.04)²
A = 0.005024 m²
Inductance is given as


N = 248