Explanation:
a) Given in the y direction (taking down to be positive):
Δy = 50 m
v₀ = 0 m/s
a = 10 m/s²
Find: t
Δy = v₀ t + ½ at²
50 m = (0 m/s) t + ½ (10 m/s²) t²
t = 3.2 s
b) Given in the x direction:
v₀ = 12 m/s
a = 0 m/s²
t = 3.2 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (12 m/s) (3.2 s) + ½ (0 m/s²) (3.2 s)²
Δx = 38 m
I think the correct answer from the choices listed above is option B. When calculating the power bill, power companies use kilowatt-hours. This unit is a derived unit of energy equal to 3.6 MJ. If energy is being transmitted or used at a constant rate (power) over a period of time, the total energy in kilowatt-hours is the product of the power in kilowatts and the time.
Answer:
F = 200 N
Explanation:
Given that,
The mass suspended from the rope, m = 20 kg
We need to find the resultant force acting on the rope. The resultant force on the rope is equal to its weight such that,
F = mg
Where
g is acceleration due to gravity
Put all the values,
F = 20 kg × 10 m/s²
F = 200 N
So, the resultant force on the mass is 200 N.
Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,

Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀
=A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values

b) 

, where
is time period of damped SHM
⇒
let
be number of oscillations made
then, 
⇒