To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.
The extension of the spring due to the weight of the object on Earth is 0.3m, then
The extension of the spring due to the weight of the object on Moon is a value of , then
Recall that gravity on the moon is a sixth of Earth's gravity.
We have that the displacement at the earth was , then
Therefore the displacement of the mass on the spring on Moon is 0.05m
B) 6.22 N
<em>Is your answer hun! have a great day:3</em>
Answer:
f = [ 1/ 2L] √ [ T /u]
u is the mass per unit length.
u = 0.008 kg / 5m= 0.0016 kg/m
T/u = 39.24 / 0.0016 =24525
√ [ T /u] = 156.61
f = [ 1/ 2L] √ [ T /u] = [1/ 2*5 ] *156.61
f = 15.66 Hz
Complete Question
The complete question is shown on the first uploaded image
Answer:
The distance which the car skid is
Explanation:
From the question we are told that
The initial velocity of the car is
The coefficient of kinetic friction is
According to the law of energy conservation
The initial Mechanical Energy = The final Mechanical Energy
The initial mechanical energy is mathematically represented as
where KE is the initial kinetic energy which is mathematically represented as
And PE is the initial potential energy which is zero given that the car is on the ground
Now
Where is the work which friction exerted on the car which is mathematically represented as
Where is the distance covered by the car before it slowed down
=>
Answer:
a)v = 476.28 m / s
, b) T = 6.69 10⁵ N
, c) λ = 0.486 m
, d) λ = 0.35 m
Explanation:
a) The speed of a wave on a string is
v = √T /μ
also all the waves fulfill the relationship
v = λ f
they indicate that the fundamental frequency is f = 980 Hz.
The wavelength that is fixed at its ends and has a maximum in the center
L = λ / 2
λ = 2L
we substitute
v = 2 L f
let's calculate
v = 2 0.243 980
v = 476.28 m / s
b) The tension of the rope
T = v² μ
the density of the string is
μ = m / L
T = v² m / L
T = 476.28² 0.717 / 0.243
T = 6.69 10⁵ N
c) λ = 2L
λ = 2 0.243
λ = 0.486 m
d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air
v = λ f
λ= v / f
λ = 343/980
λ = 0.35 m