The correct answer is:
the distance of the orbiting object to Earth.
In fact, we know that the gravitational force that keeps the object in circular motion around the Earth is equal to the centripetal force, so we can write:
If we re-arrange the equation, we find an expression for the tangential speed of the object:
and we see that it depends on 3 quantities: G, M (the mass of the Earth) and r (the distance of the object from the Earth).
Answer:
at T = 0ºC the change of state is from the solid state to the gaseous state
Explanation:
In this exercise we are asked about the changes of state, from the data we will assume that the material is water.
Water can exist in three solid states, liquid and gas, in a graph of pressure ℗ against temperature (T) there is a point called triple at T = 0.01ºC, below this point the curve has two states at high pressure solid and low pressure gas.
As a result of the previous ones at T = 0ºC the change of state is from the solid state to the gaseous state
Answer:
both
Explanation:
because when it is hot in summer 5hat is the air and the sund u can warm things up and then it get hot
Answer:
The density of the sample is 36 g/cm³
Explanation:
m= 972g
l=3cm
V = l³ = 3³ = 27 cm³
density = mass/volume
= 972/27
= 36 g/cm³
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
