Answer:
The velocity is 19.39 m/s
Solution:
As per the question:
Mass, m = 75 kg
Radius, R = 19.2 m
Now,
When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.
where
v = velocity
g = acceleration due to gravity
Answer:
C) three
Explanation:
Let gram of gold required be m . Let temperature change in both be Δ t .
heat absorbed = mass x specific heat x change in temperature
for copper
heat absorbed = 1 x .385 x Δt
for gold
heat absorbed = m x .129 x Δt
So
m x .129 x Δt = 1 x .385 x Δt
m = 2.98
= 3 g approximately .
Highest frequency EM waves: cosmic rays and gama rays
Lowest frequency EM waves:
Radio and Tv waves
a una velocidad de
22 m/s, quien lo golpea y devuelve en la misma
dirección con una velocidad de 14 m/s. Si el
tiempo de contacto del balón con la jugadora es
de 0,03 s, ¿con qué fuerza golpeó la jugadora el
balón?
21 Una bala de 0,8 g, está en la recámara de un rifl e
cuando se g
Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.
An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).
The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:
ag=G(MEarth+MMoon)/r2
Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:
acentr=(4 pi2 r)/T2
Where T is the period. Since the two accelerations have to be equal, we obtain:
(4 pi2 r) /T2=G(MEarth+MMoon)/r2
Which implies:
r3/T2=G(MEarth+MMoon)/4 pi2=const.
This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.
This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.
