If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc
1 answer:
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Answer:
(a) 37.5 kg
(b) 4
Explanation:
Force, F = 150 N
kinetic friction coefficient = 0.15
(a) acceleration, a = 2.53 m/s^2
According to the newton's second law
Net force = mass x acceleration
F - friction force = m a
150 - 0.15 x m g = m a
150 = m (2.53 + 0.15 x 9.8)
m = 37.5 kg
(b) As the block moves with the constant speed so the applied force becomes the friction force.
Answer:
See below.
Explanation:
Cubit is a unit of length based on the length of the forearm from the elbow to the tip of the middle finger and usually equal to about 18 inches (46 centimeters).
It is an ancient unit of length used in ancient Egypt and is also known as "ancient Egyptian royal cubit."
Other similar units of measurements are displayed in the image below. Thanks!
(a) 4.06 cm
In a simple harmonic motion, the displacement is written as
(1)
where
A is the amplitude
is the angular frequency
is the phase
t is the time
The displacement of the piston in the problem is given by
(2)
By putting t=0 in the formula, we find the position of the piston at t=0:
(b) -14.69 cm/s
In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:
(3)
Differentiating eq.(2), we find
And substituting t=0, we find the velocity at time t=0:
(c) -101.13 cm/s^2
In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:
Differentiating eq.(3), we find
And substituting t=0, we find the acceleration at time t=0:
(d) 5.00 cm, 1.26 s
By comparing eq.(1) and (2), we notice immediately that the amplitude is
A = 5.00 cm
For the period, we have to start from the relationship between angular frequency and period T:
Using and solving for T, we find