components of the speed of the coin is given as
now the time taken by the coin to reach the plate is given by
now in order to find the height
so it is placed at 1.52 m height
The peak luminosity of a white dwarf supernova is around 1010 Lsun, and it remains brighter than 108 Lsun for about 150 days. In
comparison, the luminosity of a bright Cepheid variable star is about 10,000 Lsun. The Hubble Space Telescope is sensitive enough to make accurate measurements of apparent brightness for Cepheid variable stars up to a distance of about 100 million light-years. Estimate the distance of a fading white dwarf supernova of luminosity 108 Lsun whose apparent brightness is comparable to that of a bright Cepheid variable star 100 million light-years from Earth. How does the distance of that supernova compare to the size of the observable universe?
1 answer:
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The coefficient of friction is missing and it has a value of μ = 0.4
Answer:
a = 3.924 m/s²
Explanation:
I've attached the kinematic free body diagram.
Taking the sum of all upward and downward forces,
EFy = 0;
N1 + N2 - m_p•g - m_v•g = 0
N1 + N2 = m_p•g + m_v•g
Where;
N1 and N2 are the normal reactions at the wheels
m_p is the mass of the driver
m_v is the mass of the vehicle
g is the acceleration due to gravity.
Plugging in the relevant values in the question,we obtain;
N1 + N2 = (385 + 75) x 9.81
N1 + N2 = 4512.6N - - - (eq1)
Now, taking sum of all horizontal forces;
EFx = (m_p + m_v) x a
So,
μ(N1 + N2) = (mp + mv) x a
Thus,
0.4(N1 + N2) = (385 + 75)a
0.4(N1 + N2) = 460a
N1 + N2 = 1150a
From eq(1),N1 + N2 = 4512.6N
Thus,
1150a = 4512.6N
a = 4512.6/1150
a = 3.924 m/s²
Therefore, the acceleration, a = 3.924 m/s²
