Answer:
Elliptical galaxies
Explanation:
Edwin Hubble classified galaxies into three categories
Elliptical
Spiral
Lenticular
The elliptical galaxies have an elipsoidal shape roughly. They have stars which are old and the primary light source of the galaxy. The formation of new stars is very limited. This increases the brightness of the galaxy. The mass of the stars are low. So, far the percentage elliptical galaxies is low compared to other galaxies.
Answer:
μk = 0.26885
Explanation:
Conceptual analysis
We apply Newton's second law:
∑Fx = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
a= -0.9 m/s²,
g = 9.81 m/s² : acceleration due to gravity
W= 75 N : Block weight
W= m*g
m = W/g = 75/9.8= 7.65 kg : Block mass
Friction force : Ff
Ff= μk*N
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W-25 = 0
N = 75
+25
N= 100N
∑Fx = m*ax
20-Ff= m*ax
20-μk*100
= 7.65*(-0.90 )
20+7.65*(0.90) = μk*100
μk = ( 20+7.65*(0.90)) / (100)
μk = 0.26885
Answer:
The acceleration of the crate is 
.
Explanation:
Given that,
Force, F = 750 N
Mass of the crate, m = 250 kg
The coefficient of friction is 0.12.
We need to find the acceleration of the crate. The net force acting on the crate is given by :
f is frictional force, 
So, the acceleration of the crate is . Hence, this is the required solution.
We call "terminal velocity" the constant speed of a falling body
when it is no longer accelerating.
We know that if a body is not accelerating, then the net force
on it is zero.
From the question, we know that the downward force of gravity
on the skydiver is 800 N.
If the 800 N downward plus the air resistance upward add up
to zero, then the air resistance upward must also be 800 N.
