This question involves the concepts of the law of conservation of momentum.
The magnitude of the final momentum of the eight ball is "0.22 N.s".
According to the law of conservation of momentum:
where,
= initial momentum of the cue ball = 0.23 N.s
= initial momentum of the eight ball = 0 N.s (since ball is initially at rest)
= final momentum of the cue ball = 0.01 N.s
= final momentum of the eight ball = ?
Therefore,
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Answer:
75.6J
Explanation:
Hi!
To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,
Given the above we have the following equation.
Q=(m)(h2)-(m)(h1)
where
m=mass=1.3×10−3kg.
h2= entalpy at 37C
h1= entalpy at -20C
Q=m(h2-h1)
remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference
Q=mCp(T2-T1)
Cp= specific heat of air = 1020 J/kg⋅K
Q=(1.3×10−3)(1020)(37-(-20))=75.6J
Answer:
range of movement is 1.49 mm
Explanation:
given data
focal length = 45 mm
distance = 1.4 m
distance from the lens = 35 mm
distance from infinity down = 1.4 m =
to find out
range of movement
solution
we will apply here lens equation that is
1/f = 1/p + 1/q
here f = 45 and p = infinity to 1400 mm
we find here image distance that is q
1/45 = 0 + 1/q ......1
q = 45 mm
and
1/45 = 1/1400 + 1/q ......2
q = 46.49
so range of movement
that is 46.49 - 45
range of movement is 1.49 mm
Answer:
Maximum force will be equal to 720 N
Explanation:
We have given that spring constant 
Maximum stretch of the spring x = 6 cm = 0.06 m
We have to find the maximum force on the spring
We know that spring force is given by
So the maximum force which is necessary to relaxed the spring will be eqaul to 720 N
Answer:
<h2>a) 496N</h2><h2>b) 50.56kg</h2><h2>c) 1.90m/s²</h2>
Explanation:
According to newton's secomd law, ∑F = ma
∑F is the summation of the force acting on the body
m is the mass of the body
a is the acceleration
Given the normal force when the elevator starts N1 = 592N
Normal force after the elevator stopped N2 = 400N
When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)
When moving up;
N1 - Fg = ma
N1 = ma + Fg ...(1)
Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;
N2 - Fg = -ma
N2 = -ma+Fg ...(2)
Adding equation 1 and 2 we will have;
N1+N2 = 2Fg
592N + 400N = 2Fg
992N 2Fg
Fg = 992/2
Fg = 496N
The weight of the person is 496N
<em>\b) To get the person mass, we will use the relationship Fg = mg</em>
g = 9.81m/s
496 = 9.81m
mass m = 496/9.81
mass = 50.56kg
c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;
N1-N2 = 2ma
592-400 = 2(50.56)a
192 = 101.12a
a = 192/101.12
a = 1.90m/s²
