Answer:
70.6 mph
Explanation:
Car A mass= 1515 lb
Car B mass=1125 lb
Speed of car B is 46 miles/h
Distance before locking, d=19.5 ft
Coefficient of kinetic friction is 0.75
Initial momentum of car B=mv where m is mass and v is velocity in ft/s
46 mph*1.46667=67.4666668 ft/s
Initial momentum of car A is given by
where
is velocity of A
Taking East as positive and west as negative then the sum of initial momentum is
The common velocity is represented as
hence after collision, the final momentum is
From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence
The acceleration of two cars
From kinematic equation
hence
Substituting the value of
in equation
A. it is <span>located at a distance of 2.6 million light years from earth</span>
Answer:
W = 3.1 N
Explanation:
moments about any convenient point will sum to zero.
I choose summing about the knife edge mark and will assume the ruler of weight W is of uniform construction.
I will assume the ruler weight makes a positive moment
W[55 - 50) - 0.040(9.8)[ 95 - 55] = 0
5W = 15.68
W = 3.136
Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s