The Speed of the car is 4m/s.
What is the speed?
The speed of an object can be defined as the total distance traveled by it in a particular interval of time. This can be determined by dividing the total distance moved by the object by the time taken to move.
It is the ratio of distance in m and time taken by the object to travel in seconds.
Hence: Distance D=28m
Time T=7s
Speed S=D/T
S=28m/7S
S=4m/S
Therefore car's speed is 4m/S
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Explanation :
Speed of car, 
Kinetic energy of the car, 
Speed of car, 
Let
is the kinetic energy of the car when it is moving with 100 km/h.




Initial velocity of both cars are 0. Using third equation of motion :
So,
and 
t is same. So,



So, the braking distance at the faster speed is twice the braking distance at the slower speed.
The work done onto the car is 506,250 J
The work done on a system implies an increase in the internal energy of the system as a result of some forces acting on the system from the outside.
From the parameters given:
- The mass of the car = 1500 kg
- The initial speed = 30 m/s
- The final speed = 15 m/s
The work done onto the car refers to the change in the kinetic energy (i.e. ΔK.E)



= 506,250 J
Therefore, we can conclude that the work done on the car is 506,250 J
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Answer:
Work, W = F * d, and
Work = change in kinetic energy, so W=deltaKE.
Hence,
deltaKE=F * d
(1/2)*m*v^2 =F * d
d=[(1/2)*m*v^2]/F
d=[(1/2)*0.6*20^2]/5
d=24 m.
Explanation:
Work = change in kinetic energy, so W=deltaKE.
For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively
(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s
<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>