Explanation:
Given that,
Initial speed of a car, u = 60 km/h = 16.67 m/s
Acceleration, a = 2m/s²
Final speed, v = 120 km/h = 33.33 m/s
We need to find the distance traveled and the time taken to make the distance.
acceleration = rate of change of velocity

let the distance be d.

Hence, the distance traveled and the time taken to make the distance is 208.25 m and 8.33 seconds respectively.
Answer:
-20000 kgm/s
Explanation:
Impulse: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is kgm/s.
Mathematically, impulse can be expressed as
I = m(v-u).............. Equation 1.
Where I = impulse applied to the car to bring it to rest, m = mass of the car, u = initial velocity of the car, v = final velocity of the car.
Given: m = 1000 kg, u = 20 m/s, v = 0 m/s ( to rest)
Substitute into equation 1
I = 100(0-20)
I = 1000(-20)
I = -20000 kgm/s
Hence the impulse applied to the car to bring it to rest = -20000 kgm/s
Digital media<span> are any </span>media<span> that are encoded in machine-readable formats. </span>
Answer:
The new force is 1/4 of the previous force.
Explanation:
Given
---- 
--- 
Required
Determine the new force
Let the two particles be q1 and q2.
The initial force F1 is:
--- Coulomb's law
Substitute 2 for r1


The new force (F2) is

Substitute 4 for r2



Substitute 


The new force is 1/4 of the previous force.
Answer:
η = 58.8%
Explanation:
Work is defined as the force applied by the distance traveled by the body.

where:
W = work [J] (units of joules)
F = force = 294 [N]
d = distance = 5 [m]
![W = 294*5\\W = 1470 [J]\\](https://tex.z-dn.net/?f=W%20%3D%20294%2A5%5C%5CW%20%3D%201470%20%5BJ%5D%5C%5C)
Efficiency is defined as the energy required to perform an activity in relation to the energy actually added to perform some activity. This can be better understood by means of the following equation.
