Answer:
amount of work done, W = 549.36 kJ
Given:
mass of a car engine, m = 2500 kg
initial velocity, u = 45 mph
final velocity, v = 65 mph
1 mile = 1609
Solution:
We know that 1 hour = 3600 s
Now, velocities in m/s are given as:
u = 45 mph = = 20.11 m/s
v = 65 mph = = 29.05 m/s
Now, the amount of work done, W is given by the change in kinetic energy of the car and is given by:
W =
W =
W =
W = 549.36 kJ
Explanation:
(a) Displacement of an object is the shortest path covered by it.
In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.
0.4 miles = 0.64 km
displacement = 0.7-0.3+0.64 = 1.04 km
(b) Average velocity = total displacement/total time
t = 15 min = 0.25 hour
Hence, this is the required solution.
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
Answer:
Explanation:
Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as
now we can say that
now electric potential is given as
now work done to bring a small charge from infinite to the surface of this sphere is given as
here we know that
now the total energy of the sphere is given as