Answer:
a) W = 46.8 J and b) v = 3.84 m/s
Explanation:
The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy
W = ΔK = 
-K₀
a) work is the scalar product of force by distance
W = F . d
Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.
W = F d cos θ
W = 39.0 1.20 cos 0
W = 46.8 J
b) zero initial kinetic language because the package is stopped
W -
= -K₀
W - fr d= ½ m v² - 0
W - μ N d = ½ m v
on the horizontal surface using Newton's second law
N-W = 0
N = W = mg
W - μ mg d = ½ m v
v² = (W -μ mg d) 2/m
v = √(W -μ mg d) 2/m
v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3
]
v = √(31.63 2/4.3)
v = 3.84 m/s
Answer:
The speed of the cart and clay after the collision is 50 cm/s .
Explanation:
Given :
Mass of lump , m = 500 g = 0.5 kg .
Velocity of lump , v = 30 cm/s .
Mass of cart , M = 1 kg .
Velocity of cart , V = 60 cm/s .
We know by conservation of momentum :
Here , 
is the speed of the cart and clay after the collision .
Putting all value in above equation .
We get :
Hence , this is the required solution .
Answer:
0 J
Explanation:
given,
mass of the ball = 5 kg
radius of the horizontal circle = 0.5 m
tension in the string = 10 N
Work done = ?
Work done by the tension in the circular path will be equal to zero.
This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.
so, work done = F s cos θ
θ = 90°,
work done = F s cos 90° ∵ cos 90° = 0
Work done = 0 J
The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.
<h3>Change in energy level of the electron</h3>
When photons jump from a higher energy level to a lower level, they emit or radiate energy.
The change in energy level of the electrons is calculated as follows;
ΔE = Eb - Ef
ΔE = -2.68 eV - (-5.74 eV)
ΔE = 3.06 eV
Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.
Learn more about energy level here: brainly.com/question/14287666
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Answer:
The fractional kinetic energy will be lost if the collision is inelastic. In inelastic collision, the kinetic energy is converted into other forms of energy.
The lost energy became heat and sound energy.
Explanation:
During inelastic collision, the kinetic energy of a moving object does not conserve. It changes into another form of energy such as sound energy and heat energy etc.
For example, when a moving car hit another car or wall etc, the kinetic energy is converted into sound and heat energy. This type of collision is inelastic collision.
