Answer:
The work is -67.76 J
Explanation:
The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.
This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.
In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.
So, the loss of kinetic energy is 
You know:
- mass=m=0.22 kg
- Initial velocity of the ball:
Final velocity of the ball: 
Replacing:
= -67.76 J
Friction work is always negative because friction is always against displacement.
<u><em>The work is -67.76 J</em></u>
Answer:
The guitarist should increase the tension of the string.
Explanation:
The frequency of a vibrating string is determined by fn=(n/(2L))√T/μ. So if the tension in the string increased, the rate at which it vibrates (the frequency) will also increase.
Therefore it is advisable that she increase the tension of the string.
I hope it helps, please give brainliest if it does
Explanation:
Given:
v₀ = 0 m/s
a = 2.50 m/s²
t = 4 s
Find: v
v = at + v₀
v = (2.50 m/s²) (4 s) + 0 m/s
v = 10 m/s
Answer:
Two estimates
Explanation:
There are mainly two estimates used in the calculation of depreciation such as the useful life and the salvage value of an asset. The salvage value is defined as the predicted amount that will be obtained by a company from an asset when it is disposed at the end of the useful life of the particular asset. On the other hand, the useful life commonly refers to the estimation of how long the asset is useful for the company. This is different from the lifespan of the asset.
Answer:
-4.1μC is the final charge on the third sphere
Explanation:
From the given data, q1 and q2 are brought into contact as they are both conductors, as such there will be evenly distribution of charges.
a) charge on each sphere(Q) = q1 + q2 / 2
= +3.8 μC + (- 2.6 μC) / 2 = 1.2μC/2 = 0.6μC
b) Now, one of those two spheres is brought into contact with the third sphere ; Q is brought into contact with q3 = Q + q3 / 2
= 0.6μC - 8.8 μC /2 = -8.2 μC/2
= -4.1μC is the final charge on the third sphere.
