Given:
mass: 100 kg
height: 500 m
1 kJ = 1000 J
gravity = 9.8 m/s²
velocity before impact: v = √2gh ; v = √2 * 9.8 m/s² * 500 m ; v = 98.99494 m/s
KE = 1/2 m v²
KE = 1/2 * 100 kg * (98.99494 m/s)²
KE = 490,000 J
Pls. see attachment.
Answer:
A damped oscillation means an oscillation that fades away with time while Forced oscillations occur when an oscillating system is driven by a periodic force that is external to the oscillating system.
Explanation:
Damping is the reduction in amplitude (energy loss from the system) due to overcomings of external forces like friction or air resistance and other resistive forces. ... When a body oscillates by being influenced by an external periodic force, it is called forced oscillation.
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Answer:
Explanation:
a )
Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .
v = u + a t
20 = 0 + 2 t
t = 20 /2 = 10 s .
Total time = 10 s + 20 s + 5 s = 35 s .
b) Average velocity = Total distance travelled / total time
Distance travelled in first 10 s
S₁ = ut + 1/2 a t²
= 0 + .5 x 2 x 10²
= 100 m
Distance travelled in next 20 s
S₂= 20s x 20 m/s = 400 m
Distance travelled in last 5 s .
deceleration in last 5 s
v = u + at
0 = 20 m/s + a x 5
a = - 4 m/s²
v² = u² - 2 a s
0 = (20 m/s)² - 2 x 4 m/s² x s
s = 50 m
S₃ = 50 m
Total distance = S₁ + S₂ + S₃
= 100 m + 400 m + 50 m
= 550 m .
Average velocity = 550 m / 35 s
= 15.71 m /s .
Answer:
Option (A) , (b) and (d) are correct option
Explanation:
According to Coulomb's law electric force between two charges is given by
From the relation we can say that force is directly proportional to magnitude of charges and inversely proportional to distance between them '
So if we increase the distance then force will decrease
Increase if any of the charge get larger
If force is attractive then both the charge will be of different sign and is force is repulsive then both the charges of same sign
From above conclusion we can say that (a), (b) and (d) are correct option
Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.
<h3>What is electrical force?</h3>
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The given data in the problem is
q₁ is the negative charge = 6 µC=6×10⁻⁶ C
q₂ is the positive charge = 3 µC=3×10⁻⁶ C
r is the distance between the charges=0.002 m
is the electric force =?
The value of electric force will be;
Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.
To learn more about the electrical force refer to the link;
brainly.com/question/1076352
