Answer:
Explanation:
At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation
Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.
Simplifying
v = 
Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get
v = ![[tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B2%5Ctimes%20%281.2%5Ctimes10%5E%7B-5%7D%29%5E2%282182-1.225%29%7D%7B9%5Ctimes%201.8%5Ctimes10%5E%7B-5%7D%7D)
[/tex]
v = 387 x 10⁻⁵ m/s
Terminal velocity = 387 x 10⁻⁵ m/s
Time taken to fall a distance of 100 m
= 
= 2.6 x 10⁴ s.
The answer is A. <span>Some work input is used to overcome friction. </span>
(a) 392 N/m
Hook's law states that:
(1)
where
F is the force exerted on the spring
k is the spring constant
is the stretching/compression of the spring
In this problem:
- The force exerted on the spring is equal to the weight of the block attached to the spring:
- The stretching of the spring is
Solving eq.(1) for k, we find the spring constant:
(b) 17.5 cm
If a block of m = 3.0 kg is attached to the spring, the new force applied is
And so, the stretch of the spring is
And since the initial lenght of the spring is
The final length will be
It must have a medium. It must travel in empty space. Mechanical waves are waves which needs medium of propogation.
Answer:
L= 0.059 mH
Explanation:
Given that
R = 855 Ω and C = 6.25 μF
V= 84 V
Frequency
ω = 51900 1/s
We know that
L=Inductance
C=Capacitance
ω =angular Frequency
ω² L C =1
(51900)² x L x 6.25 x 10⁻⁶ = 1
L= 5.99 x 10⁻⁵ H
L= 0.059 mH
