The pressure of the air at the way its blowing
<h3>Answer</h3>
(A) Resistance is directly related to length.
<h3>Explanation</h3>
Formula for resistance
R = p(length) / A
where R = resistance
p = resistivity(material of wire)
A = cross sectional area
So it can be seen that resistance depends upon 3 factors that are length of wire , resistivity of wire and the cross sectional area of the wire.
If two of the factors, resistivity and cross sectional area, are kept constant then the resistance is directly proportional to the length of wire.
<h3> R ∝ length</h3>
This means that the resistance of the wire increases with the increase in length of the wire and decreases with the decrease of length of the wire.
Answer:
a
The orbital speed is 
b
The escape velocity of the rocket is 
Explanation:
Generally angular velocity is mathematically represented as
Where T is the period which is given as 1.6 days = 
Substituting the value
At the point when the rocket is on a circular orbit
The gravitational force = centripetal force and this can be mathematically represented as
Where G is the universal gravitational constant with a value 
M is the mass of the earth with a constant value of 
r is the distance between earth and circular orbit where the rocke is found
Making r the subject
![r = \sqrt[3]{\frac{GM}{w^2} }](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7BGM%7D%7Bw%5E2%7D%20%7D)
![= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B6.67%2A10%5E%7B-11%7D%20%2A%205.98%2A10%5E%7B24%7D%7D%7B%284.45%2A10%5E%7B-5%7D%29%5E2%7D%20%7D)
The orbital speed is represented mathematically as
Substituting value
The escape velocity is mathematically represented as
Substituting values
Answer:
Explanation:
Given that,
Spring constant k=200N/m
Compression x = 15cm = 0.15m
Attached mass m =2kg
Coefficient of kinetic friction uk= 0.2
The energy in the spring is given as
U =½kx²
U = ½ × 200 × 0.15²
U = 2.25J
Force in the spring is given by Hooke's law
F = ke
F = 200×0.15
F = 30N
The weight of body which is equal to the normal is give as
W = mg
W = 2 × 9.81
W = 19.62N
W = N = 19.62 Newton's 2nd Law
From law of friction,
Fr = uk•N
Fr = 0.2 × 19.62
Fr = 3.924
Using newton second law again
Fnet = F - Fr
Fnet = 30 - 3.924
Fnet = 26.076
Work done by net force is given as
W = Fnet × d
W = 26.076d
Then, the work done by this net force is equal to the energy in the spring
W = U
26.076d = 2.25
d = 2.25/26.076
d = 0.0863m
Which is 8.63cm
So the box will slide 8.63cm before stopping
