When an object does not move even on pushing , static frictional force acts on in opposite direction of the applied force to stop the object from moving. static frictional force is a self adjusting force and it adjust its value according to the applied force if the applied force is smaller than the maximum value of static frictional force. The object starts moving once the applied force on it becomes greater than the maximum static frictional force. hence the statement is true.
Answer:
Explanation:
velocity of projection, vo = 381 m/s
angle of projection, θ = 73.5°
The formula for the range is
R = 8067.4 m
Range in shorten by 34.1 %
So, the new range is
R' = 8067.4 - 34.1 x 8067.4/100
R' = 5316.4 m
Answer:
<h2>
6.36 cm</h2>
Explanation:
Using the formula to first get the image distance
1/f = 1/u+1/v
f = focal length of the lens
u = object distance
v = image distance
Given f = 16.0 cm, u = 24.8 cm
1/v = 1/16 - 1/24.8
1/v = 0.0625-0.04032
1/v = 0.02218
v = 1/0.02218
v = 45.09 cm
To get the image height, we will us the magnification formula.
Mag = v/u = Hi/H
Hi = image height = ?
H = object height = 3.50 cm
45.09/24.8 = Hi/3.50
Hi = (45.09*3.50)/24.8
Hi = 6.36 cm
The image height is 6.36 cm
Answer:
A Fan Cart Initially Has An Acceleration Of 1.6m/s/s When It's Fan Is Directed Straight Backwards. If You Rotate The Fan By 45o, By What Percentage Do You Expect The Fan Cart's Thrust To Decrease? (Answer Should Be In Units Of 96)
a. 45%
b. 29%
c. 71%
d. 50%
The correct answer is d.
d. 50%
Explanation:
Fan cart acceleration = 1.6 m/s²
Thrust = 0.25×π×D²×ρ×v×Δv
where Δv = acceleration component and all factors remaining cconstant, when the fan is rotated by 45 ° the diameter changes to D₂ = sin 45 ×D
or 0.707×D. The thrust becomes 0.25×π×(0.707×D)²×ρ×v×Δv
=0.25×π×0.5×D²×ρ×v×Δv or 0.5(0.25×π×D²×ρ×v×Δv)
That is the thrust reduces by 50 %
To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.
Kepler's third law tells us that the period is defined as
The given data are given with respect to known constants, for example the mass of the sun is
The radius between the earth and the sun is given by
From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun
Therefore:
Substituting in Kepler's third law:
Therefore the period of this star is 3.8years
