Answer:
D. 2^(3/2)
Explanation:
Given that
T² = A³
Let the mean distance between the sun and planet Y be x
Therefore,
T(Y)² = x³
T(Y) = x^(3/2)
Let the mean distance between the sun and planet X be x/2
Therefore,
T(Y)² = (x/2)³
T(Y) = (x/2)^(3/2)
The factor of increase from planet X to planet Y is:
T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)
T(Y) / T(X) = (2)^(3/2)
Answer:
0.004 m
Explanation:
For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by

where
is the wavelength
D is the distance of the screen from the slit
d is the width of the slit
Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):

where we have
is the wavelength
D = 2.0 m is the distance of the screen
is the width of the slit
Substituting, we find

Answer:
58.32 N
Explanation:
Area of a circle = 

where r is the radius of the circle.
The cylinder has a radius of 0.02 m, its area is;
= 

=
x 
=
x 0.0004
= 1.2571 x 
Area of the cylinder is 0.0013
.
The safety valve has a radius of 0.0075 m, its area is;
= 

=
x 
=
x 5.625 x 
= 1.7679 x 
Area of the valve is 0.00018
.
From Hooke's law, the force on the safety valve can be determined by;
F = ke
= 950 x 0.0085
= 8.075 N
Minimum force,
, required can be determined by;
= 
= 
= 
= 58.32
The minimum force that must be exerted on the piston is 58.32 N.
Answer:
Angular velocity of an ultra centrifuge is 17146.42 rad/s.
Explanation:
Given that,
Acceleration of an ultra centrifuge, 
Distance from axis, r = 2 cm = 0.02 m
We need to find the angular speed. We know that the relation between angular speed and angular acceleration is given by :

So, the angular velocity of an ultra centrifuge is 17146.42 rad/s.
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