The moon has a mass of 7.4 × 1022 kg and completes an orbit of radius 3.8×108 m about every 28 days. The Earth has a mass of 6 ×
1 answer:
To solve this problem it is necessary to apply the kinematic equations of movement description, where the speed of a body is defined as the distance traveled in a given time, that is to say
where,
x = Displacement
t = time
Since the moon is making a path equal to that of a circle, we know by geometry that the perimeter of a circle is given by
At the same time we have that time is equal to
Using the equation for velocity we have finally that
Therefore the speed of the Moon in its orbit is 987.92m/s
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Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A =
where d is the diameter of the wire
1.613 x 10^-7 =
6.448 x 10^-7 = 3.142 x
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
Explanation:
The electric field at a distance r from the charged particle is given by :
k is electrostatic constant
if r = 2 m, electric field is given by :
If r = 1 m, electric field is given by :
Dividing equation (1) and (2) we get :
So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.