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Answer:
F = 4.47 10⁻⁶ N
Explanation:
The expression they give for the strength of the tide is
F = 2 G m M a / r³
Where G has a value of 6.67 10⁻¹¹ N m² / kg² and M which is the mass of the Earth is worth 5.98 10²⁴ kg
They ask us to perform the calculation
F = 2 6.67 10⁻¹¹ 135 5.98 10²⁴ 13 / (6.79 10⁶)³
F = 4.47 10⁻⁶ N
This force is directed in the single line at the astronaut's mass centers and the space station
Answer:
(a) I=0.01 kg.m²
(b) I=0.03 kg.m²
Explanation:
Given data
Mass of disk M=2.0 kg
Diameter of disk d=20 cm=0.20 m
To Find
(a) Moment of inertia through the center of disk
(b) Moment of inertia through the edge of disk
Solution
For (a) Moment of inertia through the center of disk
Using the equation of moment of Inertia

For (b) Moment of inertia through the edge of disk
We can apply parallel axis theorem for calculating moment of inertia
Answer:
The magnitude of each charge is 
Explanation:
Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.
We need to calculate the magnitude of each charge
Using formula of force

Where, q = charge
r = separation
Put the value into the formula




Hence, The magnitude of each charge is 
Answer:
cart displacement is 66 m
Explanation:
given data
velocity = 5 m/s
acceleration = 2 m/s²
time = 6 s
to find out
What is the
magnitude of cart displacement
solution
we will apply here equation of motion to find displacement that is
s = ut + 0.5×at² .............1
here s id displacement and u is velocity and a is acceleration and time is t here
put all value in equation 1
s = ut + 0.5×at²
s = 5(6) + 0.5×(2)×6²
s = 66
so cart displacement is 66 m