Answer:
The third drop is 0.26m
Explanation:
The drop 1 impacts at time T is given by:
T=sqrt(2h/g)
T= sqrt[(2×2.4)/9.8]
T= sqrt(4.8/9.8)
T= sqrt(0.4898)
T= 0.70seconds
4th drops starts at dT=0.70/3= 0.23seconds
The interval between the drops is 0.23seconds
Third drop will fall at t= 0.23
h=1/2gt^2
h= 1/2×9.81×(0.23)^2
h= 0.26m
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:
c they obey inverse square law
Answer:
Force, 
Explanation:
Given that,
Mass of the bullet, m = 4.79 g = 0.00479 kg
Initial speed of the bullet, u = 642.3 m/s
Distance, d = 4.35 cm = 0.0435 m
To find,
The magnitude of force required to stop the bullet.
Solution,
The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :
Finally, it stops, v = 0
F = -22713.92 N
So, the magnitude of the force that stops the bullet is 
