Answer:
(a) 41.75m/s
(b) 4.26s
Explanation:
Let:
Distance, D = 89m
Gravity,
= 9.8 m/
Initial Velocity,
= 0m/s
Final Velocity,
= ?
Time Taken,
= ?
With the distance formula, which is
D =
+ 
and by substituting what we already know, we have:
89 =
×9.8×
With the equation above, we can solve for
:

Now that we have solved
, we can use the following velocity formula to solve for
:
, where
is also equals to
, so we have

By substituting
,
, and
,
We have:

100N describes the weight of the sandbag, while 100kg is the mass of the sandbag.
To calculate acceleration, divide your weight by the mass, thus the accleration is:
Answer:
wavelenght
Explanation:
The wavelength is the spatial period of a wave, analogous to the temporal period, it is the distance between two consecutive points with maximum amplitude that are repeated in space . In the waves of the sea, the wavelength is easily observed in the separation between two consecutive ridges.
Bbhjjjhfffgggtffffrttyyhhhhhhhh
If the distance between two charges is halved, the electrical force between them increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:

where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:

the magnitude of the force changes as follows:

so, the force increases by a factor 4.