Complete question:
A 0.50 kilogram frog is at rest on the bank surrounding a pond of water. As the frog leaps from the bank, the magnitude of the acceleration of the frog is 4.0 meters per second^2. Calculate The magnitude of the net force exerted on the frog as it leaps.
Answer:
2.0N
Explanation:
Given that,
Mass, m of the frog = 0.5 kg
The acceleration of the frog = 4.0 m/s².
We have been asked To find,
The magnitude of the net force exerted on the frog as it leaps.
So
We calculate this using the formula below :
F = ma
When we insert the values into the formula, we have:
F = 0.5 kg × 4 m/s²
F = 2.0 N
Therefore, the magnitude of net force is 2.0 N.
A proton repelling another proton
Like charges of the protons would repel one another.
Answer:
Height of mirror 0.075 m
width of mirror 0.325 m
Explanation:
given data
wide = 1.3 m
high = 0.30 m
driver’s eyes = 0.50 m
rear window = 1.50 m
solution
we take here height / width of the mirror is
height / width =
.................1
and
height /width of the window is
height /width =
.................2
and
distance of eye / window by the mirror is
distance of eye / window =
.................3
so here
θ = θi = θr ....................4
and tanθ for vertical is
tanθ =
tanθ =
....................5
so
h =
....................6
put here value and we get
h = 0.075 m
and
when we take here tanθ for horizontal than it will be
tanθ =
tanθ =
.......................7
so
....................8
put here value and we get
w = 0.325 m