Answer:
<h2>
6.36 cm</h2>
Explanation:
Using the formula to first get the image distance
1/f = 1/u+1/v
f = focal length of the lens
u = object distance
v = image distance
Given f = 16.0 cm, u = 24.8 cm
1/v = 1/16 - 1/24.8
1/v = 0.0625-0.04032
1/v = 0.02218
v = 1/0.02218
v = 45.09 cm
To get the image height, we will us the magnification formula.
Mag = v/u = Hi/H
Hi = image height = ?
H = object height = 3.50 cm
45.09/24.8 = Hi/3.50
Hi = (45.09*3.50)/24.8
Hi = 6.36 cm
The image height is 6.36 cm
Answer:
1000 N
Explanation:
First, we need to find the deceleration of the running back, which is given by:
where
v = 0 is his final velocity
u = 5 m/s is his initial velocity
t = 0.5 s is the time taken
Substituting, we have
And now we can calculate the force exerted on the running back, by using Newton's second law:
so, the magnitude of the force is 1000 N.
The acceleration of the crate after it begins to move is 0.5 m/s²
We'll begin by calculating the the frictional force
Mass (m) = 50 Kg
Coefficient of kinetic friction (μ) = 0.15
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (N) = mg = 50 × 10 = 500 N
<h3>Frictional force (Fբ) =?</h3>
Fբ = μN
Fբ = 0.15 × 500
<h3>Fբ = 75 N</h3>
- Next, we shall determine the net force acting on the crate
Frictional force (Fբ) = 75 N
Force (F) = 100 N
<h3>Net force (Fₙ) =?</h3>
Fₙ = F – Fբ
Fₙ = 100 – 75
<h3>Fₙ = 25 N</h3>
- Finally, we shall determine the acceleration of the crate
Mass (m) = 50 Kg
Net force (Fₙ) = 25 N
<h3>Acceleration (a) =?</h3>
a = Fₙ / m
a = 25 / 50
<h3>a = 0.5 m/s²</h3>
Therefore, the acceleration of the crate is 0.5 m/s²
Learn more on friction: brainly.com/question/364384
Answer:
dsin∅ = m×
λ
so, dsin∅red = 3(670nm)
also, dsin∅? =5λ?
however ,if they overlap then dsin∅red = dsin∅?
3(670nm) /5 =402nm
∴λ = 402nm
Explanation:
