We need to considerate only the horizontal component of the motion of the toy car.
The formula for the distance in a decelerated motion is:
s = s₀ + v₀·t - 1/2·a·t²
where:
s₀ = initial position = 0
v₀ = initial velocity = 1.21 m/s
t = time elapsed = 0.342 s
a = deceleration = 0.131 m/s²
Plugging in numbers:
s = 0 + 1.21×0.342 - 0.5×0.141×(0.342)²
= 0.406 m
Hence, the toy car traveled a distance of about 41 cm.
Using F = Ke
F = 39200×(25÷100) = 39200×0.25
F= 9800N
Just read the back of the cake box
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
