Apoint source of light illuminates an aperture 2.00 m away. a 15.0 cm -wide bright patch of light appears on a screen 2.00 m behind the aperture. how wide is the aperture?

wide is the aperture is 7.5 cm

Explanation:

Given data

aperture away =  2.00 m

wide bright = 15 cm = 0.15 m

screen behind the aperture =  2.00 m

to find out

How wide is the aperture

solution

we know here we have a triangle shape that height will be

2 + 2 = 4 m and base 0.15 m

so we now easy find w by triangle

w / 2 = 0.15 / 4

so w = 0.15 / 4 × 2

so w = 0.075 m

wide is the aperture is 7.5 cm

coil wire, increase current, put iron inside the coil

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s