Answer:
Mass of the box = 0.9433 kg
Explanation:
Mass of racket-ball
= 0.00427 kg
Velocity of racket-ball before collision
= 22.3 m/s
Velocity of racket-ball after collision with box
= -11.5 m/s
[Since ball is bouncing back, so velocity is taken negative.]
Velocity of the box before collision
= 0 m/s
<em>[Since the box is stationary, so velocity is taken zero]</em>
Velocity of box moving forward after collision
= 1.53 m/s
To find the mas of the box
.
By law of conservation of momentum we have:
Momentum before collision = Momentum after collision
This can be written as:


We can plugin the given value to find 


Adding both sides by 0.4911


Dividing both sides by 1.53.


∴
kg
Mass of the box = 0.9433 kg (Answer)
Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.
<h3>How to find Wavelength of the light?</h3>
When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.
This is a straightforward situation in which we can apply the
Fraunhofer single slit diffraction equation:
y = mλD/a
Where:
y = Displacement from the center line for minimum intensity = 1.35 mm
λ = wavelength of the light.
D = distance
a = width of the slit = 0.75
m = order number = 1
Solving for λ
λ = y + a/ mD
Changing the information that the issue has provided:
λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2
=5.0625 *10^-7 = 506.25
so
Wavelength of the light 506.25.
To learn more about Wavelength of the light refer to:
brainly.com/question/15413360
#SPJ4
It becomes a different element
Answer:
86.51° North of West or 273.49°
Explanation:
Let V' = velocity of boat relative to the earth, v = velocity of boat relative to water and V = velocity of water.
Now, by vector addition V' = V + v'.
Since v' = 6.10 m/s in the north direction, v' = (6.10 m/s)j and V = 100 m/s in the east direction, V = (100 m/s)i. So that
V' = V + v'
V' = (100 m/s)i + (6.10 m/s)j
So, we find the direction,Ф the boat must steer to from the components of V'.
So tanФ = 6.10 m/s ÷ 100 m/s
tanФ = 0.061
Ф = tan⁻¹(0.061) = 3.49°
So, the angle from the north is thus 90° - 3.49° = 86.51° North of West or 270° + 3.49° = 273.49°