Answer:
skateboard b
Explanation:
p=mv
skateboard a
p=(60kg)(1.5m/s)=90kg*m/s
skateboard b
p=(50kg)(2m/s)=100kg*m/s
Answer: F = 131.7N
Explanation:
You are given the following parameters.
Mass M = 30 kg
Coefficient of static friction μ = 0.5
Ø = 30 degrees
When the person is trying to drag the box with force F, the static frictional force Fs will be acting in the opposite direction.
From the figure attached, resolve all forces into horizontal component and vertical component.
Horizontal component:
Fs - F cosØ = 0
Fs = F cosØ
F cosØ = μN ...... (1)
Vertical component:
N + F SinØ - mg = 0
N = Mg - F SinØ ..... (2)
Substitutes m, g and Ø into the equation 2
N = (30 × 9.8) - F × sin30
N = 294 - 0.5F
Substitute N and coefficient of friction into the equation (1)
F cos30 = 0.5 (294 - 0.5F)
Open the bracket
0.8660F = 147 - 0.25F
Collect the like terms
1.116025F = 147
F = 147/1.116025
F = 131.7 N
Therefore, the minimum force the person needs to have to move the box along the floor is 131.7 N
Answer:
= 625 nm
Explanation:
We now that for
for maximum intensity(bright fringe) d sinθ=nλ n=0,1,2,....
d= distance between the slits, λ= wavelength of incident ray
for small θ, sinθ≈tanθ= y/D where y is the distance on screen and D is the distance b/w screen and slits.
Given
d=1.19 mm, y=4.97 cm, and, n=10, D=9.47 m
applying formula
λ= (d*y)/(D*n)
putting values we get
on solving we get
= 625 nm