Answer:
d = 52 μm
Explanation:
given,
wavelength of the light source (λ)= 550 nm
distance to form interference pattern(D) = 1.5 m
y = 1.6 cm = 0.016 m
width of the slits = ?
now, using displacement formula
for the first maxima, m = 1
d = 5.2 x 10⁻⁶ m
d = 52 μm
hence, the width of her slits is equal to d = 52 μm
Answer:
a)Yes will deform plastically
b) Will NOT experience necking
Explanation:
Given:
- Applied Force F = 850 lb
- Diameter of wire D = 0.15 in
- Yield Strength Y=45,000 psi
- Ultimate Tensile strength U = 55,000 psi
Find:
a) Whether there will be plastic deformation
b) Whether there will be necking.
Solution:
Assuming a constant Force F, the stress in the wire will be:
stress = F / Area
Area = pi*D^2 / 4
Area = pi*0.15^2 / 4 = 0.0176715 in^2
stress = 850 / 0.0176715
stress = 48,100.16 psi
Yield Strength < Applied stress > Ultimate Tensile strength
45,000 < 48,100 < 55,000
Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.
Answer:
Explanation:
Given
mass of laptop m=2 kg
The velocity of car u=8 m/s
The coefficient of static friction is
The coefficient of kinetic friction is
As the car is moving, so the coefficient of kinetic friction comes into play
deceleration offered by friction
Using the equation of motion
insert the values
Vf = 0 + 3.5•8.7
= 30.45 m/s