The correct answer would be that destructive interference is happening. In this interference, the crest of a wave meets a trough of another wave resulting to an amplitude that is lower. The opposite is called the constructive interference. Hope this answers the question.
Answer:
fcosθ + Fbcosθ =Wtanθ
Explanation:
Consider the diagram shown in attachment
fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)
Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)
Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)
sum of x-direction forces = 0
fx+ Fbx=Wx
fcosθ + Fbcosθ =Wtanθ
Answer:
3.964 s
Explanation:
Metric unit conversion:
1 miles = 1.6 km = 1600 m.
1 hour = 60 minutes = 3600 seconds
75 mph = 75 * 1600 / 3600 = 33.3 m/s
22.5 mph = 22.5 * 1600/3600 = 10 m/s
Let g = 9.81 m/s2
Friction is the product of coefficient and normal force, which equals to the gravity
The deceleration caused by friction is friction divided by mass according to Newton 2nd law.
So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is
Answer:
7 deg
Explanation:
= mass of the rod = 
= weight of the rod = 
= spring constant for left spring = 
= spring constant for right spring = 
= stretch in the left spring
= stretch in the right spring
= length of the rod = 0.75 m
= Angle the rod makes with the horizontal
Using equilibrium of force in vertical direction for left spring
Using equilibrium of force in vertical direction for right spring
Angle made with the horizontal is given as
Options:
A. The luminosity of Star 1 is a factor of 100 less than the luminosity of Star 2.
B. Star 1 is 100 times more distant than Star 2.
C. Without first knowing the distances to these stars, you cannot draw any conclusions about how their true luminosities compare to each other.
D. Star 1 is 10 times more distant than Star 2.
E. Star 1 is 100 times nearer than Star 2.
Answer:
D. Star 1 is 10 times more distant than star 2
Explanation:
For two stars of identical size and temperature, the closer one to us will appear brighter. The relationship between the distance and luminosity of stars is an inverse- square relationship.
Luminosity, L = 1/r²
Where r is the distance of the star to the earth
Since star 1 is dimmer in brightness than star 2 by a factor of 100,
L₁/L₂ = 1/100
i.e. L₁ = 1, L₂=100
L₁ = 1/r₁² ............(1)
1 = 1/r₁²
L₂ = 1/r₂²
100 = 1/r₂² .........(2)
divide equation (2) by equation (1)
100/1 = ( 1/r₂² )/ (1/r₁²)
100 = (r₁/r₂)²
r₁/r₂ = √100
r₁/r₂ = 10
r₁ = 10r₂
