Answer:
d = 380 feet
Explanation:
Height of man = perpendicular= 130 feet
Angle of depression = ∅ = 70 °
distance to bus stop from man = hypotenuse = d = 130 sec∅
As sec ∅ = 1 / cos∅
so d = 130 sec∅ or d = 130 / cos∅
d = 130 / cos(70°)
d = 380 feet
Answer:
Friction acts in the opposite direction to the motion of the truck and box.
Explanation:
Let's first review the problem.
A moving truck applies the brakes, and a box on it does not slip.
Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.
The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.
156.8 Joules of energy is in the box's gravitational potential energy store
<u>Explanation</u>:
<em>Given:</em>
Mass of the box Dane is holding = 8 Kilograms
Height at which Dane is holding the box above the ground= 2 metres
<em>To Find:</em>
Gravitational potential energy in the box=?
<em>Solution:</em>
gravitational potential energy is the work done per mass on a object to move that object from one fixed location to to another location against gravity.Its unit is joules or J
Thus Gravitational potential energy is represented as,
where
is the gravitational potential energy
m is the mass
h is the height
g is the gravitational force( 9.8 
)
Now substituting the given values,
72s for 24 complete oscillations.
Thus, a complete oscillation takes 72/24=3s
Answer: period T=3s
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
