Answer:
Explanation:
The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or . For our case we want to use
, which for our values is equal to:
Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the <em>least</em> number of significant figures):
Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :
Work done on the wagon is given by the product of net force and displacement. It is given by :
W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.
a) 2.64 s
We can solve this part of the problem by using the following SUVAT equation:
where
s is the displacement of the stone
u is the initial velocity
t is the time
a is the acceleration
We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:
- s (displacement) negative, since it is downward: so s = -75.0 m
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)
Substituting into the equation,
Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.
b) 10.4 m/s
The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:
where again, we must be careful to the signs of the various quantities:
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2
Substituting t = 2.64 s, we find the final velocity of the stone:
where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.
c) 4.11 s
In this case, we can use again the equation:
where
s is the displacement of the package
u is the initial velocity
t is the time
a is the acceleration
We have:
s = -105 m (vertical displacement of the package, downward so negative)
u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)
a = g = -9.8 m/s^2
Substituting into the equation,
Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after
t = 4.11 seconds.