A ray in oil (n = 1.52) reaches a boundary with water (n = 1.33) at 55.9 deg. Does it reflect internally or refract into the air
1 answer:
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Answer:
The object will travel 675 m during that time.
Explanation:
A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.
In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.
In this case, the position is calculated using the expression:
x = xo + vo*t + ½*a*t²
where:
- x0 is the initial position.
- v0 is the initial velocity.
- a is the acceleration.
- t is the time interval in which the motion is studied.
In this case:
- x0= 0
- v0= 0 because the object is initially stationary
- a= 6
- t= 15 s
Replacing:
x= 0 + 0*15 s + ½*6 *(15s)²
Solving:
x=½*6 *(15s)²
x=½*6 *225 s²
x= 675 m
<u><em> The object will travel 675 m during that time.</em></u>
Answer:
0.767
Explanation:
The work done on the truck by the frictional drag force is given by
where
F is the magnitude of the frictional force
d = 38.0 m is the maximum displacement allowed for the truck
The negative sign is due to the fact that the force of friction is opposite to the motion of the truck
The force of friction can also be written as:
where
is the coefficient of kinetic friction between the truck and the lane
m is the mass of the truck
g is the acceleration of gravity
So we can rewrite the work done as
(1)
According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:
(2)
where
v = 0 is the final velocity of the truck
u = 23.9 m/s is the initial velocity of the truck
By combining (1) and (2) we get
And solving for , we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:
Answer:
a) 1.22 s
b) 9.089 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
Time taken by the ball to reach the highest point is 1.22 seconds
The maximum height the ball will reach above the ground is 1.75+7.339 = 9.089 m