Answer:
The light bends away from the normal
Explanation:
We can solve the problem by using Snell's law:
where:
is the index of refraction of the first medium
is the index of refraction of the second medium
is the angle of incidence (angle between the incoming ray and the normal to the interface)
is the angle of refraction (angle between the outcoming ray and the normal to the interface)
We can rearrange the equation as
In this problem, light travels from an optically denser medium to an optically rarer medium, so
Therefore, the term 
is greater than 1, so
which means that the angle of refraction is greater than the angle of incidence, and so the light will bend away from the normal.
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed
, and an accelerated motion on the y-axis, with initial speed
and acceleration
:
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).
To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
Therefore:
which has two solutions:
is the time of the beginning of the motion,
is the time at which the projectile hits the ground.
Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
<span>Even in space, there is still presence of gravity. The
cause of weightlessness is not how far above the earth the space shuttle is but
rather how fast it is travelling. The shuttle is in free fall causing
weightlessness, but it is travelling fast enough to miss the earth as it falls.
Similarly, the airplane could also provide weightlessness if it went free fall
as well. However, that ends as the plane hits the ground. </span>
Answer:
B. A collision scene
Explanation:
It could have been a parade ceremony, but, if you notice the vehicle's hazard lights or an emergency vehicle ahead, it is common sense to figure that they is a collision scene nearby.
Answer:
Therefore the ratio of diameter of the copper to that of the tungsten is
Explanation:
Resistance: Resistance is defined to the ratio of voltage to the electricity.
The resistance of a wire is
- directly proportional to its length i.e
- inversely proportional to its cross section area i.e
Therefore
ρ is the resistivity.
The unit of resistance is ohm (Ω).
The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m
The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m
For copper:
......(1)
Again for tungsten:
........(2)
Given that 
and 
Dividing the equation (1) and (2)
[since and ]
Therefore the ratio of diameter of the copper to that of the tungsten is
