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nata0808 [166]
3 years ago
15

A light ray passing from medium 1 to medium 2 is bent away from the perpendicular normal to the boundary surface. What happens t

o the velocity of the refracted wave as it enters medium 2?
The velocity increases.


The velocity decreases.


The velocity becomes zero.


The velocity does not change.
Physics
2 answers:
Gre4nikov [31]3 years ago
8 0

Answer: Velocity Increases

Explanation: Plato

SVETLANKA909090 [29]3 years ago
6 0

Answer:

The answer is a for Plato users.

Explanation:

Since the angle of the refracted ray moves away from the normal, it must be traveling in a faster medium.

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A woman is running around a playground while keeping an eye on her child who is on a swing set. The child is going back and fort
Nutka1998 [239]

Answer:

(a) 1500 m

(b) 2827.43m

Explanation:

Given that the time for one cycle of the swing is 1 s

The radius of the swing, R= 3.0m

The angle covered, \theta, by each swing is a quarter of the circle. i.e.

\Theta=\frac {\pi}{2}

Speed of running of women =5 m/s.

Time of running = 5 minutes= 5 x 60 secondes= 300 s.

(a) As distance= (speed) x (time)

So, the required distance= 5 x 300 m= 1500 m.

(b) As there are two swings in one cycle, so, the distance covered in one swing is the length of the circular shown in the figure.

As arc length, l= \theta R, where \theta is the angle, in radian, subtended by the arc at the center, and R is the radius of curvature of the arc.

So, the distance covered by the child in 1 swing = \frac {\pi}{2}\times 3 m=\frac{3\pi}{2}m.

In 1 cycle, there are 2 swings, so distance covered in 1 cycle = 3 \pi m.

Now, in 1 second there is 1 cycle, so in 5 minutes there will be 300 cycles.

So, the total distance covered by the child

= 3\pi\times 300 m= 2827.43 m.

3 0
4 years ago
A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc
liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s,    a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

           v₀ₓ =  10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

           v_y = v_{oy} - g t

           v_{y}^2 = v_{oy}^2 - 2 g (y - y_o)

at the highest point of the trajectory the vertical speed must be zero

           v_y = 0

therefore the velocity of the body is

          v = (10 i ^ + 0j ^) m / s

the acceleration is

          a = (0 i ^ - g j⁾

          a = (0i ^ - 9.8 j ^) m / s²

5 0
3 years ago
contractor will use a package emulsion having a specific gravity of 1.25 and relative bulk strength of 130 to open an excavation
tester [92]

Answer:

The burden distance is 7 ft

Solution:

As per the question:

Specific gravity of package emulsion, SG_{E} = 1.25

Specific gravity of diabase rock, SG_{R} = 2.76

Diameter of the packaged sticks, d = 3 in

Now,

To calculate the first trail shot burden distance, B:

B = [\frac{2SG_{E}}{SG_{R}} + 1.5]\times d

B = [\frac{2\times 1.25}{2.76} + 1.5]\times 3 = 7.22

B = 7 ft

5 0
3 years ago
How long does it take the moon to go from full-moon phase to new-moon phase
Amiraneli [1.4K]
A complete cycle of phases is 29.531 days.
From full-moon to new-moon is half of that.
4 0
3 years ago
A garbage truck has a mass of 2100kg it starts from rest and travels 75m in 15 seconds the truck uniformly accelerates the entir
Sedbober [7]
F=ma 

Velocity is Distance over time so Vf = 75/15 = 5m/s

Find acceleration V=Vo+at. plugging in the values you know, you get
0.33m/s^2

F=(2100)(0.33)=693N
3 0
3 years ago
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