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nata0808 [166]
2 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]2 years ago
8 0

Answer: Velocity Increases

Explanation: Plato

SVETLANKA909090 [29]2 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 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os
Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}

\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\

<u>k = 3.5 N/m</u>

4 0
3 years ago
A spring is stretched to a displacement of 3.4 m from equilibrium. Then the spring is released and allowed to recoil to a displa
Stells [14]
Answer to A spring<span> is </span>stretched<span> to a </span>displacement<span> of </span>3.4 m<span> from </span>equilibrium<span>. </span>Then<span> the </span>spring<span> is</span>released<span> and ... </span>Then<span> the </span>spring<span> is </span>released<span> and </span>allowed<span> to </span>recoil<span> to a </span>displacement<span> of </span>1.9 m<span> from</span>equilibrium<span>. The </span>spring constant<span> is </span>11 N/m<span>. What </span>best describes<span> the </span>work involved<span> as the </span>spring recoils<span>? A)87 J of </span>work<span> is performed ...</span>
5 0
3 years ago
Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up
Mademuasel [1]

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

v = \sqrt{2as}

v = \sqrt{2*18.9*391}

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

V=  \sqrt{v^2 + 2aS'}

V = \sqrt{121.57^2+ 2 (-9.92)(332)}

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6 0
3 years ago
What type of tide occurs when the sun, moon, and earth form a right triangle in space?
stiks02 [169]
Neap tide is your answer
7 0
3 years ago
A wheel has a constant angular acceleration of 4.5 rad/s2. during a certain 5.0 s interval, it turns through an angle of 128 rad
dalvyx [7]
The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2 
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration 
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1 
w1 = wo + ãt 
w1 - final angular velocity 
wo - initial angular velocity 
18 = 0 + 4.5t t = 4 s
3 0
3 years ago
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