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3 years ago

12

Light rays travel from one medium into another and refract away from the boundary. What changes about the light to cause this re

fraction?
A. Its speed increases.
B. Its frequency increases.
C. Its frequency decreases.
D. Its speed decreases.

2 answers:

vaieri [72.5K]3 years ago

8 0

Answer:

a

Explanation:

I think its a its speed increases.....

Paladinen [302]3 years ago

6 0

B it’s frequency increases

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Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow

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The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.

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A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

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3 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

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