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Natali [406]
3 years ago
12

Monochromatic light passes through two parallel narrow slits and forms an interference pattern on a screen. As the distance betw

een the two slits is decreased, the distance between light bands in the pattern on the screen will
decrease.

increase.

remain the same.
Physics
2 answers:
In-s [12.5K]3 years ago
8 0
As the distance between the two slits is <span>decreased, the distance between light bands in the pattern on the screen will increase. The correct option among all the options that are given in the question is the second option or the penultimate option. I hope that this is the answer that has actually helped you.</span>
Virty [35]3 years ago
4 0

The option you are looking for is the second one, have a great day!

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Will give brainliest! 50 points!!!
tatiyna

Answer:

C

Im not totally sure but that's what my science teacher taught me sorry if is wrong

6 0
3 years ago
Any fracture or system of fractures along which Earth moves is known as a
azamat
The answer is B.) Stress 
7 0
3 years ago
2. Tomas is hanging from a tree limb, that is inclined at a 65° angle. The force
LUCKY_DIMON [66]

Answer:

57 N

Explanation:

Were are told that the force

of gravity on Tomas is 57 N.

And it acts at an inclined angle of 65°

Thus;

The vertical component of the velocity is; F_y = 57 sin 65

While the horizontal component is;

F_x = 57 cos 65

Thus;

F_y = 51.66 N

F_x = 24.09 N

The net force will be;

F_net = √((F_y)² + (F_x)²)

F_net = √(51.66² + 24.09²)

F_net = √3249.0837

F_net = 57 N

4 0
3 years ago
The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc
qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

R = \frac{6}{2.5} - 2.4 ohm

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

\frac{3R}{R+3} \times 2.5 = 6

R= 12 ohm

2) circuit Q

I\times (R+0.1) =V

R+0.1 =\frac{6}{2.5}

R = 2.3 ohm

5 0
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.
5 0
3 years ago
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