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photoshop1234 [79]
3 years ago
9

Imagine the door to a theatre is opened and you are standing outside. How is it possible that you can hear the sound of the movi

e or play even if you cannot see the screen or stage ?
A) reflection only happens with sound waves not waves of light so the sound waves bounce out to your ears.

B) the sound waves refract as they move from one medium to another

C) the darkness in the theatre absorbed light but not sound so you can hear

D) the sound waves diffract much more than the light waves so try bend and spread as they move through the doorway.
Physics
1 answer:
meriva3 years ago
8 0
Imagine the door to a theatre is opened and you are standing outside. You can hear the sound of the movie or play even if you cannot see the screen or stage because <span>the sound waves diffract much more than the light waves so try bend and spread as they move through the doorway. The answer is letter D</span>
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I need to submit my homework in an hour, this is only question i cant solve. someone pls help!!
Sauron [17]

Hope this helps.......☺

8 0
3 years ago
A 64 kg swimmer jumps, with a velocity of 4.2 m/s, off the front of a 25 kg kayak when the kayak is moving forward at a velocity
Crank

Answer:

3.88m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and 2 are the initial velocities

v is the final velocity

Given

m1 = 64kg

u1 = 4.2m/s

m2 = 25kg

u2 = 3.2m/s

Required

Final velocity v

Substitute the given values into the formula

64(4.2)+25(3.2) = (65+25)v

268.8+80 = 90v

348.8 = 90v

v = 348.8/90

v = 3.88m/s

Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s

8 0
2 years ago
In case of collision of objects in two dimensions which statement is true after the collision?
Grace [21]

The correct answer to the question is : C) The horizontal momentum and the vertical momentum are both conserved. 

EXPLANATION :

Before coming into any conclusion, first we have to understand the law of conservation of momentum.

As per the law of conservation of momentum, the total linear as well as angular momentum of an isolated system is always conserved . The law of conservation of energy is a universal fact.

Hence, during any type of collision, the total momentum is always conserved.

Hence, the total horizontal momentum as well as total vertical momentum are always conserved during both elastic as well as inelastic collision.



5 0
3 years ago
Read 2 more answers
A 2.0-kg mass is projected from the edge of the top of a 20-m tall building with a velocity of 24 m/s at some unknown angle abov
Digiron [165]

Ox:vₓ=v₀

x=v₀t

Oy:y=h-gt²/2

|vy|=gt

tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g

y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°

cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s

Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ

3 0
3 years ago
A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much
Lostsunrise [7]

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

8 0
3 years ago
Read 2 more answers
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