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2 years ago

Help with ALL PLZZZz

Physics

1 answer:

Vika [28.1K]2 years ago

4 0

I can't read the ones on the top but, 7. Is D which you put lol and I believe what you put for 9. is right and 10. I believe your answer is H aka C lol Hope this helps!!! :D

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The most frequent compulsion that is exhibited in obsessive-compulsive disorder is

Vinil7 [7]

The most frequent compulsion that is exhibited in obsessive compulsive disorder is cleansing

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2 years ago

Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state

topjm [15]

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

Given

Process 1 ---> 2

The relation of the process P = constant

Pressure of point (1) P1 = 10 bar = P2

Volume of point (1) V1 = 1 m^3

Volume of point (2) V2 =4 m^3

The relation of the process V = constant

Process 2 ---> 3

The relation of the process V = constant

V3 = V2

Pressure of point (3) P3 = 10 bar

Volume of point (3) V3 = 4 m^3

Process 3 ---> 1 

The relation of the process PV = constant

Required

Sketch the processes on the PV coordinates

The work for each process in kJ

Solution

The work is defined by

W= $\int\limits^a_b {x} \, dx$

a=V2

b=V1

x=P

dx=dV

Process 1 ---> 2

P3 = P4 = 5 bar

W= $\int\limits^a_b {x} \, dx$

a=V3

b=V2

x=4

dx=dV

putting the value of a, b, x, dx in above integral

W=400 kJ

Process 2 ---> 3 

V = constant Then there is no change in the volume,hence W = 0 kJ

Process 3 ---> 1

By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1

W= $\int\limits^a_b {x} \, dx$

a=V1

b=V3

x=1V^-1

dx=dV

putting the value of a, b, x, dx in above integral

W=| ln V | limit a and b

= -160.944 KJ

5 0

3 years ago

Pls pls pls pls pls help

Rzqust [24]

Answer:

what grade are you in so i can help

Explanation:

8 0

3 years ago

A magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that a

Reika [66]

Answer:

2.62A

Explanation:

Given

V = 0.43 V

I = 3.1 A

Then, V = IR, R = V/I

R = 0.43/3.1

R = 0.14 Ω

The induced emf = dB/dt * A

So that, dB/dt = emf/A

Since dB/dt is constant then Emf/A(circle) = Emf/A square

So Emf (square)/Emf (circle) = A square / A circle

A circle = πr². The perimeter of the square is 2πr which also is the circumference of the square.

Since the perimeter is 2πr, then each side would be πr/2. Thus, the area of the square would be, (πr/2)² = π²r²/4

So A square/Acircle = (π²r²/4) / πr² = π/4 = 0.79

this means that, emf square = emf circle * 0.79

emf square = 0.43*0.79 = 0.34V

I = V/R

I = 0.34/0.13

I = 2.62A

3 0

3 years ago

A ray of light strikes a mirror with an angle of incident of 48º. Calculate: i) the angle of reflection and ii). the angle of de

Arada [10]

Answer:

I) angle of reflection = 48 degrees

II) angle of deviation = 0

Explanation:

Given that the:

angle of incident = 48º.

According to law of reflection which state that the angle of incident is equal to the angle of reflection.

I.) Therefore,

angle of reflection = 48º.

II.) There can only be deviation if the mirror has a diffused surface. But for a smooth plane mirror, there can never be any deviation.

Therefore ,

The angle of deviation = 0

8 0

3 years ago