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Oksana_A [137]
2 years ago
5

The temp of some water with a mass of 10.0 g and a specific heat of 4.2 changed from 25.0 ' C to 35.0 ' C. How much heat was abs

orbed?
A) 100 J
B) 230 J
C) 420 J
D) 600 J
Physics
2 answers:
Stella [2.4K]2 years ago
6 0
The correct answer is A
Marat540 [252]2 years ago
6 0

Actually, I just took this quiz and the answer is C) 420 J

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9 0
3 years ago
A solenoid with an inductance of 8 mH is connected in series with a resistance of 5 Ω and an EMF forming a series RL circuit. A
monitta

Answer:

induced EMF = 240 V

and by the lenz's law  direction of induced EMF is opposite to the applied EMF

Explanation:

given data

inductance = 8 mH

resistance = 5 Ω

current = 4.0 A

time t = 0

current grow = 4.0 A to 10.0 A

to find out

value and the direction of the induced EMF

solution

we get here induced EMF of induction is express as

E = - L \frac{dI}{dt}    ...................1

so E = - L \frac{I2 - I1}{dt}

put here value we get

E = - 8 × 10^{-3} \frac{10 - 4}{0.2*10^{-3}}

E = -40 ×  6

E = -240

take magnitude

induced EMF = 240 V

and by the lenz's law we get direction of induced EMF is opposite to the applied EMF

5 0
3 years ago
Use the drop-down menu to complete the statement. Based on the field lines, the electric charges indicated by the question marks
zheka24 [161]

Answer: The same

Explanation: I just did it on Edg

5 0
3 years ago
Read 2 more answers
Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s
Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375  m / s

3 0
3 years ago
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