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Mumz [18]
4 years ago
11

In the following diagram, the voltage is 1.5 volts and the resistance is 7.5 ohms. Use Ohm's law to determine the current in the

circuit.
11.3 amps
5.0 amps
6.0 amps
0.2 amps

Physics
2 answers:
Basile [38]4 years ago
5 0

I believe it is 0.2 amps  Hope this helps sorry if wrong

alina1380 [7]4 years ago
4 0

I believe it’s the last one 0.2 amps

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Explanation:

Check out the picture I drew for a minute before reading this...

B. Distance [the red line] is a scalar quantity reflecting how far an object has traveled. Displacement [the green line] is a vector quantity reflecting how far an object has moved from a point. The key difference is that distance can be any sort of path while displacement is always a vector (or a straight line) between a starting point and a finishing point. Sometimes distance and displacement are equal to one another. Sometimes you have a distance traveled, but zero displacement overall; which is what's going on in your question.

A. The distance that the racecar traveled is indeed 500m. But at the end of the lap, it is right back where it started. So overall, it has been displaced 0m.

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3 years ago
The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

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Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

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Answer:

Its B trust me

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