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sdas [7]
3 years ago
8

Brad has three beakers of water. Each beaker contains 500 mL of water. The temperature of the water in the first beaker is 40°C.

The water in the second beaker is at 100°C, and the water in the third beaker is at 60°C. As part of a lab experiment he will carefully combine the water in all three beakers. He predicts that the combined water will have a final temperature of 50°C.
Evaluate Brad’s claim and clearly state whether he is correct or not. Explain your answer to support your evaluation.
Physics
1 answer:
dexar [7]3 years ago
7 0

Explanation:

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What would be the distance moved if we had a 70 n force and work done is 8j
Lostsunrise [7]

Answer:

0.1143m

Explanation:

W=f×s

8=70s

make s the subject of the formula

s=8/70

=0.1143m

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6 A textbook sitting on a table weighs 3.6 N. If you push straight down on the book with a force of
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On flat ground, a 70-kg person requires about 300 W of metabolic power to walk at a steady pace of 5.0 km/h (1.4 m/s). Using the
Darina [25.2K]

Answer:

C 350W

Explanation:

Given power output to walk on a flat ground to be 300W, h = 0.05x, v = 1.4m/s

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x = horizontal distance covered

Total energy used = potential energy used in climbing and the energy used in a walking the horizontal distance.

E = mgh + 300t

Where t is the time taken to cover the distance

x = vt and h = 0.05vt

So

E = mg×0.05×vt + 300t

Substituting respective values

E = 70×9.8×0.05×1.4t +300t = 348t

P = E/t = 348W ≈ 350W.

4 0
4 years ago
I just need an answer ASAP
nikdorinn [45]

Answer: c

Explanation: hope this helps :)

8 0
3 years ago
Read 2 more answers
A bar magnet whose magnetic dipole moment is 21 A·m2 is aligned with an applied magnetic field of 3.6 T. How much work must you
iren [92.7K]

Answer:

The amount of work must be do to rotate the bar magnet is 151.2 J

Explanation:

Given:

Magnetic moment \mu = 21 A. m^{2}

Magnetic field B = 3.6 T

To find work do to rotate the bar magnet,

From the formula of work done in case of magnetic field,

    U = \mu .B \cos 0 -\mu .B \cos 180

Here \theta changes 0 to 180

But \cos 180 = -1

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Therefore, the amount of work must be do to rotate the bar magnet is 151.2 J

3 0
3 years ago
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