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3 years ago

5. While a gas does 500J of work on its surroundings, 800J of heat is added to the gas. The

Physics

1 answer:

andreev551 [17]3 years ago

4 0

Explanation:

a) ΔU = -500 J + 800 J = 300 J

b) ΔU = mCΔT

300 J = (0.3 kg) (1000 J/kg/°C) ΔT

ΔT = 1 °C

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Could someone explain to me how to got the answer B, thank you very much

Mnenie [13.5K]

Answer:

since -6 lasted for 5 seconds, multiplying both would result in -30

3 lasted for 10 seconds, so multiplying both would give +30

average = ( 30 + (-30) ) / 2

30 -30 is already equal to zero, so the answer should be 0

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3 years ago

I need help by today guys plz help

NISA [10]

radio waves bc they have the longest wave lenthgs in a magnetic spectrum

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3 years ago

A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from th

Alik [6]

when a hole is made at the bottom of the container then water will flow out of it

The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.

By Bernoulli's equation we can write

$Po + \frac{1}{2}\rho v_1^2 + \rho g h = Po + \frac{1}{2}\rho v_2^2 + \rho g *0$

Now by equation of continuity

$A_1v_1 = A_2v_2$

$\pi (0.2)^2 v_1 = \pi (0.01)^2 v_2$

from above equation we can say that speed at the top layer is almost negligible.

$v_1 = 0$

now again by equation of continuity

$\rho g h = \frac{1}{2} \rho v^2$

$v = \sqrt{2 g h}$

here we have

$h = h_1 - h_2$

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now speed is given by

$v = \sqrt{2* 9.8 * 0.47}$

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3 years ago

Which is an example of an unstructured activity that promotes resistance training?

aleksandrvk [35]

One possible unstructured activity that promotes resistance training would be climbing playground equimpent - A.

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3 years ago

(1 pt) A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 35 meters above the ground. This takes 14

Elenna [48]

Answer:

w = 5832.372 Joules

Explanation:

Mass of water, m = 20 kg

The water was pulled up to a height of 35 meters, i.e. h = 35 m

It takes 14 minutes to pull up the water through the height, 35 m

speed = distance/ time = 35/14 = 2.5 m/min

The bucket's height, y = speed * time = 2.5t meters

6 kg of water drips out of the bucket throughout the 14 minutes

The rate at which the water drips drips out = (6/14) = 0.4286 kg/min

Mass of water that drips out in time, t = 0.4286t kg

The mass of water remaining = (20 - 0.4286t) kg

Change in Workdone, Δw = mgΔy

Δy = 2.5 Δt

Δw = mg * 2.5 Δt

dw = (20 - 0.4286t)g2.5 dt

integrating both sides

dw = (50g - 1.07gt)dt

$w = \int\limits^a_b {(50g-1.07gt)} \, dx$ where b = 0, a = 14

w = 50gt - 1.07g(t²)/2 g = 9.8 m/s²

w = 490t - 5.243t²

w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)

w = 6860 - 1027.628

w = 5832.372 Joules

3 0

3 years ago