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2 years ago

Guys helppppp pleaseeeeeeeee

Physics

1 answer:

yan [13]2 years ago

4 0

2 pencils, because that would equal left grams then the others

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In the first law of Thermodynamics ΔE = Q - W, what does ΔE stand for???

Alexxx [7]

ΔE = q + w

q = heat (quantity of)

q and w can be positive or negative depending on if work/heat is being absorbed/done on the system or released/done by the system

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2 years ago

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Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water

tangare [24]

Answer:

$T_b=107.3784\ ^{\circ}C$

Explanation:

Given:

thickness of the base of the kettle, $dx=0.52\ cm=5.2\times 10^{-3}\ m$

radius of the base of the kettle, $r=0.12\ m$

temperature of the top surface of the kettle base, $T_t=100^{\circ}C$

rate of heat transfer through the kettle to boil water, $\dot Q=0.409\ kg.min^{-1}$

We have the latent heat vaporization of water, $L=2260\times 10^3\ J.kg^{-1}$

and thermal conductivity of aluminium, $k=240\ W.m^{-1}.K^{-1}$

So, the heat rate:

$\dot Q=\frac{0.409\times 2260000}{60}$

$\dot Q=15405.67\ W$

From the Fourier's law of conduction we have:

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}$

where:

$A=$ area of the surface through which conduction occurs

$T_b=$ temperature of the bottom surface

$15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}$

$T_b=107.3784\ ^{\circ}C$ is the temperature of the bottom of the base surface of the kettle.

6 0

3 years ago

How much work is done in holding a 20 N sack of potatoes while waiting in line at the

Yuki888 [10]

Answer: A

Explanation:

honestly, it sounded the best

8 0

2 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

Here are some energy resources. List the renewable energy resources.

Artemon [7]

Solar wind geothermal

3 0

3 years ago

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