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2 years ago

How much energy is required to take 500g of ice at -40 degrees to water at room temperature (22 degrees Celsius)? Show work pls

Physics

1 answer:

Minchanka [31]2 years ago

3 0

Answer: Given that:

Mass (m) = 20 g = 0.02 Kg,

temperature (T₁) = -10°C = -10+273 = 263 K

temperature (T₂) = 10°C = 10+273 = 283 K

Specific heat of water (Cp) = 4.187 KJ/Kg k

We know that Heat transfer (Q) = m. Cp.( T₂ - T₁)

= 0.02 × 4.187 × (283-263)

Q = 1.67 KJ

Heat transferred is 1.67 KJ

Explanation:

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The planet Mars has much less gravitational force that Earth. What would happen if you went to Mars?

vovangra [49]

gravitational force of planet exerted on its object near the surface is known as weight

so here we know that gravitational force of mars is much less than the gravitational force of Earth

So on the surface of mars the Weight of objects must be much less than the weight of object on surface of Earth

so here correct answer must be

<em>D. Your weight would decrease.</em>

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3 years ago

Pls help

belka [17]

Answer:

2156 J

Explanation:

From the question,

Work done = Combined mass of the bucket and water×height×gravity.

W = (M+m)hg............................. Equation 1

Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.

Given: M = 20 kg, m = 2 kg, h = 10 m

Constant: g = 9.8 m/s²

Substitute these value into equation 1

W = (20+2)×10×9.8

W = 22×98

W = 2156 J

4 0

2 years ago

I need help on this.

4vir4ik [10]

It brings cold water from the bottom of the ocean.

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2 years ago

Read 2 more answers

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

3 years ago

Which is not a type of driving distraction

Yuliya22 [10]

No grooming your snail while driving. Just ask Spongebob. Oh by the way, what are the options?

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2 years ago