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3 years ago

A cooking pot is sitting in a kitchen for several hours unused. Why does it feel cold when you touch it?

Physics

1 answer:

masha68 [24]3 years ago

8 0

Answer:

I would go with 2

Explanation:

But i would also not go with my answer. Lol

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Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the

zlopas [31]

Answer:

$\dfrac{I_1}{I_2}=\dfrac{4}{9}$

Explanation:

c = Speed of wave

$\rho$ = Density of medium

A = Area

$\nu$ = Frequency

$\nu_1=\dfrac{2}{3}\nu_2$

Intensity of sound is given by

$I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2$

So,

$I\propto \nu^2$

We get

$\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}$

The ratio is $\dfrac{I_1}{I_2}=\dfrac{4}{9}$

8 0

2 years ago

You calculate the density of a block of aluminum to be 2.68 g/cm3. You look up the density of a block of aluminum at room temper

Hunter-Best [27]

Answer:

Systematic errors.

Explanation:

The density of the aluminium was calculated by a human and this is not natural but can be due to errors in the calibration of the scale for measuring the weight or taking readings from the measuring cylinder.

Random errors are natural errors. Random errors in experimental measurements are caused by unknown and unpredictable changes in the experiment. Systematic errors are due to imprecision or problems with instruments.

3 0

3 years ago

Edward with it gathering physical evidence at the scene of a crime. He finds a fingerprint pressed into the wax of the candle. W

muminat

Answer:trace

Explanation:

3 0

2 years ago

Read 2 more answers

student built a simple heat engine that could lift masses. If the heat engine takes in 300 J of heat and loses 50 J to the envir

k0ka [10]

Answer:

So it will lift the mass by h = 17 m

Explanation:

As per energy conservation we know that

$Q_1 = W + Q_2$

here we know that

$Q_1 = 300 J$

$Q_2 = 50 J$

now we have

$W = 300 - 50$

$W = 250 J$

so work done by the engine is 250 J

now we have

$W = mgh$

$250 = 1.5 \times 9.81 \times h$

$h = 17 m$

4 0

3 years ago

Please help me with my physics I do not understand! Please provide work thank you.

Studentka2010 [4]

Fe=K Q1/Q2/d2
Q1 is the first charge
Q2 is the second charge
d is the distance
K= 9x10^9 NM^2/C2
Now let’s plug the numbers
Fe=9x10^9NM^2/C2 (2x10^-4C)(8x10^-4C) / (0.3m^2) you notice we took away the negative charges when we plugged the charges
Ok now we notice that we have C2 which is C to the power 2 we can write it as C^2 and we have two CSU’s beside each one of the charges we can get rid of them all by curtailment
And we can curtailment the M^2and the other M^2
Now we left with only 9x10^9N (2x10^-4)(8x10^-4)/ 0.3
Let’s multiply the (9)(2)(8)=144
And add the exponents (9)+(-4)+(-4)=1
So now we got 144x10N divide by the distance which is 0.3
144x10N / 0.3 = 4800N
Hope it helps u understand :)

3 0

3 years ago