Pain in the pleura

A 214 g aluminum cup holds and is in thermal equilibrium with 892 g of water at 76°C. The combination of cup and water is cooled

givi [52]

Answer: rate at which energy is being removed is 58.9J/s or 58.9W

Explanation:

Given;

Mass of water Mw =892g = 0.892kg

Mass of aluminium Ma = 214g = 0.214kg

Rate of change in temperature ∆T = 0.9°C per minute

Specific heat capacity of water Cw = 4186J/kgC

Specific heat capacity of aluminium Ca = 900J/kgC

Since both the water and aluminium are at thermal equilibrium and the decrease with the same rate, the energy removed per minute is given as;

Q = (MwCw + MaCa)∆T

Q = (0.892×4186 + 0.214×900)×0.9

Q = 3533.8608J per minute

To determine the energy being removed in watts

P = Q/t

P = 3533.8608J/60s

P = 58.89768W

P = 58.9W

8 0

4 years ago

Read 2 more answers

How much work is done when you push a crate horizontally with 150 N across a 13-m floor

Mila [183]

Work = (force) x (distance)

When a force of 150 N pushes through a distance of 13 meters,
it does

Work = (150 N) x (13 m) = 1,950 joules .

6 0

3 years ago

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0

3 years ago

A tin can whirled on the end of a string moves in a circle because

Ilya [14]

Answer:

There is an inward force acting on the can

Explanation:

This inward force is known as Centripetal force and it is responsible for making the can whirl on the end of a string in circle and it is also directed towards the center around which the can is moving.

8 0

3 years ago

A container with volume 1.83 L is initially evacuated. Then it is filled with 0.246 g of N2. Assume that the pressure of the gas

alisha [4.7K]

Answer:

The pressure is $P = 1652 \ Pa$

Explanation:

From the question we are told that

The volume of the container is $V = 1.83 \ L = 1.83 *10^{-3 } \ m^3$

The mass of $N_2$ is $m_n = 0.246 \ g = 0.246 *10^{-3} \ kg$

The root-mean-square velocity is $v = 192 \ m/s$

The root -mean square velocity is mathematically represented as

$v = \sqrt{ \frac{3 RT}{M_n } }$

Now the ideal gas law is mathematically represented as

$PV = nRT$

=> $RT = \frac{PV}{n }$

Where n is the number of moles which is mathematically represented as

$n = \frac{ m_n }{M }$

Where M is the molar mass of $N_2$

So

$RT = \frac{PVM_n }{m _n }$

=> $v = \sqrt{ \frac{3 \frac{P* V * M_n }{m_n } }{M_n } }$

=> $v = \sqrt{ \frac{ 3 * P* V }{m_n } } }$

=> $P = \frac{v^2 * m_n}{3 * V }$

substituting values

=> $P = \frac{( 192)^2 * 0.246 *10^{-3}}{3 * 1.83 *10^{-3} }$

=> $P = 1652 \ Pa$

3 0

3 years ago