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3 years ago

7

Physical therapists know as you soak tired muscles in a hot tub, the water will cool down as you heat up. If a 67.9 kg person at

37.1 oC immerses in 50.2 kg of water at 40.5o C, the equilibrium temperature is 38.7 oC.
What is the specific heat of the person?
Use 4.186 kJ/kgoC for the specific heat of water.

1 answer:

Lubov Fominskaja [6]3 years ago

4 0

Answer:

c = 5032.8J/kg°C

Explanation:

the transfer of heat is given by

$Q_1=-Q_2\\\\Q=cm(T_f-T_i)$

c is the specific heat, m the mass and T the temperatures at equilibrium an at the beginning.

By replacing we obtain

$c_1m_1(T_f-T_{1i})=-c_2m_2(T_f-T_{2i})\\\\c_2=-\frac{c_1m_1(T_f-T_{1i})}{m_2(T_f-T_{2i})}=-\frac{(67.9kg)(4186J/kg\°C)(38.7\°C-37.1\°C)}{(50.2kg)(38.7\°C-40.5\°C)}\\\\c_2=5032.8\frac{J}{kg\°C}$

hope this helps!!

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You have a tungsten sphere (emissivity ε = 0.35) of radius 25 cm at a temperature of 25°C. If the sphere is enclosed in a room w

egoroff_w [7]

Answer:

Explanation:

Stefan's formula for emission of radiation is

E = e σ A ( T⁴ - T₀⁴ )

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2 years ago

A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150

Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

$\rho \mathrm{Vg}=2 \mathrm{mg} \\$

$V=\frac{2 m}{\rho} \\$

$V=\frac{2 \times 150}{1000} \\$

$\mathrm{~V}=0.3 \mathrm{~m}^3$

Area of pool $=\frac{\pi}{4} d^2$

$&=\frac{\pi}{4} \times 6^2 \\$

$&=28.27 \mathrm{~m}^2$

Height of the water rise

$h &=\frac{V}{A} \\$

$h &=\frac{0.3}{28.27} \\$

$& \mathrm{~h}=0.0106 \mathrm{~m}$

Pressure increased

P = ρ g h

P = 1000 x 10 x 0.0106

P = 106.103 Pa

To learn more about absolute pressure refer to:

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1 year ago

What is the difference between chemical change and a physical change?

NeX [460]

Answer:

Explanation:

There are several differences between a physical and chemical change in matter or substances. A physical change in a substance doesn't change what the substance is. In a chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.

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3 years ago

In this model, the velocity of the spacecraft at position 2 is the velocity of the craft at position 4. at position 1, the direc

Free_Kalibri [48]

1. The velocity of the spacecraft at position 2 is greater than the velocity of the craft at position 4.

This is due the gravity field of the Earth is used to accelerate the craft. This is true when in a specific point the direction of the movement of the craft is the same direction of the movement of the planet.

In this case the craft will be “catched” by the Earth’s gravitational field, making the craft to enter a circular orbit.

2. At point 1, the direction of the spacecraft changes because of the gravitational force between earth and the spacecraft.

As explained in the first answer, this is the exact point where the trajectory of the spacecraft enters into a circular orbit because of the attraction due gravity of the Earth and therefore changes its direction.

3. Position 3 represents the orbital path of Earth

Being this the orbital path of the Earth and considering the trajectory of the craft, the condition of accelerating the craft is accomplished. If the orbital path of the Earth were the opposite, the effect on the craft would be braking.

Note all of these is related to the gravitational assistance, this consists in a maneuver in which the energy of the gravitational field of a planet or satellite is used to obtain an acceleration or braking of the probe or craft, changing its trajectory.

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1 year ago

Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t

katrin2010 [14]

Answer:

$1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

23.4843749996 m

Yes

Explanation:

E = Electric field = $2\times 10^5\ N/C$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of proton= $1.67\times 10^{-27}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

Acceleration is given by

$a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g$

The acceleration is $1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m$

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0

3 years ago