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dezoksy [38]
3 years ago
6

Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at

a rate of 20,000 Btu/ hr., what is the temperature of the water leaving the radiator?
Physics
1 answer:
beks73 [17]3 years ago
8 0

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

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Train cars are coupled together by being bumped into one another. Suppose two loaded cars are moving toward one another, the fir
tia_tia [17]

Answer:

7560 Joules

Explanation:

m_1 = Mass of first car = 1.5\times 10^5\ kg

m_2 = Mass of second car = 2\times 10^5\ kg

u_1 = Initial Velocity of first car = 0.3 m/s

u_2 = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s

Energy lost is

\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J

The Energy lost in the collision is 7560 Joules

7 0
3 years ago
PLS HEEEEEELPPPPPPPPP!!!!!!!!​
Sveta_85 [38]

txt now you need to add to 14 to those two to get it between the 14 and 19 and basically you will just be able to do this the liquid is 45 as you know that's all I got to say for you is that you have to answer the phone for you equational to pay

7 0
3 years ago
A 3.0 kg object is loaded into a toy spring gun, and the spring has a spring constant 750N/m. The object compresses the spring b
kiruha [24]

Answer:

Explanation:

mass of object, m = 3 kg

spring constant, K = 750 n/m

compression, x = 8 cm = 0.08 m

angle of gun, θ = 30°

(a) As the ball is launched, it has some velocity due to the compression in the spring, so it has some kinetic energy.

(b) Let v be th evelocity of ball at the tim eof launch.

by using the conservation of energy

1/2 Kx² = 1/2 mv²

750 x 0.08 x 0.08 = 3 x v²

v = 1.265 m/s

By use of the formula of maximum height

h = \frac{v^{2}Sin^{2}\theta}{2g}

h = \frac{1.265^{2}Sin^{2}30}{2\times 9.8}

h = 0.02 m

h = 2 cm  

4 0
3 years ago
Consider a wave along the length of a stretched slinky toy, where the distance between coils increases and decreases. What type
GrogVix [38]

Answer:

"Longitudinal wave" is the appropriate answer.

Explanation:

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3 0
2 years ago
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A beryllium-9 ion has a positive charge that is double the charge of a proton, and a mass of 1.50 ✕ 10−26 kg. At a particular in
seropon [69]

Answer:

Magnetic force, F = 3.52\times 10^{-13}\ N

Explanation:

Given that,

A beryllium-9 ion has a positive charge that is double the charge of a proton, q=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C

Speed of the ion in the magnetic field, v=5\times 10^6\ m/s

Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.

The magnitude of the field is 0.220 T.

We need to find the magnitude of the magnetic force on the ion. It is given by :

F=qvB\\\\F=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\\\\F=3.52\times 10^{-13}\ N

So, the magnitude of magnetic force on the ion is 3.52\times 10^{-13}\ N.

3 0
3 years ago
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