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monitta
3 years ago
9

Some homes that use baseboard heating use copper tubing. hot water runs through and heats the copper tubing, which in turn heats

aluminum fins. it is actually the aluminum fins that heat the air rising through the fins. how much energy would it take to heat a section of the copper tubing that weighs about 645.0 g , from 13.20 âc to 28.22 âc ? copper has a specific heat of 0.3850 (j/g)ââc.
Physics
1 answer:
AlekseyPX3 years ago
7 0
When you heat a certain substance with a difference of temperature \Delta T the heat (energy) you must give to it is
E(=Q) =mc\Delta T
where c is the specific heat of that substance (given in J/(g*Celsius))
In this case
E=645*0.3850*(28.22-13.20) =3729.8 (Joule)

Observation: the specific heat of a substance is given in J/(g*Celsius) or J/(g*Kelvin)  because on the temperature scale a difference of 1 degree Celsius = 1 degree Kelvin
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Two blocks are connected by a massless cable which goes through the center of a rotating turntable. The blocks have masses M1 =
Airida [17]

Answer:

If the final question is; at what velocity will the first block start to move outward in m/s?

v = 3.5596 \frac{m}{s}

Explanation:

The motion have the velocity that will make the block move using:

F_{1}*F_{r}+ F_{2}= Ec \\Ec= \frac{M*v^{2} }{r}

m_{1} = 1.2 Kg

m_{2} = 2.8 Kg

r = 0,45 m

μs= 0,54

Resolving:

\frac{m_{1}*v^{2}  }{r} = us*m_{1}* g + m_{2} * g

v^{2} = \frac{((us *m_{1} + m_{2})*g )* r}{m_{1} }

v^{2} = \frac{((0.54 *1.2 + 2.8)*9.8 )* 0.45}{1.2}

v^{2} = 12.6714 \frac{m^{2} }{s^{2} }

v = 3.559 \frac{m}{s}

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3 years ago
Which of the following is not a symptom associated with hypertension
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3 years ago
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A type of light bulb is labeled having an average lifetime of 1000 hours. It’s reasonable to model the probability of failure of
SashulF [63]

Answer:

0.2592 \ or \ 25.92\%

Explanation:

The exponential density function is given as

f(t)=\left \{ {{0} \atop {ce^{ct}}} \right\\0,t

\mu=\frac{1}{c}\\c=\frac{1}{\mu}\\\\=\frac{1}{1000}=0.001\\\\f(t)=0.001e^{-0.001t}

To find probability that bulb fails with the first 300hrs, we integrate from o to 300:

P(0\leq X\leq 300)=\int\limits^{300}_0 {f(t)} \, dt\\\\=\int\limits^{300}_0 {0.001e^{-001t}} \, dt\\ =|-e^{-0.001t}|  \ 0\leq t\leq 300

P(0\leq X\leq 300)=-0.7408+1\\=0.2592

Hence probability of bulb failing within 300hrs is 25.92% or 0.2592

3 0
3 years ago
A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp
ch4aika [34]

Answer:

Part(a): The equation of motion is \bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}.

Part(b): The equation of motion is  \bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}.

Explanation:

If 'm' be the mass of the object, 'k' be the force constant and '\beta' be the damping constant, then the equation of motion of the particle can be written as

\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)

Given m = 1 Kg, k = 14 N~m^{-1}, \beta = 9. Substituting these values in equation (I),

\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0

Taking a trial solution x(t) = e^{mt}, the auxiliary equation can be written as

m^{2} + 9m + 14 = 0............................................................(II)

and its solutions are m_{1} = -2~and~m_{2} = -7, resulting the general solution

x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)

The velocity at any instant of time of the mass is

v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)

Part(a):

Given x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}. Substituting these values in equation (III) and (IV),

&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)

Solving equations (V) and (VI), we have

C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}

So the equation of motion is

x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}

Part(b):

Given x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}. Substituting these values in equation (III) and (IV),

&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)

Solving equations (V) and (VI), we have

C_{1} = -1~and~C_{2} = \dfrac{11}{5}

So the equation of motion is

x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}

3 0
3 years ago
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