I’m a god, a planet, and I measure heat. What am I?

Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr

nirvana33 [79]

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

cr = 3.7 J/gm degree K^-1

we know that power is given as

Power P = VI

But P =E/t

so equating both side we have

$\frac{E}{t} = VI$

solving for t

$t = \frac{E}{VI}$

$t = \frac{m_w C_w \Delta T}{VI}$

$t = \frac{51 \times 4.18 \times (5 -0)}{24\times 0.1}$

t = 444.125 sec

5 0

4 years ago

Read 2 more answers

Which factors affect heat transfer between a warm and a cool substance?

Furkat [3]

1. the conductivity of the material

2. temprature differences

3.thickness of the material

4. area of the material

4 0

4 years ago

Read 2 more answers

A Keystone Pipeline has a diameter of 36 inches and a design flow rate of 590,000 barrels per day of crude oil at 40ºC. Estimate

Sav [38]

Answer:

Power = 606.83 hp/mile

Explanation:

Given data:

diameter of pipeline is 36 inches = 3ft

flow rate is 590,000 barrels per day = 28.755 ft^3/sec

temperature of pipe = 46 degree celcius

length of pipe is 1.9 mile = 10032 ft

velocity of flow $= \frac{28.755}{\frac{\pi}{4} 3^2} = 4.068 ft/s$

applying bernouli eq at two point in pipeline

P1 = P2

v1 = v2

z1 = z2 so we have

$h_{in} = h_f$

$h_{in} = \frac{ flv^2}{2gD}$

$h_{in} = \frac{0.0155 \times 10032 \times 4.068^2}{2\times 32.174 \times 3}$

$h_{in} = 13.32 ft$

POwer $P = \rho gQ h_{in}$

$P = 53.71 \times 32.174 \times 27.755 \times 13.32 = 634141.29 ldf ft/sec$

Power P = 1152.99 hp 0

power needed for each mile $= \frac{1152.99}{1.9} = 606.83 hp/mile$

4 0

3 years ago

a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and roll

algol13

Answer:

$h = 0.0362\,m$

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

$U_{g,A} + K_{A} = U_{g,B} + K_{B}$

$U_{g,B} - U_{g,A} = K_{A} - K_{B}$

$m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}$

The moment of inertia of the coin is:

$I = \frac{1}{2}\cdot m \cdot r^{2}$

After some algebraic handling, an expression for the maximum vertical height is derived:

$m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}$

$h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}$

$h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }$

$h = 0.0362\,m$

6 0

3 years ago

You lift a 45-newton bag of mulch 1.2 meters and carry it a distance of 10 meters to the garden. How much

zysi [14]

To lift the bag straight up takes (F · D) = (45 · 1.2) = 54 joules of energy (work).

Moving the bag horizontally 'across' gravity requires no work.
It doesn't matter how far.

4 0

4 years ago