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4 years ago

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how much heat flows per second through a well lagged bar of alummium of thermal cond7ctivity 230W ,the bar has length of 300mm a

nd a cross sectional area of 100 mm² with a temorature difference of 40°c between the ends of the bar

1 answer:

Mars2501 [29]4 years ago

6 0

Answer:

The amount of heat flows per second in given aluminium bar is 3.066 $JS^{-1}$ .

Explanation:

We know that , amount of heat flowing per second is given by

$\frac{\mathrm{d} Q}{\mathrm{d} t} = \frac{KA\Delta T}{L}$

Where A = area = $300 mm^{2} = 300\times 10^{-6} m^{2}$

$\Delta T = 40^{o} c = 40 K$

K = Thermal conductivity = $230 W/m-k$

L = Length = $300mm = 300 \times 10^{-3}$

Upon substituting these values in above equation we get ,

Heat flowing per second = $3.066 JS^{-1}$

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Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

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Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

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3 years ago

A boat has a mass of 4040 kg. Its engines generate a drive force of 4660 N due west, while the wind exerts a force of 880 N due

maxonik [38]

Answer:

Explanation:

Given:

Mass of the boat, m = 4040 kg

The driving force of engine, FB = 4660 N in west = + 4660 N

The force of wind, Fwi = 880 N in east = -880 N

The force of water, Fwa = 1400 N in east = -1400N

Total three forces are acting on the boat

Fnet= Fb+fwi+Fwa

Fnet= 4660 - 880 - 1400

Fnet= +2380N

Acceleration (a) = Force/mass

= 2380/4040

= 0.59m/s2

6 0

3 years ago

Read 2 more answers

An 20-cm-long bicycle crank arm, with a pedal at one end, is attached to a 25-cm-diameter sprocket, the toothed disk around whic

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Answer:

0.033815 m/s²

12.48810875 m

Explanation:

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$\omega_f$ = Final angular velocity = 95 rpm

$\omega_i$ = Initial angular velocity = 64 rpm

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{95\times \frac{2\pi}{60}-64\times \frac{2\pi}{60}}{12}\\\Rightarrow \alpha=0.27052\ rad/s^2$

Tangential acceleration is given by

$a_t=\alpha r\\\Rightarrow a_t=0.27052\times 0.125\\\Rightarrow a_t=0.033815\ m/s^2$

The tangential acceleration of the pedal is 0.033815 m/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{\left(95\times\frac{2\pi }{60}\right)^2-\left(64\times\frac{2\pi}{60}\right)^2}{2\times 0.27052}\\\Rightarrow \theta=99.90487\ rad$

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4 years ago

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For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
Fx = Fcos(∅)
We can multiply this by 2 to find the resultant of both horses.
Fx = 2Fcos(∅)
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3 years ago

you push with an 18-N horizontal force on a 5-kg box of coffee resting on a on a horizontal surface. the force of friction on th

KatRina [158]

The acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.

Answer:

Explanation:

As per Newton's second law of motion, acceleration of any object is directly proportional to the net external unbalanced force acting on that object and inversely proportional to the mass of the object.

Since there are two forces acting on the box in opposite direction, the net force will be the difference of horizontal and frictional force acting on the object.

Net force = Horizontal force - Frictional force = 18 N - 8 N = 10 N.

Now, from second law of motion, $Acceleration = \frac{Net force}{Mass}$

So, acceleration = 10 N /5 kg = 2 m/s².

Since, acceleration exerted by the box is found to be 2 m/s², we can determine the final velocity of the object after 10 seconds using the first equation of motion.

v = u + at, Here v is the final velocity and u is the initial velocity which is zero for the present case. Other parameters like a is found to be 2 m/s² and time is 10 seconds.

So Final velocity v = 0+(2×10)=20 m/s.

And the final position can be determined using the second equation of motion.

s = ut+1/2at²

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So the final position is 100 m.

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3 years ago