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lys-0071 [83]
3 years ago
11

What form of energy is a bonfire and a bunsen burner?

Physics
1 answer:
xenn [34]3 years ago
8 0

Answer:

heat and light energy

Explanation:

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A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp
VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir T_L=281K

We know that efficiency of Carnot engine is given by

\eta =1-\frac{T_L}{T_H}

0.26 =1-\frac{281}{T_H}

T_H=379.72K

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so T_H=379.72K

So 0.35=1-\frac{T_L}{379.72}

T_L=246.818K

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0
3 years ago
B. Find the velocity of a rider on the ride.
Stells [14]

D

Explanation:

Mutiply and alagrab to subterc

5 0
3 years ago
A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo
Tasya [4]

There are 3 forces acting on the stoplight:

• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward

• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0

We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to

2<em>T</em> sin(<em>θ</em>) = <em>W</em>

2 (1000 N) sin(<em>θ</em>) = 100 N

sin(<em>θ</em>) = 0.05

<em>θ</em> ≈ 2.87°

If <em>y</em> is the vertical distance between the stoplight and the ground, then

tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)

Solve for <em>y</em> :

tan(2.87°) = (15 m - <em>y</em>) / (100 m)

<em>y</em> = 15 m - (100 m) tan(2.87°)

<em>y</em> ≈ 9.99 m

3 0
3 years ago
A certain superconducting magnet in the form of a solenoid of length 0.72 m can generate a magnetic field of 3.5 T in its core w
ira [324]

Answer:

55407

Explanation:

we have given that magnetic field B=3.5 T

current through the coil=90 A

Length of solenoid =0.72 m

we know the formula of magnetic field

B=\frac{\mu _0NI}{l}

so N=\frac{Bl}{\mu _0I}=\frac{3.5\times 0.72}{4\pi \times 10^{-6}\times 90}=55407.277=55407\ turns

so the number of turn in solenoid will be 55407

3 0
3 years ago
If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
kaheart [24]

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

\Delta T is the temperature of difference that we want to find

\Delta x=2.5 cm =0.025 m represent the thickness of the material

If we solve \Delta T in absolute value from the equation (1) we got:

\Delta T =\frac{Q \Delta x}{Ak}

First we convert 3KW to W and we got:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

And we have everything to replace and we got:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

5 0
3 years ago
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