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klio [65]
3 years ago
9

.A 100.0-W lightbulb is 22 percent efficient. This means that 22 percent of the electrical energy is transformed to radiant ener

gy.
a. How many joules does the lightbulb transform into radiant energy each minute it is in operation?

b. How many joules of thermal energy does the lightbulb output each minute?
Physics
1 answer:
algol [13]3 years ago
6 0

Answer:

Explanation:

100 W bulb is using energy of 100 J in one second.

22 percent of the electrical energy is transformed to radiant energy.

a )

So , electrical energy is transformed to radiant energy per second

= 100 x .22 = 22 J

energy transformed in one minute = 22 x 60  J

= 1320 J

b )

electrical energy is transformed to heat  energy per second

= 100 x .78 = 78 J

energy transformed in one minute = 78 x 60  J

= 4680  J

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The maximum gauge pressure in a hydraulic lift is 18.0 atm what is the largest size vehicle
victus00 [196]
There is a missing part in the question. Found the complete text on internet:
"<span>What is the largest size vehicle (kg) it can lift if the diameter of the output line is 28.0 cm? "

Solution
The diameter of the piston is 28.0 cm, this means its radius is 14.0 cm (half the diameter), so the area of the piston is
</span>A=\pi r^2 = \pi (0.14 m)^2 =6.15 \cdot 10^{-2} m^2
<span>
The maximum pressure of the lift is
</span>p=18.0 atm = 1.82 \cdot 10^6 Pa
<span>
Therefore the maximum force the piston can lift is
</span>F=pA=(1.82 \cdot 10^6 Pa)(6.15 \cdot 10^{-2} m^2)=1.12 \cdot 10^5 N
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And the size (the mass) of the vehicle is
</span>m= \frac{F}{g}= \frac{1.12 \cdot 10^5 Pa}{9.81 m/s^2} =1.14 \cdot 10^4 kg<span>
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3 0
3 years ago
Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o
VikaD [51]

Answer:

For the point inside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

For the point outside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

<u>x = -x0:</u>

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}

We have to integrate it over the ring, which is an angular integration.

E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta  = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

<u>x = +x0: </u>

We will only change the boundaries of the last integration.

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

6 0
3 years ago
Basketball player Darrell Griffith is on record as
Gre4nikov [31]

Explanation:

1.

We use the equation

h = \frac{gt^2}{2}, where

h is the height traveled,

g is the acceleration due to gravity and

t is the time taken to reach height h.

We can now calculate t to be

\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }

= 0.495 s

Let v be the initial velocity of the player.

The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.

v = 9.81 m/s^2 x 0.495 s = 4.85 m/s

2.

The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is

4.85 m/s / 0.3 s = 16.2 m/s^2

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3 years ago
A 1.0-cm-diameter pipe widens to 2.0, then narrows to 0.5. Liquid flows through the first segment at a speed of 4.0. What is the
uysha [10]

Answer:

3.14 ×  10⁻⁴  m³  /s

Explanation:

The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.

Q = Area x velocity

Given:

Diameters of 3 sections of the pipe are given as  

d1  =  1.0  cm,  d2  =  2.0  cm  and  d3  =  0.5  cm.

Speed in the first segment of the pipe is  

v1  =  4  m/s.

From the equation of continuity the flow rate through different cross-sections remains the same.

Flow  rate  =  Q  =  A1  v1  =  A2  v2  =  A3  v3.

Q = A1v1

   =π/4  d²1  v1  =  π/4  * 0.01² ×4.0 m³/s  =  3.14 ×  10⁻⁴  m³  /s

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If it takes 20n to move a box, how much power will be needed to move the box a diatance
solmaris [256]

The power applied to move the box anywhere is

   (20 n) x (distance moved) / (time to move the distance) .
 
3 0
4 years ago
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