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Alex777 [14]
2 years ago
10

A 60-W light bulb emits spherical electromagnetic waves uniformly in all directions. If 50% of the power input to such a light b

ulb is emitted as electromagnetic radiation, what is the radiation intensity at a distance of 2.00 m from the light bulb?
A) 15 W/m2
B) 4.8 W/m2
C) 2.4 W/m2
D) 0.60 W/m2
E) 1.2 W/m2
Physics
1 answer:
makkiz [27]2 years ago
4 0

Answer:

a

Explanation:

A 60-W light bulb emits spherical electromagnetic waves uniformly in all directions. If 50% of the power input to such a light bulb is emitted as electromagnetic radiation, what is the radiation intensity at a distance of 2.00 m from the light bulb    

so i did 60(50%)=30÷2=15

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Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2

\hat{ECD} = tan^{-1} 2 = 63.43^o

This is also the angle ACB, so we can find the length AB:

tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}

2 = \frac{AB}{234}

AB = 2*234 = 468 m

So the height of the building is 468m

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Which of the following occurs with both a cold front and a mountain breeze?
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Explanation:

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One problem with weight training as a way to improve overall health is that the results of a weight-training program are not mea
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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

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Answer: Option (C) is the correct answer.

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As both left and right spheres are touching each other so, the electrons will move towards the right sphere. As a result, there will be too many electrons (negative charge) present on the right sphere and very less electrons present in the left sphere.

Thus, we can conclude that the statement right sphere is negatively charged, another is charged positively, is true.

7 0
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