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3 years ago

The efficiency of a device such as a lamp can be calculated using this equation:

Physics

2 answers:

loris [4]3 years ago

6 0

efficiency = (useful energy transferred ÷ energy supplied) × 100

It's easy to use this formula, but we have to know both the useful energy and the energy supplied. The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out. I see two ways to do it:

Way #1:

We all know about the law of conservation of energy. So we know that the total energy coming out must be 250J, because that's how much energy is going in. The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.

(useful energy) / (energy supplied) = (175J) / (250J) = 70% efficiency

================================

Way #2:

How much of the energy is wasted ? . . . 75J wasted

What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted

30% of the input energy is wasted. That leaves the other 70% to be useful energy.

jek_recluse [69]3 years ago

3 0

Answer:

70%

Explanation:

I did the USA Test Prep.

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4 years ago

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A soccer ball is kicked horizontally. What is the average speed if its distance is 21.0 m after 4.00 s?

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3 years ago

A piece of bismuth with a mass of 4.06 g 4.06 g gains 423 J 423 J of heat. If the specific heat of bismuth is 0.123 J / ( g ° C

Sholpan [36]

Answer: 846°C

Explanation:

The quantity of Heat Energy (Q) required to heat bismuth depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that:

Q = 423 joules

Mass of bismuth = 4.06g

C = 0.123 J/(g°C)

Φ = ?

Then, Q = MCΦ

423 J = 4.06g x 0.123 J/(g°C) x Φ

423 J = 0.5J/°C x Φ

Φ = (423J/ 0.5g°C)

Φ = 846°C

Thus, the change in temperature of the sample is 846°C

4 0

4 years ago

The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:

kow [346]

Answer:

$\frac{d_{1}}{d_{2}}=0.36$

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

$\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]}$ (1)

So, the temperature of the first and the second star will be:

$T_{1}=3866.7 K$

$T_{2}=6444.4 K$

Now the relation between the absolute luminosity and apparent brightness is given:

$L=l\cdot 4\pi r^{2}$ (2)

Where:

L is the absolute luminosity
l is the apparent brightness
r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

$\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}$

So the relative distance between both stars will be:

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}}$ (3)

The Boltzmann Law says, $L=A\sigma T^{4}$ (4)

σ is the Boltzmann constant
A is the area
T is the temperature
L is the absolute luminosity

Let's put (4) in (3) for each star.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}$

As we know both stars have the same size we can canceled out the areas.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=0.36$

I hope it helps!

5 0

3 years ago

Two forces are acting on an object. The first force has magnitude F1=33.4 N and is pointing at an angle of θ1=23.8 clockwise fro

marishachu [46]

Answer:

Fe= 28.2 N : Magnitude of the equilibrant (Fe)

β = 18.34° , clockwise from the positive x axis

Explanation:

Concept of the equilibrant

It is called equilibrant to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero

To solve this problem we decompose the forces given into x-y components to find the resulting force:

Look at the attached graphic

F₁= 33.4 N , θ₁=23.8° clockwise from the positive y axis (y+)

F₁x= 33.4 *sin23.8° = 13.48 N

F₁y= 33.4 *cos23.8° =30.6 N

F₂=46.1 N , θ₂=28.8 counterclockwise from the negative x axis (x-)

F₂x= -46.1 *cos28.8° = -40.4 N

F₂y= -46.1 *sin28.8° = -22.2 N

Components of the resultant in x-y R(x,y)

Rx= 13.48 N -40.4 N = - 26.92 N

Ry= 30.6 N -22.2 N = + 8.4 N

Components of the equilibrant in x-y Fe(x,y)

Fex= +26.92 N

Fey= - 8.4 N

Magnitude of the equilibrant (Fe)

$F_{e} = \sqrt{(F_{ex})^{2}+{(F_{ey})^{2} }$

$F_{e} = \sqrt{(26.92)^{2}+(8.4)^{2} }$

Fe= 28.2 N

Angle the equilibrant makes with the x axis ( β)

$\beta = tan^{-1} (\frac{F_{ey} }{F_{ex} } )$

$\beta = tan^{-1} (\frac-8.4 }{26.92 } )$

β = -18.34°

β = 18.34° , clockwise from the positive x axis

8 0

3 years ago