Ask question

Ask question

All categories

3 years ago

7

At what distance does a 100 Watt lightbulb deliver the same power per unit surface area as a 75 Watt lightbulb produces 10 m awa

y from the bulb? (Assume both have the same efficiency for converting electrical energy in the circuit into emitted electromagnetic energy.). Recall that Watts = Joules/second = power = energy per unit time. Assume that the power of the electromagnetic waves spreads uniformly in all directions (i.e. spreads out over the area of a sphere) and use the formula for the surface area of a sphere.

1 answer:

ch4aika [34]3 years ago

4 0

Answer:

At 11.5 m

Explanation:

The power per unit area corresponds to the intensity, which is given by

$I=\frac{P}{4\pi r^2}$

where

P is the power

$4\pi r^2$ is the area irradiated at a distance r from the source (it corresponds to the surface area of a sphere of radius r)

Here we want the intensity of the two light bulbs to be the same, so

$I_1 = I_2\\\frac{P_1}{4 \pi r_1^2}=\frac{P_2}{4\pi r_2^2}$

where we have

P1 = 100 W is the power of the first light bulb

P2 = 75 W is the power of the second light bulb

r2 = 10 m is the distance from the second light bulb

Solving for r1, we find

$r_1 = r_2 \sqrt{\frac{P_1}{P_2}}= (10 m) \sqrt{\frac{100 W}{75 W}} = 11.5 m$

You might be interested in

Insulators have very high .

Vlad1618 [11]

Answer:

Resistance to electrical currents

Explanation:

Conductors have low resistance to electrical currents, and are used to "conduct" the flow of electricity.

Insulators have very high resistance and are used to protect us from the flow of electricity.

5 0

3 years ago

What two factors are a part of thermohaline circulation

Margarita [4]

Answer:

These deep-ocean currents are driven by differences in the water's density, which is controlled by temperature (thermo) and salinity (haline). This process is known as thermohaline circulation.

Explanation:

7 0

3 years ago

Which of the following statements are true about the international system of measurement?

anygoal [31]

The International System Units or the SI units is scientific method of expressing the magnitudes or quantities of important natural phenomena. There are seven base units in the system, from which other units are derived. This system was formerly called the meter-kilogram-second (MKS) system.

8 0

3 years ago

A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc

Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a (→)

F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m

⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)

⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0

3 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago