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3 years ago

What is the maximum value of the electric field at a distance2.5 m from a 100-W light bulb?

Physics

1 answer:

Alborosie3 years ago

4 0

Answer

given,

Power of bulb,P = 100 W

distance from bulb, r = 2.5 m

Intensity of bulb,

$I = \dfrac{P}{A}$

P is the power of bulb

A is the area

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

$I = \dfrac{100}{4\pi\times 2.5^2}$

I = 1.274 W/m²

The relation between intensity I and E_max

$I = \dfrac{E_{max}^2}{2\mu_0 c}$

where c is the speed of light

μ₀ = 4 π x 10⁻⁷

$E_{max}=\sqrt{I(2\mu_0 c)}$

$E_{max}=\sqrt{1.274\times (2\times 4\pi \times 10^{-7}\times 3 \times 10^8)}$

E_{max} = 30.99 V/m

now, maximum magnetic field

$B_{max}=\dfrac{E_{max}}{c}$

$B_{max}=\dfrac{30.99}{3\times 10^8}$

$B_{max}= 1.033 x 10⁻⁷\ T$

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A scientist travelled 80 kilometres each day in the buggy.

7nadin3 [17]

Explanation:

In one day a scientist can travel = 80 km

In 10 days. a scientist can travel = 80* 10

= 800 km.

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4 years ago

Read 2 more answers

Two force vectors are oriented such that the angle between their directions is 38 degrees and they have the same magnitude. If t

Usimov [2.4K]

Answer:

Sum of the forces will be equal to 3.479 N

Explanation:

We have given two same forces are oriented at an angle of 38°

Magnitude of each force is given $F_1=F_2=1.84$

We have to find the sum of the forces

Sum of the forces will be equal to

$F=\sqrt{F_1^2+F_2^2+2F_1F_2cos\Theta }$

So sum of the forces will be equal to $F=\sqrt{1.84^2+1.84^2+2\times 1.84\times 1.84\times cos38^{\circ} }=3.479$

So sum of the force will be equal to 3.479 N

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3 years ago

on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker

Aleksandr-060686 [28]

B4 the tackle:

The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north 

and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east 

After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle 

The vector triangle is right angled: 

magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s 

so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] 

v(f) = 5.6 m/s (to 2 sig figs) 

direction of v(f) is the same as the direction of the final momentum 

so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) 

so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E 

btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it

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3 years ago

efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

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3 years ago

How can a mass be hot enough to flow but still act like a solid

Lunna [17]

The inner core is solid because it is made of very dense, or heavy, materials like iron and nickel. Even though it is very hot, these materials don't

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2 years ago