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vivado [14]
3 years ago
5

How much power is required to light a lightbulb at 100V of voltage when the lightbulb has a resistance of 500 Ohms?

Physics
1 answer:
salantis [7]3 years ago
3 0

Answer:

Power = 20 Watts

Explanation:

Given the following data;

Voltage = 100 V

Resistance = 500 Ohms

To find the power that is required to light a lightbulb;

Mathematically, power can be calculated using the formula;

Power = \frac {Voltage^{2}}{resistance}

Substituting into the formula, we have;

Power = \frac {100^{2}}{500}

Power = \frac {10000}{500}

Power = 20 Watts

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The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan
ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

I = \frac{P}{A}

The area of a sphere is given by

A = 4\pi r^2

So replacing we have to

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

Since the question tells us to find the proportion when

r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}

So considering the two intensities we have to

I_1 = \frac{P_1}{4\pi r_1^2}

I_2 = \frac{P_2}{4\pi r_2^2}

The ratio between the two intensities would be

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

The power does not change therefore it remains constant, which allows summarizing the expression to

\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2

Re-arrange to find I_2

I_2 = I_1 (\frac{r_1}{r_2})^2

I_2 = 100*(\frac{1}{5})^2

I_2 = 4W/m^2

Therefore the intensity at five times this distance from the source is 4W/m^2

3 0
3 years ago
What is the impulse of a 3 kg object that starts from rest and moves to 20 m/s?
IrinaK [193]

Answer:

The impulse on the object is 60Ns.

Explanation:

Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.

F = m a

F = m(\frac{v_{2}  - v_{1} }{t})

⇒     Ft = m(v_{2} - v_{1})

where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object, v_{1} is its initialvelocity and v_{2} is the final velocity of the object.

Therefore,

impulse = Ft = m(v_{2} - v_{1})

From the question, m = 3kg, v_{1} = 0m/s and v_{2} = 20m/s.

So that,

Impulse = 3 (20 - 0)

             = 3(20)

             = 60Ns

The impulse on the object is 60Ns.

8 0
2 years ago
Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

7 0
3 years ago
A piece of rope is pulled by two people in a tug-of-war. each exerts a 400-n force. what is the tension in the rope?
Leokris [45]
Newtons 3.law: Action = Reaction

If a body exerts a force on a rope of 400 N the rope exerts a force on the body of 400N also. So the tension in the rope is 400N. See pictures below.

3 0
3 years ago
Read 2 more answers
PLEASE HELP!!!! The displacement vectors A and B, when added together, give the resultant vector R, so that R = A + B. Use the d
GuDViN [60]

Answer:

Ax = 0

Ay = 6 m

Bx = 8 cos phi = cos 34 = 6.63 m

By = 8 sin phi = 8 sin (-34) = -4.47 m

Rx = Ax + Bx = 0 + 6.63 = 6.63 m

Ry = Ay + By = 6 - 4.47 = 1.53 m

R = (6.63^2 + 1.53^2)^1/2 = 6.80 m

tan theta = Ry / Rx = 1.53 / 6.8 = ,225

theta = 12.7 deg

7 0
3 years ago
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