PHYSICS HIGH SCHOOL

The resistance of 100 W bulb is less than resistance of 40 W bulb. Explain the reason.

Answers

Answer 1
Answer:

Ok so we know that electric power is:

If we express resistance.

Now if you have 100W of power you will probably get a bigger resistance than with 40W of power.

However here it says that the resistance of 40W bulb is bigger than 100W bulb. Which means the statement is incorrect.

Hope this helps.

r3t40


Related Questions

COLLEGE

The following forces act on an object: 20 N north, 40 N south, and 40 N west. What is the magnitude of the net force?

Answers

The magnitude of the resulting force is around 44.72 N. The net force in the y direction is -20N and in the x direction -40N. You then find the magnitude of the force by |F| = (x^2 + y^2)^1/2
COLLEGE

Two electrons are separated by one cm. What is the ratio of the electric force to the gravitational force between them? Note that Me = 9.11 x 10^-31kg, Ke=8.99 x 10^9 N m^2/ C^2 , G= 6.67 x 10^-11 N m^2/kg^2, e=1.6 x 10^-19 C, and Fg = G m1xm2/r^2

a)2.3 x 10^2
b) 1.3 x 10^20
c)3.1 x 10^22
d)4.2 x 10^42

Answers

The ratio of the electric force to the gravitational force between them is d. 4.2 × 10⁴²

The electric force between the two electrons according to Coulomb's law is F = ke²/r² where

  • k = electric constant = 8.99 × 10⁹ Nm²/C²,
  • e = electron charge = 1.6 × 10⁻¹⁹ C
  • r = distance between electrons = 1 cm = 0.01 m.

Also, the gravitational force between the electrons according to Newton's law of gravitation is f = GMe²/r² where

  • G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²,
  • Me = electron mass = 9.11 × 10⁻³¹ kg
  • r = distance between electrons = 1 cm = 0.01 m.

Since we want to find the ratio between the electric force and the gravitational force, we have F/f = ke²/r² ÷  GMe²/r²

= ke²/GMe²

Substituting the values of the variables into the equation, we have

ke²/GMe² = 8.99 × 10⁹ Nm²/C² × (1.6 × 10⁻¹⁹ C)²/[6.67 × 10⁻¹¹ Nm²/kg² × (9.11 × 10⁻³¹ kg)²]

= 8.99 × 10⁹ Nm²/C² × 2.56 × 10⁻³⁸ C²/[6.67 × 10⁻¹¹ Nm²/kg² × 82.9921 × 10⁻⁶² kg²]

= 23.0144 × 10⁻²⁹ Nm²/553.557307 × 10⁻⁷³ Nm²

= 0.0416 × 10⁴⁴

= 4.16 × 10⁴²

≅ 4.2 × 10⁴²

So, the ratio of the electric force to the gravitational force between them is d. 4.2 × 10⁴².

Learn more about ratio of electric to gravitational force here:

brainly.com/question/8994535

Answer:

d)

Explanation:

We are given that

By using 100cm=1 m

Electric force=

Using the formula

Gravitational force,

Using the formula

Option d is true.

COLLEGE

During test-out procedures, a 239Pu-fueled thermal reactor is operated for a time at a power of 1 megawatt.The power is then to be increased to 100 megawatts in 8 hours. (a) On what stable period should the reactor be placed? (b) What reactivity insertion is required?

Answers

The stable placed on x+7*8 the equation for your hysics will be solved high iron paid
COLLEGE

A hot-rolled steel has a yield strength of Syr-Sye-170 kpsi and a true strain at fracture of 0.55. Estimate the factor of safety for the following stress state, by using a) the maximum shear stress theory b) the distortion energy theory. (30 points) S1. Ox-30 kpsi, o,' =-15 kpsi, τ,--45 kpsi.

Answers

Answer:

a) Factor of safety ≤ 1.689

b) Factor of safety ≤ 1.944

Explanation:

Given data:

Yield strength,

True strain factor = 0.55

Now, principle stress is given as:

on substituting the values we get

or

and

a) By maximum shear stress theory

we have

on substituting the values, we get

or

Factor of safety ≤ 1.689

b) By the distortion energy theory

since no force is acting in the z- direction, thus

on substituting the values, we get

or

Factor of safety ≤ 1.944

MIDDLE SCHOOL

How does the wavelength of a sound affect the way it moves around corners

Answers

This is the diffraction. The longer the wavelength, the wider it spread. (The greater angle of the sector pattern formed)
COLLEGE

The armature of a 60 Hz ac generator rotates in a 0.12 T B-field. If the area of the coils is 0.03 m2, how many loops must the coil contain if the peak output is to be V = 270 V? What will the current & power output (not the power loss) be if the resistance of the coil is 0.4 ??

Answers

Explanation:

it is given that,

Frequency of AC generator, f = 60 Hz

Magnetic field, B = 0.12 T

Area of the coil, A = 0.03 m²

Peak out put, V = 270 V

Resistance of the coil, R = 04 ohms

1. Let N is the number of loops the coil must contain. The peak voltage of AC generator is given by :

N = 199

2. Applying Ohm's law as :

I = 675 A

3. Power,

P = 182250 watts

Hence, this is the required solution.

MIDDLE SCHOOL

Light and heat energy travel through space at a speed of A.186,000 feet per second b.186,000 miles per second C. 186,000 miles per minute D.1860 miles per hour

Answers

Answer: B. 186,00 miles per second.

Explanation:  

The speed of light in a vacuum is equal to  

1 / sqrt(e0 * u0)  

where e0 is the electric permittivity of free space, and u0 is the magnetic permeability of free space. So that's where the speed of light comes from.  

But where do the permittivity and permeability come from? Actually, nobody knows. They are seemingly arbitrary universal constants.

Answer:

b.186,000 miles per second

Explanation:

MIDDLE SCHOOL

What are the properties of ionic compounds?

Answers

There are 2 main properties that Ionic compounds share:

1) High melting and boiling points. - Ionic bonds are extremely strong. So therefore, lots heat energy is needed to break them down. So ionic compounds have high melting and boiling points. 

2) Conductive when liquid- Ions are charged particles, but ionic compounds can only conduct electricity if their ions are free to move. They are not free to move in a solid state, however they are free to move when the compound has been melted or dissolved in a liquid.

Hope this helps! :)
High melting and boiling point
ionic bonds are very strong
a lot of energy is needed to break them
conductive when liquid
ionic com. can only conduct electricity if their ions are free to move
COLLEGE

Which of the following describes the importance of cyanobacteria to early Earth atmosphere formation? A.) Cyanobacteria released oxygen into the air, creating a protective ozone layer
B.) Cyanobacteria relased nitrogen into the air, creating a protective ozone layer
C.) Cyanobacteria absorbed excess oxygen from the air, creating proper amounts of atmospheric oxygen
D.) Cyanobacteria absorbed excess carbon dioxide from the air, creating a protective ozone layer

Answers


I do believe it is a
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