All categories

3 years ago

Which of the following could be the source of resistance in a household electric circuit?

Physics

2 answers:

Tpy6a [65]3 years ago

8 0

D. all of these

all of these use electricity

Hope I helped!

exis [7]3 years ago

7 0

The answer in d (all of these)

You might be interested in

If a weightlifter used 1,200 newtons of force to lift a barbell to a height of 2.5 meters in 3 seconds, how much work did he do?

Blizzard [7]

C.) Work done = Force x displacement = 1200 x 2.5 = 3000 J. Hope this helps!

3 0

3 years ago

Consider the line graph depicting the motion of two cars. What statement regarding the cars motion is accurate?

Rom4ik [11]

Answer:

Its C

Explanation:

I did it on usatestprep

5 0

3 years ago

A father pushes his child in a cart. The cart starts to move.

LuckyWell [14K]

The potential energy of the car when it let go is 20,000 J.

The speed of the car at the bottom of the ramp is 20 m/s.

The given parameters;

mass of the car, m = 100 kg
height of the car, h = 20 m

The potential energy of the car is calculated as follows;

P.E = mgh

P.E = 100 x 10 x 20

P.E = 20,000 J

The speed of the car at the bottom of the ramp is calculated as follows;

$K.E = P.E\\\\\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2 \times 10 \times 20} \\\\v = 20 \ m/s$

Learn more here:brainly.com/question/18597080

5 0

3 years ago

Who invented the first practical incandescent light bulb?.

OverLord2011 [107]

Answer:

Edison

Explanation:

Among other notable inventions, Edison and his assistants developed the first practical incandescent lightbulb in 1879 and a forerunner of the movie camera and projector in the late 1880s.

6 0

3 years ago

What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of 1.3 cm and a uniformly di

Vlad [161]

Answer:

2.37 * 10^4 m/s

Explanation:

Constants :

Mass of electron = 9.11 * 10^(-31) kg

Electric charge of an electron = 1.602 * 10^(-19) C

Parameters given:

Radius of sphere = 1.3cm = 0.013m

Charge of sphere = 2.3 * 10^(−15) C

Using the law of conservation of energy, we have that:

K. E.(initial) + P. E.(initial) = K. E.(final) + P. E.(final)

K. E.(final) = 0, since final velocity is zero and P. E.(final) = 0 since the electron reaches a final distance of infinity.

Hence,

K. E.(initial) = P. E.(initial)

0.5mv^2 = (kqQ)/r

Where k = Coulumbs constant

Q = charge of the sphere.

r = radius of the sphere.

=> 0.5*m*v^2 = (kqQ)/r

0.5 * 9.11 * 10^(-31) * v^2 = (9 * 10^9 * 1.602 * 10^(-19) * 2.3 * 10^(-15))/0.013

4.555 * 10^(-31) * v^2 = 2550.88 * 10^(-25)

=> v^2 = 2550.88 * 10^(-25) / 4.555 * 10^(-31)

v^2 = 560 * 10^6 = 5.60 * 10^8

=> v = 2.37 * 10^4 m/s

4 0

3 years ago