1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Digiron [165]
2 years ago
12

Which type of electricity moves along a pathway to turn on a light?

Physics
1 answer:
Nina [5.8K]2 years ago
7 0

Answer:

Current electricity

Explanation:

moves along a path to turn on a light.

You might be interested in
A horizontal force of 30N is applied to a mass of 10 kg causing it to accelerate. If the coefficient of friction is 0.20 what is
Blizzard [7]

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

In this question ,

surface vertical force = Weight of the object

Thus ;

svf = ( mass ) × ( gravity acceleration )

_________________________________

If gravity acceleration is 10 :

svf = 10 × 10 = 100 N

So ;

frictional force = 100 × 0.20

frictional force = 20 N

##############################

If gravity acceleration is 9.8 :

svf = 10 × 9.8 = 98 N

So ;

frictional force = 98 × 0.20

frictional force = 19.6 N

_________________________________

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

5 0
3 years ago
Will give correct answer brainliest
sineoko [7]

200 \times 80\%

<h2><em>calculate</em></h2>

<em>200 \times  \frac{80}{100}</em>

<h2><em>reduce </em><em>the </em><em>numbers</em></h2>

<em>2 \times 80</em>

<h2><em>multiply</em></h2>

<em>= 160</em>

<h2><em>there </em><em>for </em><em>we </em><em>have </em><em>a </em><em>solution</em><em> to</em><em> the</em><em> </em><em>equation</em></h2>

<em>hope </em><em>it</em><em> helps</em>

<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em> </em><em>on</em><em> learning</em>

<em>mark </em><em>me</em><em> as</em><em> brainlist</em><em> plss</em>

6 0
3 years ago
Read 2 more answers
A ball rolls up a ramp with a velocity of 8.0 m/s. How high up the ramp does it travel?
Flura [38]

The greatest height the ball will attain is 3.27 m

<h3>Data obtained from the question</h3>
  • Initial velocity (u) = 8 m/s
  • Final velocity (v) = 0 m/s (at maximum height)
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Maximum height (h) =?

The maximum height to which the ball can attain can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 8² – (2 × 9.8 × h)

0 = 64 – 19.6h

Collect like terms

0 – 64 = –19.6h

–64 = –19.6h

Divide both side by –19.6

h = –64 / –19.6h

h = 3.27 m

Thus, the greatest height the ball can attain is 3.27 m

Learn more about motion under gravity:

brainly.com/question/13914606

5 0
2 years ago
_____ are small buildings containing transformers and electrical equipment
Vika [28.1K]
This room is called Substation
6 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
Other questions:
  • How does Photosynthesis use solar energy?
    6·1 answer
  • How is kinetic energy related to the distance
    7·2 answers
  • 10 points and brainliest to correct answer plz
    6·1 answer
  • Solve 13.004+3.09+112.947
    9·1 answer
  • Which is hotter, Canopus or Vega? How much brighter?
    11·1 answer
  • How did the transcontinental railroad impact travel?
    11·1 answer
  • The electric potential ( relative to infinity ) due to a single point charge Q is 400 V at a point that is 0.6 m to the right of
    15·1 answer
  • What is true of the water cycle?
    12·2 answers
  • A 20-cm-long stick of m = 0.600 kg is lifted by a rope tied 5.0 cm from the upper end. The other end touches a smooth floor. The
    15·1 answer
  • Which would most likely form a homogenous mixture?
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!