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
Law Incorporation [45]
2 years ago
6

Electricity flows from what to what

Physics
2 answers:
DerKrebs [107]2 years ago
5 0
Electricity flows from positive to negative
SVEN [57.7K]2 years ago
5 0

Answer:

electric current refers to electrons and protons flow in the opposite direction. current is flow electrons ,but current and electron flow in the opposite direction. current flows from positive to negative and electron flows from negative to positive.

You might be interested in
Illustrates an Atwood's machine. Let the masses of blocks A and B be 7.00 kg and 3.00 kg , respectively, the moment of inertia o
Harman [31]

Answer:  

A) 1.55  

B) 1.55

C) 12.92

D) 34.08

E)  57.82

Explanation:  

The free body diagram attached, R is the radius of the wheel  

Block B is lighter than block A so block A will move upward while A downward with the same acceleration. Since no snipping will occur, the wheel rotates in clockwise direction.  

At the centre of the whee, torque due to B is given by  

{\tau _2} = - {T_{\rm{B}}}R  

Similarly, torque due to A is given by  

{\tau _1} = {T_{\rm{A}}}R  

The sum of torque at the pivot is given by  

\tau = {\tau _1} + {\tau _2}  

Replacing {\tau _1} and {\tau _2} by {T_{\rm{A}}}R and - {T_{\rm{B}}}R respectively yields  

\begin{array}{c}\\\tau = {T_{\rm{A}}}R - {T_{\rm{B}}}R\\\\ = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R\\\end{array}  

Substituting I\alpha for \tau in the equation \tau = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R  

I\alpha=\left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R  

\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right  

The angular acceleration of the wheel is given by \alpha = \frac{a}{R}  

where a is the linear acceleration  

Substituting \frac{a}{R} for \alpha into equation  

\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right we obtain  

\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right  

Net force on block A is  

{F_{\rm{A}}} = {m_{\rm{A}}}g - {T_{\rm{A}}}  

Net force on block B is  

{F_{\rm{B}}} = {T_{\rm{B}}} - {m_{\rm{B}}}g  

Where g is acceleration due to gravity  

Substituting {m_{\rm{B}}}a and {m_{\rm{A}}}a for {F_{\rm{B}}} and {F_{\rm{A}}} respectively into equation \frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right and making a the subject we obtain  

\begin{array}{c}\\{m_{\rm{A}}}g - {m_{\rm{A}}}a - \left( {{m_{\rm{B}}}g + {m_{\rm{B}}}a} \right) = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g - \left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right)a = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)a = \left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g\\\\a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}\\\end{array}  

Since {m_{\rm{B}}} = 3kg and {m_{\rm{B}}} = 7kg  

g=9.81 and R=0.12m, I=0.22{\rm{ kg}} \cdot {{\rm{m}}^2}  

Substituting these we obtain  

a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}  

\begin{array}{c}\\a = \frac{{\left( {7{\rm{ kg}} - 3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {7{\rm{ kg}} + 3{\rm{ kg}} + \frac{{0.22{\rm{ kg/}}{{\rm{m}}^2}}}{{{{\left( {0.120{\rm{ m}}} \right)}^2}}}} \right)}}\\\\ = 1.55235{\rm{ m/}}{{\rm{s}}^2}\\\end{array}

Therefore, the linear acceleration of block A is 1.55 {\rm{ m/}}{{\rm{s}}^2}

(B)

For block B

{a_{\rm{B}}} = {a_{\rm{A}}}

Therefore, the acceleration of both blocks A and B are same

1.55 {\rm{ m/}}{{\rm{s}}^2}

(C)

The angular acceleration is \alpha = \frac{a}{R}

\begin{array}{c}\\\alpha = \frac{{1.55{\rm{ m/}}{{\rm{s}}^2}}}{{0.120{\rm{ m}}}}\\\\ = 12.92{\rm{ rad/}}{{\rm{s}}^2}\\\end{array}

(D)

Tension on left side of cord is calculated using

\begin{array}{c}\\{T_{\rm{B}}} = {m_{\rm{B}}}g + {m_{\rm{B}}}a\\\\ = {m_{\rm{B}}}\left( {g + a} \right)\\\end{array}

\begin{array}{c}\\{T_{\rm{B}}} = \left( {3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} + 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 34.08{\rm{ N}}\\\end{array}

(E)

Tension on right side of cord is calculated using

\begin{array}{c}\\{T_{\rm{A}}} = {m_{\rm{A}}}g - {m_{\rm{A}}}a\\\\ = {m_{\rm{A}}}\left( {g - a} \right)\\\end{array}

\begin{array}{c}\\{T_{\rm{A}}} = \left( {7{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} – 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 57.82{\rm{ N}}\\\end{array}

6 0
3 years ago
A stone is dropped from the top of a tower. What is its velocity after 3.0 seconds?
kap26 [50]

For purposes of completing our calculations, we're going to assume that
the experiment takes place on or near the surface of the Earth. 

The acceleration of gravity on Earth is about 9.8 m/s², directed toward the
center of the planet.  That means that the downward speed of a falling object
increases by 9.8 m/s for every second that it falls.

3 seconds after being dropped, a stone is falling at (3 x 9.8) = 29.4 m/s. 

That's the vertical component of its velocity.  The horizontal component is
the same as it was at the instant of the drop, provided there is no horizontal
force on the stone during its fall.

5 0
3 years ago
Explain why average velocity in one dimension can be positive or negative.
Softa [21]
Imagine an object is moving in one dimension on a number line, and for this we'll say that the numbers on the line are a metre apart. If the object moves from 2 m to 7 m, the change in position is 7-2=+5 metres. But if the object moves back from 7 m to 2 m, the change in position is 2-7=-5 metres. since velocity =  \frac{change in position}{time}, and time is always positive, velocity will be positive in one direction and negative in the other direction.
3 0
3 years ago
Read 2 more answers
You are driving on the highway, and you come to a steep downhill section. As you roll down the hill, you take your foot off the
Vinvika [58]

Answer:

Weight of the car, normal force, drag force

Explanation:

The forces acting on the car are:

  • The normal force which acts perpendicularly to the downhill plane
  • The weight of the car which acts vertically downwards
  • The drag force due to air resistance which acts in opposition to the motion of the car

Friction is ignored, so the force due to friction is assumed negligible

6 0
3 years ago
In 2014, about how far in meters would you have to travel on the surface of the Earth from the North Magnetic Pole to the Geogra
pogonyaev

Answer:

267.07 km

Explanation:

We have given the radius of the earth = 6378.1 km

In 2014 the difference between the magnetic north pole and geographical north pole is 2.40°

2.40°=2.40\times \frac{\pi }{180}=0.041866 radian

We know that linear distance is given by S=R\Theta =6378.1\times 0.041866=267.07km

So we have to travel 267.07 km in going from magnetic north pole to geographic north pole

3 0
3 years ago
Other questions:
  • Temperature is a measure of the average ____________ energy of an object's particles. light mechanical potential kinetic
    7·2 answers
  • If the moon were 2 times closer to earth than it is now, the gravitational force between earth and the moon would be
    12·1 answer
  • Which of the following depend upon the amount of matter in an object?
    10·1 answer
  • How many hydrogen atoms are there in the following compound: 4CHA (2 points)
    7·2 answers
  • If the force applied to an object is not greater than the starting friction, what will happen to the object? Which answer is cor
    12·1 answer
  • An empty paper cup is the same temperature as the air in the room. A student fills the cup with cold water. Which of the followi
    15·1 answer
  • While traveling on a dirt road, the bottom of a car hits a sharp rock anda small hole develops at the bottom of its gas tank. If
    11·1 answer
  • Yurem is pulling a wagon across the playground with a force of 10 N. He asks Elianna to help. She agrees and pushes the back of
    11·2 answers
  • What is the final speed of an object that starts from rest and accelerates uniformly at 3.0 meters per second2 over a distance o
    8·1 answer
  • 39. A dog runs on a waxed floor at an initial speed of 2 m/s. It slides to a stop with an
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!