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harina [27]
3 years ago
12

If you could see the electric field coming off a proton you would notice that electric field lines are what?

Physics
1 answer:
Readme [11.4K]3 years ago
3 0
The electric field generated by a proton is a radial field, so we would see the field lines going radially from the proton toward every direction. I've attached a picture representing the electric field generated by a positive charge, such as a proton, and we can see the field lines going radially from the charge toward every direction. 

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Imagine that Kevin can instantly transport himself between Planet X and Planet Y. Which statement could be said about Kevin in t
Over [174]
What are the choices ? 

Without some directed choices, I'm, free to make up any
reasonable statement that could be said about Kevin in this
situation.  A few of them might be . . .

-- Kevin will have no trouble getting back in time for dinner.

-- Kevin will have no time to enjoy the scenery along the way.

-- Some simple Physics shows us that Kevin is out of his mind.
He can't really do that.

           -- Speed = (distance covered) / (time to cover the distance) .

If time to cover the distance is zero, then speed is huge (infinite).

           -- Kinetic energy = (1/2) (mass) (speed)² .

If speed is huge (infinite), then kinetic energy is huge squared (even more).
There is not enough energy in the galaxy to push Kevin to that kind of speed.

         -- Mass = (Kevin's rest-mass) / √(1 - v²/c²)

-- As soon as Kevin reaches light-speed, his mass becomes infinite.
-- It takes an infinite amount of energy to push him any faster.
-- If he succeeds somehow, his mass becomes imaginary.
-- At that point, he might as well turn around and go home ...
     if he ever reached Planet-Y, nobody could see him anyway.
8 0
2 years ago
Read 2 more answers
A sheet of metal is 2mm wide 10cm tall and 15cm long. it was 4g. what is the density? <br>​
Hoochie [10]

Answer:

Ro = 133 [kg/m³]

Explanation:

In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

Ro = m/V

where:

m = mass [kg]

V = volume [m³]

We will convert the units of length to meters and the mass to kilograms.

L = 15 [cm] = 0.15 [m]

t = 2 [mm] = 0.002 [m]

w = 10 [cm] = 0.1 [m]

Now we can find the volume.

V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]

And the mass m = 4 [gramm] = 0.004 [kg]

Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]

3 0
2 years ago
A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0
3 years ago
The following chart provides the pH of some common foods. How many would be considered basic?
Yuri [45]

The ones with pH greater than 7

Explanation:

A pH scale is a scale for expressing the level of acidity or alkalinity of aqueous solutions.

This scale is called a pH or pOH scale. The pH is the common one.

pH of a solution is the negative logarithm to base 10 o the hydrogen ion concentration of the solution.

  • The scale ranges from 1 through 14.
  • An acidic solution has a pH value less than 7. Neutral solutions have pH of 7 and basic solutions have pH greater than 7.

learn more:

acidity brainly.com/question/5121777

#learnwithBrainly

6 0
2 years ago
Problems - Show all work.
Fittoniya [83]

Answer:

21s

Explanation:

Given parameters;

Radius  = 10m

Speed or velocity  = 3m/s

Unknown:

Period  = ?

Solution:

To solve this problem, use the expression:

      v  = \frac{2\pi r}{T}  

r is the radius

T is the unknown

           Input the parameters and solve for T;

    3  = \frac{2 x \pi  x 10}{T}  

    62.84 = 3T

         T  = 21s

4 0
2 years ago
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