All categories

3 years ago

What is a scalar?

Physics

2 answers:

Tcecarenko [31]3 years ago

3 0

a quqntity that doea not indict a direction

lana66690 [7]3 years ago

3 0

Answer:C

Explanation:

A scale has magnitude without direction

You might be interested in

The distance between a charge and the source of an electric field changes from 3 mm to 6 mm. as a result of the change, the elec

STatiana [176]

The electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.

<h3>What is electric potential energy?</h3>

Electric potential energy can be defined as the energy needed to move a charge against an electric field.

It is calculated using the formula;

U = Kq1 q2 ÷ r

Where Q = electric potential energy

k = Coulombs constant

q1 and q2 = charges

r = distance of separation

Electric potential energy is inversely proportional to the distance of separation of the charges.

If the distance of the charges changes from 3mm to 6mm, then the electric potential energy of the charges is reduced because it decreases with increase in the distance of the charges.

Therefore, the electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.

Learn more about electric potential energy here:

brainly.com/question/14812976

#SPJ1

4 0

2 years ago

A battery is used to charge a parallel-plate capacitor, after which it is disconnected. Then the plates are pulled apart to twic

Igoryamba

Answer: C.

Explanation:

For a parallel-plate capacitor where the distance between the plates is d.

The capacitance is:

C = e*A/d

You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.

Now, the stored energy can be written as:

E = (1/2)*Q^2/C

Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:

E = (1/2)*Q^2*d/(e*A)

e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.

Then we can define:

K = (1/2)*Q^2/(e*A)

And now we can write the energy as:

E = K*d

Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.

8 0

3 years ago

PLEASE HELP!

Anna [14]

Answer:

Explanation:

At constant pressure , work done by gas = P x ΔV where P is pressure and ΔV is change in volume

ΔV = 9.2 - 5.6 = 3.6 L

3.6 L = 3.6 x 10⁻³ m³

ΔV = 3.6 x 10⁻³ m³

P = 3.7 x 10³ Pa

So work done

= 3.7 x 10³ x 3.6 x 10⁻³ J

= 13.32 J .

( c ) is the answer , because work is done by the gas so it will be positive.

5 0

3 years ago

Read 2 more answers

Which of the following cabinet departments protests us from terrorist attacks?

MariettaO [177]

Answer: A

Explanation:

5 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago