All categories

3 years ago

Which one of the following is described fully by its magnitude?

Physics

1 answer:

kompoz [17]3 years ago

5 0

Energy (C) is fully described by its magnitude.
There's nothing else about it to measure.

The other four things on the list are all vectors.
They need directions to go along with their magnitudes.

You might be interested in

If a ski lift raises 150 passengers averaging 590 N in weight to a height of 244 m in 96.3 s, at constant speed, what average po

SVEN [57.7K]

Answer:

224236.76 W

Explanation:

Power: This can be defined as the rate of change of energy. The S.I unit of power is Watt(W).

From the question,

P = E/t ......................... Equation 1

Where P = power, E = Energy or work, t = time.

But,

E = Wt×d.................. Equation 2

Where Wt = Total weight of the passengers, d = distance.

Substitute equation 2 into equation 1

P = (Wt×d)/t.............. Equation 3

Given: Wt = 590×150 =88500 N, d = 244 m, t = 96.3 s.

Substitute into equation 3

P = (88500×244)/96.3

P = 224236.76 W

3 0

3 years ago

Read 2 more answers

The part of the circuit that converts electrical energy into other forms

konstantin123 [22]

Answer:

Load

Explanation:

The load in an electric circuit is any device that converts electrical energy into another form of energy.

5 0

2 years ago

A convex lens of focal length 15 cm produces a magnification of +4. The

BartSMP [9]

$\sf \huge \pink {Answer : - }$

Focal length = 15cm

Magnification = +4

Type of lens : convex

We have to find distance of object..

★ Lateral Magnification :

m = v/u

v denotes distance of image
u denotes distance of object

4 = v/u

v = 4u

★ Lens Formula :-

→ 1/v - 1/u = 1/f

Focal length of convex lens is taken +ve and that for concave lens is taken -ve.

→ 1/4u - 1/u = 1/15

→ -3/4u = 1/15

→ u = -45/4

→ u = -11.25 cm

∴ (D) is the correct answer

4 0

2 years ago

Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i

Kobotan [32]

The magnitude of electrical force on charge $q_{3}$ due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector position of each particle respect to origin are described below:

$\vec r_{1} = (-0.500, 0)\,[m]$

$\vec r_{2} = (+0.500, 0)\,[m]$

$\vec r_{3} = (0, +0.500)\,[m]$

Then, distances of the former two particles particles respect to the latter one are found now:

$\vec r_{13} = (+0.500, +0.500)\,[m]$

$r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{13} =\frac{\sqrt{2}}{2}\,m$

$\vec r_{23} = (-0.500, +0.500)\,[m]$

$r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{23} =\frac{\sqrt{2}}{2}\,m$

The resultant force is found by Coulomb's law and principle of superposition:

$\vec R = \vec F_{13}+\vec F_{23}$ (1)

Please notice that particles with charges of same sign attract each other and particles with charges of opposite sign repeal each other.

$\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13} +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23}$ (2)

Where:

$k$ - Electrostatic constant, in newton-square meters per square Coulomb.
$q_{1}$ , $q_{2}$ , $q_{3}$ - Electric charges, in Coulombs.
$r_{13}$ , $r_{23}$ - Distances between particles, in meters.
$\vec u_{13}$ , $\vec u_{23}$ - Unit vectors, no unit.

If we know that $k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{1} = 2\times 10^{-6}\,C$ , $q_{2} = 2\times 10^{-6}\,C$ , $q_{3} = 2\times 10^{-6}\,C$ , $r_{13} =\frac{\sqrt{2}}{2}\,m$ , $r_{23} =\frac{\sqrt{2}}{2}\,m$ , $\vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} \right)$ and $\vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$ , then the vector force on charge $q_{3}$ is:

$\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$