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AysviL [449]
3 years ago
14

The (nonconservative) force propelling a 1.50 × 103-kg car up a mountain road does 4.70 × 106 J of work on the car. The car star

ts from rest at sea level and has a speed of 27.0 m/s at an altitude of 2.00 × 102 m above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are non conservative forces
Physics
1 answer:
aliina [53]3 years ago
6 0

Answer:

be the mehcanic for once and see what the answer is

Explanation:

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In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs
bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

V=\frac{I}{nAq}

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

\frac{I}{q} = 1.2*10^{18}

A= 1.3*10^{-8}m^2

n=6.3*10^{28} e/m^3

\omicron{O} = 1.2*10^{-4}(m/s)(N/c) Mobility

We can find the drift velocity replacing,

V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}

V= 1.465*10^-3m/s

The electric field is given by,

E= \frac{V}{\omicron{O}}

E=\frac{1.465*10^-3}{1.2*10^{-4}}

E=12.2V/m

7 0
3 years ago
As a wave travels through a medium, it displaces particles in a direction parallel to the motion of the wave. We can conclude th
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We can conclude that it is a longitudinal wave because the wave is traveling through a medium displacing particles<span>
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In Linnean taxonomy, the most unique level is _____.<br><br> species<br> genus<br> kingdom<br> breed
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The answer is species, I hope this helps!
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3 years ago
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You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

F=\frac{kQq}{r^2}

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

#SPJ4

3 0
1 year ago
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