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2 years ago

Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of

1 answer:

5 0

I don't know if you need to complete this question or do it otherwise, however, I managed to find on the Internet on several places this completion of your sentence:
<span>Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of q = 2 x 10</span>⁶ W/m³.
I'm not sure whether that is the answer you were looking for, but that's what I found.

You might be interested in

When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some poi

eimsori [14]

Answer: The force of kinetic friction is smaller than that of static friction, but F_g remains the same.

Explanation:

The situation is same as when a book is pushed with an increasing force on a table; When the force is low, book doesn't move, until that under a given force starts moving, and then it goes on movement even if the force decreases a bit.

The physical explanation for this, that friction force adopts any value needed to avoid to move the object, till a limit value is achieved, called static friction force, equal to the normal force times the static friction coefficient.

Once in movement, the kinetic friction coefficient replaces the static one , and in general is lower than the static one, so the force diminishes.

In the case of the box sliding down the board, the force that tries to move the object down the board, is the component of the weight parallel to the board, that can be showed that being equal to the weight times the sinus of the angle of the board with the horizontal, as follows:

F_g = m g sin θ

When θ increases, F_g does the same, so friction force always has the same magnitude than F_g (but opposite direction) so the box doesn't move, till that θ takes a value that produces a F_g equal to static friction force.

Beyond this limit angle, F_g (remaining the same for a given angle) is greater than the kinetic friction force, and the box slides.

In the limit, when θ=90º, sin θ =1⇒ F_g = mg, so the object is in free fall.

6 0

3 years ago

HELP URGENT WILL GIVE BRAINLIEST

frozen [14]

The answer for this question is A

3 0

2 years ago

A lawn roller in the form of a thin-walled, hollow cylinder with mass M is pulled horizontally with a constant horizontal force

olga nikolaevna [1]

Answer:

$a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}$

Explanation:

Given the mass as M, the rotational inertia of the mower is;

$I_{cm}=MR^2$

-The roller doesn't slip while rolling;

$v_{cm}=wR, a_{cm}=\alpha R$

$\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M} \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \ \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}$

6 0

3 years ago

In SI units, what is the magnitude the net force acting on a 1,152 kg car that accelerates uniformly along a straight line from

geniusboy [140]

Answer:

Fnet = 14515.2 Newton

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

$Fnet = Fapp + Fg$

Where;

Fnet is the net force.
Fapp is the applied force.
Fg is the force due to gravitation.

Given the following data;

Mass = 1,152 kg

Initial velocity, u = 3m/s

Final velocity, v = 17m/s

Time, t = 5 seconds

To find the magnitude of the net force;

First of all, we would determine the acceleration of the car.

Acceleration = (v-u)/t

Acceleration = (17 - 3)/5

Acceleration = 14/5

Acceleration = 2.8m/s

To find the applied force;

Fapp = mass * acceleration

Fapp = 1,152 * 2.8

Fapp = 3225.6 N

Next, we would find the force exerted on the car due to gravity.

Fg = mass * acceleration due to gravity

We know that acceleration due to gravity is equal to 9.8m/s²

Fg = 1152 * 9.8

Fg = 11289.6N

Substituting the values into the equation, we have;

Fnet = 3225.6 + 11289.6

Fnet = 14515.2 Newton

7 0

2 years ago

Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radiu

Gnesinka [82]

The orbiting velocity of the satellite is 4.2km/s.

To find the answer, we need to know about the orbital velocity of a satellite.

<h3>What's the expression of orbital velocity of a satellite?</h3>

Mathematically, orbital velocity= √(GM/r)
r = radius of the orbital, M = mass of earth

<h3>What's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?</h3>

M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶m
Orbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)

=4.2km/s

Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.

Learn more about the orbital velocity here:

brainly.com/question/22247460

#SPJ1

8 0

1 year ago