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2 years ago

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An electric motor consumes 9.00 kj of electrical energy in 1.00 min. if one-third of this energy goes into heat and other forms

of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2500 rpm?

1 answer:

liq [111]2 years ago

5 0

How do I make a question help PLZZZ

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3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in

Eduardwww [97]

Explanation:

3

i believe that they are all going at 3.2 meters each, I did 4 times 0.8

4 0

2 years ago

Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the

Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0

2 years ago

A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, t

LenKa [72]

Answer:

a) W = - 318.26 J, b) W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

sin θ = Wₓ / W

Wₓ = W sin 60

cos θ = Wy / W

Wy = W cos 60

X axis

How the body is going at constant speed

fr - Wₓ = 0

fr = mg sin 60

fr = 15 9.8 sin 60

fr = 127.31 N

Y Axis

N - Wy = 0

N = mg cos 60

N = 15 9.8 cos 60

N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

W = mg L cos θ

Where the angles are between the weight and the displacement is

θ = 60 + 90 = 150

W = 15 9.8 2.50 cos 150

W = - 318.26 J

b) The work of the normal force

From Newton's equations

N = Wy = W cos 60

N = mg cos 60

W = N L cos 90

W = 0

c) The work of the friction force

W = fr L cos 0

W = 127.31 2.50

W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

W = F L cos 0

W = 127.31 2.50

W = 318.275 J

e) the net force

F ’= fr - Wx = 0

W = F ’L cos 0

W = 0

4 0

2 years ago

Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount

Mariulka [41]

Answer:

The value is $\Delta s = 8.537 *10^{25 } \ J/K$

Explanation:

From the we are told that

The radius of the sphere is $r = 6.32 *10^{8} \ m$

The temperature is $T_x = 5350 \ K$

The average temperature of the rest of the universe is $T_r = 2.73 \ K$

Generally the change in entropy of the entire universe per second is mathematically represented as

$\Delta s = s_r - s_x$

Here $s_r$ is the entropy of the rest of the universe which is mathematically represented as

$s_r = \frac{Q}{T_r}$

Here Q is the quantity of heat radiated by the star which is mathematically represented as

$Q = 4 \pi * r^2 * \sigma * T^4_x$

Here $\sigma$ is the Stefan-Boltzmann constant with value

$\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.$

=> $Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4$

=> $Q = 2.332 *10^{26} \ J$

So

$s_r = \frac{2.332 *10^{26}}{2.73}$

=> $s_r = 8.5415 *10^{25}\ J/K$

Here $s_x$ is the entropy of the rest of the universe which is mathematically represented as

$s_x = \frac{Q}{T_x}$

=> $s_x = \frac{2.332 *10^{26} }{5350}$

=> $s_x = 4.359 *10^{22} \ J/K$

So

$\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}$

=> $\Delta s = 8.537 *10^{25 } \ J/K$

7 0

2 years ago

The lightest duty and most widely used non flexible metal conduit is

Alenkinab [10]

Aluminium, or heavier version copper.

6 0

1 year ago