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3 years ago

Electrical power companies sell electrical energy

Physics

1 answer:

mixer [17]3 years ago

4 0

Heat used by electric heater :

Q = m • c • ∆T

Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)

Q = 8.82 × 10⁶ J

Cost of electrical energy :

Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)

Cost = $ 0.3675

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Heat is best defined as

Kisachek [45]

C. The flow of energy from one point to another.

5 0

3 years ago

An infinite line of charge with charge density λ1 = -2 μc/cm is aligned with the y-axis as shown. 1) what is ex(p), the value of

Natasha2012 [34]

As we know by Gauss's law that

$\int E.dA = \frac{q}{\epsilon_0}$

so for line charge the gaussian surface is cylindrical in shape

so we will have

$E(2\pi RL) = \frac{\lambda L}{\epsilon_0}$

now by rearranging the terms

$E = \frac{\lambda}{2\pi \epsilon_0 R}$

so here we will have to find the x component of electric field so it is given by above equation

$E_x = \frac{\lambda}{2\pi \epsilon_0 x}$

here x = distance from the wire where we need to find electric field

4 0

3 years ago

A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.

LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

Mass of the person standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = $\frac{MR^2}{2}$

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia $(I_y_y )$ of the system with respect to the axis yy.

⇒ $I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g$

⇒ $I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2$

⇒ $I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2$

⇒ $I_y_y=289.088\ kg.m^2$

The moment of inertia of the system is 289.088 kg.m^2

7 0

3 years ago

An object (mass 12.5 kg) slides on a frictionless surface at 1.32 m/s. How much work must be done to bring the object to rest

mote1985 [20]

Answer:

10.89 J.

Explanation:

The following data were obtained from the question:

Mass (m) = 12.5 kg

Velocity (v) = 1.32 m/s

Work done =?

To obtain the workdone, we shall determine the kinetic energy of the object since work and energy has the same unit of measurement. This is illustrated below:

Mass (m) = 12.5 kg

Velocity (v) = 1.32 m/s

Kinetic energy (K.E) =?

K.E = ½mv²

K.E = ½ × 12.5 × 1.32²

K.E = 6.25 × 1.7424

K.E = 10.89 J

The kinetic energy of the object is 10.89 J. Hence, the workdone in bringing the object to rest is 10.89 J.

8 0

3 years ago

How long will it take a car to travel a distance of 1km if it has an average speed of 60kph

beks73 [17]

Answer:

1÷60 h

time equals distance upon speed

6 0

3 years ago