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Alexandra [31]
3 years ago
12

Which diagram represents a closed circuit with the least resistance?

Physics
2 answers:
KatRina [158]3 years ago
6 0

Answer:

Circuit B is right.

Explanation:

A P E X

solniwko [45]3 years ago
4 0

Answer:Circuit B is the answer

Explanation:

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What is the power of a crane that does 5.60 kJ of work in 2.80 s?
eimsori [14]

Answer:

I don't understand what is a power cane??

6 0
3 years ago
An energy plant produces an output potential of 1500 kV and serves a city 143 km away. A high-voltage transmission line carries
jekas [21]

Answer:

2123.55 $/hr

Explanation:

Given parameters are:

V_{plant} = 1500 KV

L = 143 km

I = 500 A

\rho = 2.4 \Omega / km

So, we will find the voltage potential provided for the city as:

V_{wire} =IR = I\rho L = 1500*2.4*143 = 514.8 kV

V_{city} = V_{plant}- V_{wire} = 1500-514.8 = 985.2 kV

Then, we will find dissipated power because of the resistive loss on the transmission line as:

P = I^2R = I^2\rho L=500^2*2.4*143 = 8.58*10^7 W

Since the charge of plant is not given for electric energy, let's assume it randomly as x =  \frac{\dollar 0.081}{kW.hr}

Then, we will find the price of energy transmitted to the city as:

Cost = P * x = 8.58*10^7 * 0.081 * 0.001 = 6949.8 $/hr

To calculate money per hour saved by increasing the electric potential of the power plant:

Finally,

I_{new} = P/V_{new} = I/1.2\\P_{new} = I_{new}^2R_{wire}\\Cost = P_{new}/1.44=6949.8/1.44 = 4826.25 $/hr

The amount of money saved per hour = 6949.8 - 6949.8/1.44 = 2123.55 $/hr

Note: For different value of the price of energy, it just can be substituted in the equations above, and proper result can be found accordingly.

3 0
3 years ago
How much energy is needed to raise the temperature of a 55g sample of aluminum from 22.4 to 94.6?
BaLLatris [955]
The heat needed is given by Mcθ , where m is the mass in Kg, c is the heat capacity of aluminium, and θ is the change in temperature. 
Specific heat capacity of aluminium is 0.9 j/g°c
thus; Heat = 55 × 0.9 × 72.2
                 = 3573.9 Joules or 3.574 kJ
3 0
4 years ago
An object, initially at rest moves 250m in 17s. What is it's acceleration?
Gekata [30.6K]
With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity

d= 250s
t= 17s

a= d/t
 =\frac{250}{17}
 = 4.7 \frac{m}{s}
8 0
3 years ago
59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determin
Daniel [21]

Explanation:

a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:

x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)

y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)

From Eqn(2), we see that

N = mg\cos 25°\;\;\;\;\;\;\;(3)

so using Eqn(3) on Eqn(1), we get

mg\sin 25° - \mu_kmg\cos 25° = ma

Solving for the acceleration, we see that

a = g(\sin 25° - \mu_k\cos 25°)

\;\;\;\;= 2.45\:\text{m/s}^2

b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation

v^2 = v_0^2 + 2ax

Since the crate started from rest, v_0 = 0. Thus our equation reduces to

v^2 = 2ax \Rightarrow v = \sqrt{2ax}

v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}

\;\;\;\;= 6.32\:\text{m/s}

6 0
3 years ago
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