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ANEK [815]
3 years ago
7

This is Formation of the Solar System Lab Report

Physics
1 answer:
Ainat [17]3 years ago
3 0

Answer:

the answer is B it is right

Explanation:

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The circuit has a 3 volt EMF and two ohm resistors. How much power in watts does this circuit draw? A) 4.5 , B) 24, C) 1.13 D) 2
dlinn [17]

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance  of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

P=\dfrac{V^2}{R}

Substitute all the values,

P=\dfrac{3^2}{2}\\\\P=4.5\ W

So, the power of this circuit is equal to 4.5 watts.

5 0
3 years ago
Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
3 years ago
A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

V=30.1m/s

C_d=0.45

A=3.3m^2

\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

F_D=807.25N

We need calculate now the energy lost through a time T, then,

W = F_D d

But we know that d is equal to

d=vt

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0
3 years ago
Can any one give me a amazon code i will repay you. plz i am trying to buy my mom somin off amazon
Alexandra [31]

Answer:i have none but good luck also this is a learning site ;-;

Explanation:

4 0
3 years ago
A 2 kg basketball has a momentum of 4 kg m/s. What is the ball's velocity?
Mumz [18]
That is the answer to the question
I hope this helps you.
Thank you for your question

4 0
3 years ago
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