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2 years ago

What is the effective resistance of this dc circuit

Physics

1 answer:

Inessa05 [86]2 years ago

5 0

You've managed somehow to post the mirror image of the circuit diagram, including the numbers and values of the resistors. I'm curious to know how you did that.

The three resistors at the left end of the diagram are 3Ω , 2Ω , and 1Ω all in series. They behave like a single resistor of (3+2+1) = 6Ω .

That 6Ω resistor is in parallel with the 2Ω drawn vertically in the middle of the diagram. That combination acts like a single resistor of 1.5Ω in that position.

Finally, we have that 1.5Ω resistor in series with 1Ω and 4Ω . That series combination behaves like a single resistor of 6.5Ω across the battery V.

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A spacecraft left Earth to collect soil samples from Mars. Which statement is true about the strength of Earth’s gravity on the

ryzh [129]

Answer:

: It Decreases.

As the spacecraft gets farther and farther from Earth, the gravitational

forces between the spacecraft and the Earth decrease.

Explanation:

5 0

3 years ago

Which of the following represents the greatest energy transition from a higher energy level to a lower one?

prohojiy [21]

You need to use Planck's law:
E = h·υ = (h·c)/λ

Without making all the calculations, a fraction is bigger than another when the denominator is smaller. Therefore you need to find the smallest wavelength (λ) which is 450nm.

You could also be helped by colors: in order of decreasing energy, you have blue - green - yellow - red.

In any case, the correct answer is a).

8 0

3 years ago

Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia

Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

Wn = 9.14 x 10¹⁷ N

3 0

3 years ago

wo parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude

melisa1 [442]

Answer:

-18896.49 V/m

Explanation:

Distance between the two plates = 10 cm = 10 x $10^{-2}$ m = 0.1 m

Also, one of the plates is taken as zero volt.

a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V

b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V

Potential field strength = -dV/dx

where dV is voltage difference between these points,

dx is the difference in distance between these points

For the first case above,

potential field strength = -393/0.0705 = -5574.46 V/m

For the second case ,

potential field strength = -393/0.0295 = -13322.03 V/m

Magnitude of the field strength across the plates will be

-5574.46 + (-13322.03) = -5574.46 + 13322.03 = -18896.49 V/m

6 0

3 years ago

The quantity of charge through a conductor is modeled as Q = (3.00 mC/s4)t4 − (2.00 mC/s)t + 9.00 mC. What is the current (in A)

Mumz [18]

Answer:

The current at time t = 4.00 s is 0.766 A.

Explanation:

Given that,

The quantity of charge through a conductor is modeled as :

$Q=(3t^4-2t+9)\ mC$

We need to find the current (in A) at time t = 4.00 s. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(3t^4-2t+9)}{dt}\\\\I=12t^3-2$

At t = 4 s

$I=12(4)^3-2=766\ mC/s\\\\I=0.766\ C/s=0.766\ A$

So, the current at time t = 4.00 s is 0.766 A. Hence, this is the required solution.

7 0

2 years ago