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anyanavicka [17]
3 years ago
8

The voltage across a resistor is found to be 1.5 V. It is also found that there is a charge of 2 Coulombs passing through the re

sistor within 10 seconds. What is the value of the resistor in ohms
Physics
1 answer:
djverab [1.8K]3 years ago
8 0

Answer:

R = 7.5 Ω

Explanation:

  • If the resistor is in the linear zone of operation, the resistance must obey Ohm's Law:

        V = I*R (1)

  • By definition, the current flowing through the resistor, is equal to the charge passing through the resistor, per unit time.
  • So, we can write the following expression for the current I:

        I =\frac{\Delta Q}{\Delta t} = \frac{2C}{10s}  = 0.2 A (2)

  • From (1) and (2) we can solve for R, as follows:

       R =\frac{V}{I} = \frac{1.5V}{0.2A} = 7.5 \Omega

  • The value of the resistor is 7.5 Ω.
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Answer:

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Explanation:

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Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

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In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

4 0
3 years ago
PLS SOMEONE HELP QUICK
SVETLANKA909090 [29]

Answer:

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Explanation:

6 0
2 years ago
Read 2 more answers
The microwaves in a certain microwave oven have a wavelength of 12.2 cm. How wide must this oven be so that it will contain five
leva [86]

Answer:

a

 l = 0.305 \  m

b

  f = 3.0*10^{11} \  Hz

Explanation:

From the question we are told that

  The  wavelength is  \lambda  =  12.2 \  cm  = 0.122 \  m

  The  number of antinodal planes of the electric field considered is n  =  5

The  width is mathematically represented as

       l  =  \frac{ n \lambda}{2}

       l = \frac{5 * 0.122 }{ 2}

      l = 0.305 \  m

Generally the  frequency the errors was made is  mathematically represented as

   f =  \frac{c}{\lamda_k}

Here c is the speed of light with value  c =  3.0*10^{8} \  m/s

     \lambda_k is the wavelength of the microwave has to be in order for there still to be five antinodal planes of the electric field along the width of the oven, which is mathematically represented as

     \lambda_k  =  \frac{ \lambda *  \frac{0.04}{2} }{n/2}

      \lambda_k  =  \frac{0.122*0.02}{5/2}

So

   f =  \frac{3.0*10^{8}}{0.000976}

    f = 3.0*10^{11} \  Hz

   

       

8 0
3 years ago
An electric vehicle starts from rest and accelerates at a rate of 2.4 m/s2 in a straight line until it reaches a speed of 27 m/s
DENIUS [597]
A) use v=u+at for both

First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.

b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.

First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.
6 0
3 years ago
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