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postnew [5]
3 years ago
5

what is the resistance of a clock if it has a current of 0.30 a and runs on a 9.0-v battery? 0.033 2.7 9.3 30

Physics
2 answers:
OleMash [197]3 years ago
6 0

Answer : Resistance of a clock is  30\ \Omega.

Explanation :

It is given that :

The current flowing in the clock, I = 0.3 A

Potential difference, V = 9 V

According to Ohm's law :

V=I\times R

So,

R=\dfrac{V}{I}

R=\dfrac{9\ V}{0.3\ A}

R=30\ \Omega

\Omega is the SI unit of the resistance given by a physicist Georg Simon Ohm.

So, the correct option is (d) i.e. 30\ \Omega

Hence, this is the required solution.

Nataliya [291]3 years ago
5 0
R = V/I =  9 / 0.3

R = 30 ohms.
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30
pickupchik [31]

Answer:

22m/s

Explanation:

lowest part on the graph (closest to x-axis)

4 0
2 years ago
A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the
blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

6 0
2 years ago
A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter
yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

  • Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

        I_{1} = \frac{V}{R_{int} +r_{L} }

  • where Rint = r and RL = r
  • Replacing these values in I₁, we have:

       I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)

  • When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

       I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r}  (2)

  • We can find the relationship between I₂, and I₁, dividing both sides, as follows:

        \frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}

  • The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.  
7 0
3 years ago
An artist wants to create a metal sculpture using a mold so that his artwork can be readily mass produced. He wants his sculptur
lukranit [14]

Answer:NO

Explanation:

No the mold should not be of the same size as that of sculpture because the material from which molds is made may shrink or expand depending upon its properties .

For example grey cast iron shrinks on cooling.

We need to make mold bigger in general so that if there is a need of finishing it can be done easily without altering the size of sculpture.

5 0
3 years ago
Which transition represents a time when water molecules are moving closer together?
slava [35]
A. Is very attractive. If it's sublimation directly from water vapor in the air to ice on the glass, then yes. But from liquid water mist to water ice, no. Ice is less dense than water. That's why cubes float in your soda. Better leave 'A' alone. . . . D. Ice pellets turn to liquid. That one's good.
8 0
3 years ago
Read 2 more answers
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