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Mekhanik [1.2K]
3 years ago
14

If a battery-operated car has a resistance of 3 Ohms and the current moving through it is 2 amps, what is the voltage of the bat

tery?
A) 6V
B) 1.5 V
C) 1 V
D) 0.67 V
Physics
2 answers:
aivan3 [116]3 years ago
6 0

Answer:

A) 6V

Explanation:

To solve this problem we need to use ohm's law

We have the resistance

Resistance=R=3ohms

and the current:

Current=I=2A

To find the voltage we use the following formula from ohm's law:

Voltage=(Current)(Resistance)

so

V=I*R

substituting the known values:

V=(2A)(3ohms)=6volts

the voltage of the battery is 6V which is option A.

Leno4ka [110]3 years ago
5 0

Answer:

For me I think it is A

Explanation:

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8 0
3 years ago
Read 2 more answers
So, RCF, or relative centrifugal force should be equal to RCF = <img src="https://tex.z-dn.net/?f=11%2C18%2Ar%2A%28RPM%2F1000%29
zysi [14]

After plugging all the data into the equation, the result of the relative centrifugal force (RCF)  is measured in terms of g.

<h3>What is relative centrifugal force?</h3>

The relative centrifugal force (RCF) or the g force is the radial force generated by the spinning rotor as expressed relative to the earth's gravitational force.

RCF = ac/g

where;

  • ac is centripetal acceleration
  • g is acceleration due to gravity

RCF = \frac{\omega ^2 r}{g} = 1.118\times 10^{-5} \ (RPM)^2 r = 11.18r\ (RPM/1000)^2

where;

  • r is radius in cm

<h3>For example, </h3>

Find the maximum RCF of the JS-4.2 rotor can be obtained from its maximum speed (4200 rpm) and its rmax (250 mm);

RCF = 11.18 \times 25\ cm \times (\frac{4200 \ RPM}{1000} )^2 = 4,930.3 \times g

Thus, after plugging all the data into the equation, the result is measured in terms of g.

Learn more about relative centrifugal force here: brainly.com/question/26887699

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6 0
2 years ago
TRUE or FALSE: During a collision between objects of different mass, there is a greater force applied to the less massive object
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If they are moving at the same speed then it would be true because the larger object would have more mass and would have more momentum then the smaller object that has less mass! Hope this helps!
4 0
3 years ago
A 2.4 kg block is dropped onto a spring and platform of negligible mass. The block is released
statuscvo [17]

The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

  • <em>mass of the block, m = 2.4 kg</em>
  • <em>height of the block, h =  5 m</em>
  • <em>compression of the spring, x = 25 cm = 0.25 m</em>

The spring constant is calculated as follows;

F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2  - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2  = mgh -   \frac{1}{2} kx^2\\\\mv^2   = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

Learn more here:brainly.com/question/14289286

8 0
3 years ago
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