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Mariulka [41]
3 years ago
7

A circuit has a resistance of 2.5 Ω and is powered by a 12.0 V battery.

Physics
1 answer:
Marizza181 [45]3 years ago
5 0

Just apply Ohm's Law:


I=V/R=12/2.5=4.8A

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What is the drawback to using superconductors?
Fittoniya [83]

Answer:

Option A

The cost of keeping the semiconductor below the critical temperature is unreasonable

Explanation:

First of all, we need to understand what superconductors are. Superconductors are special materials that conduct electrical current with almost zero resistance. This means that there is little or no need for a voltage source to be connected to them. As a matter of fact, once a superconductor is connected to a power supply, one can remove the power supply and the current will still flow.

However, most superconducts can only conduct at very low temperatures up to -200 degrees Celcius. This is because, at that temperature, their atoms and molecules are relatively settled, hence they pose little or no resistance to the flow of current.

This as you can guess is extremely difficult to do, as you will need a lot of effort to cool it to that temperature and maintain it.

This makes option a the answer:

The cost of keeping the semiconductor below the critical temperature is unreasonable.

7 0
3 years ago
What is the gravitational force between Mars and the sun? 7.43 × 1030 N 1.79 × 1026 N 1.65 × 1021 N 3.76 × 1032 N
VMariaS [17]

The gravitational force between Mars and the Sun is 1.65\cdot 10^{21} N

Explanation:

The magnitude of the gravitational force between two objects is given by  the equation:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we have:

m_1 = 1.99\cdot 10^{30} kg is the mass of the Sun

m_2 = 6.39\cdot 10^{23} kg is the mass of Mars

r=229\cdot 10^6 km = 229\cdot 10^9 m is the average distance Mars-Sun

Substituting into the equation, we find the gravitational force:

F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(6.39\cdot 10^{23})}{(229\cdot 10^9)^2}=1.62\cdot 10^{21} N

So, the closest answer is

1.65\cdot 10^{21} N

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

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4 0
4 years ago
In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the
Rufina [12.5K]

Answer:

The  angular speed is w = 5.89 \ rad/s

Explanation:

From the question we are told that

    The time taken is  t = 1.6 s

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         \theta  =  n *  2 *  \pi

substituting values

        \theta  =  1.5 *  2 *  3.142

       \theta  = 9.426 \ rad

 The angular speed is mathematically represented as

         w =  \frac{\theta }{t}

substituting values

         w =  \frac{9.426}{1.6}

          w = 5.89 \ rad/s

     

3 0
3 years ago
A meteoroid is traveling east through the atmosphere at 18. 3 km/s while descending at a rate of 11.5 km/s. What is its speed, i
Annette [7]

Answer:

The speed of meteoroid is 21.61 km/s in south-east.

Explanation:

Given that,

A meteoroid is traveling through the atmosphere at 18.3 km/s. while descending at a rate of 11.5 km/s it means 11.5 km/s in south.

We need to draw a diagram

Using Pythagorean theorem

AC^2=AB^2+BC^2

AC^2=(18.3)^3+(11.5)^2

AC=\sqrt{(18.3)^2+(11.5)^2}

AC=21.61\ km/s

Hence, The speed of meteoroid is 21.61 km/s in south-east.

6 0
3 years ago
an object has a mass of 50kg, a final height of 20m and an initial height of 8m. what is the amount of work done
Andrei [34K]

amount of work done is 5880 J

Given:

mass of object = 50kg

Final height = 20m

initial height = 8m

To Find:

amount of work done

Solution:

work is done when a force acts upon an object to cause a displacement. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

The work done by gravity is given by the formula,

W = mgh

W = 50 x 9.8 x ( 20-8)

= 5880 J

So the work done is 5880 J

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7 0
2 years ago
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