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RoseWind [281]
2 years ago
9

What potential difference is required across an 32 -Ω resistor to cause 33.72 A to flow through it?​

Physics
1 answer:
Inessa05 [86]2 years ago
6 0

\text{Voltage,}~ V = IR =33.72\times 32 =  1079.04~ \text{volt}

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When an object slides across the floor, it is slowed down by friction and the floor surface gets warmer. what energy conversion
OverLord2011 [107]
The movement of the object is considered to be kinetic energy while the object getting warmer indicates that there is thermal (heat) energy formed.

Based on this, as the object slides across the floor, friction slows down this motion and the object becomes warmer as kinetic energy is converted into thermal energy.
6 0
3 years ago
Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
3 years ago
11. Velocity-time graph for the motion of an object in a straight path is a straight line parallel to the time axis
Vilka [71]

a) uniform velocity

b) zero or no acceleration

c) (see picture)

EXPLANATION:

(see picture)

8 0
2 years ago
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