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Nataly_w [17]
2 years ago
11

Which temperature scale has no negative temperatures A. Celsius B. Joule C. Fahrenheit D. Kelvin

Physics
2 answers:
vodomira [7]2 years ago
7 0

Kelvinnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

Brilliant_brown [7]2 years ago
6 0
The Kelvin scale has no negatives on it.

Zero Kelvin is 'Absolute Zero', and nothing can get colder than that.
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Answer:

The net magnetic flux through any closed surface must always be zero.

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Um relógio de ponteiros funciona durante um mês. Qual é, em N.C, o número de voltas, aproximadamente, que o ponteiro dos minutos
Margaret [11]
I can’t understand I’m sorry but if you need help with math I can help you
7 0
3 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
2 years ago
PLEASE HELP ASAP!!
kirza4 [7]

Answer:

A. Distance over which the force is applied

Explanation:

As we know that in pulley system the mass of the car is balanced by the tension in the string

so here we will have

T = r \times F

so here in order to decrease the force needed to lift the car we have to increase Distance over which the force is applied

So here if we increase the distance over which force is applied then it will reduce the effort applied by us in this pulley system as the torque will be more if the distance is more.

7 0
3 years ago
Un libro del peso di 12 N è in equilibrio su un tavolo. Sapendo che il coefficiente di attrito statico vale 0,5, la forza di att
Tanzania [10]

Answer:

60

Explanation:

Translation -

A book weighing 12 N is balanced on a table. Knowing that the static friction coefficient is 0.5, how much is the friction force worth?

Friction force is

f = u * n

f = 0.5 * 12N

f = 60

4 0
3 years ago
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