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denpristay [2]
2 years ago
8

A electric heater that draws 13.5 a of dc current has been left on for 10 min. how many electrons that have passed through the h

eater during that time? (e = 1.60 Ã 10-19 c)
Physics
1 answer:
kicyunya [14]2 years ago
3 0
By definition, Ampere is a unit of current which is a measure of the amount of charge passing through a point in a circuit per unit  of time, with an equivalent charge of 1.602 x 10^(-19) Coulomb per electron. To determine the number of electrons passing through the heater, we use the definition of the current. We calculate as follows:

13.5 A = 13.5 C per second
Charge = 13.5 C/s (10 min) ( 60 s / 1 min)
Charge = 8100 C 

Number of electrons = 8100 C / 1.602 x 10^(-19) C per electron
Number of electrons = 5.1 x 10^22 electrons

Therefore, there are 5.1 x10^22 electrons that assed through the heater for 10 minutes.
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A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.0 g plastic bead, with a charg
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Answer:

Please find the answer in the explanation

Explanation:

Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.

Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.

We can therefore conclude that the upper plate, is positively charged

B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC

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A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p
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Answer:

a) v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s}  =3.2051 \frac{m}{s}

b) v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s}  =5.2083 \frac{m}{s}

c) a=\frac{v_{2}-v_{1}  }{t_{2}-t_{1}  } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }

Explanation:

<em><u>The knowable variables are </u></em>

d_{t}=25m

t_{1}=1.3 s

t_{2}=3.9 s

t_{3}=5.5 s

Since the three traffic signs are <u>equally spaced</u>, the <u>distance between each sign is \frac{25}{3} m</u>

a) v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s}  =3.2051 \frac{m}{s}

b) v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s}  =5.2083 \frac{m}{s}

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) a=\frac{v_{2}-v_{1}  }{t_{2}-t_{1}  } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }

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Answer:

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The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f
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Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

                 584 = Ha + Lg

an also this mean that torques acting on the body is balanced

         So,   0.410 Ha  = 0.840 Lg

    Making Lg the subject of formula in the equation above we

   Lg = 0.4881 Ha

 Considering the first equation and replacing Lg with this recent equation we have

                      584 = Ha + 0.4881 Ha

          Therefore Ha = 392.44 N

This value obtained is  for both hands for each hand we divide by 2

Therefore we have for each hand = 392.44/2 =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

             Lg = 0.4881 ×  392.44

                  = 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

                  191.22/2 = 95.8 N.

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