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Dennis_Churaev [7]
3 years ago
7

Why do clothes often cling together after tumbling in a clothes dryer?

Physics
1 answer:
nadya68 [22]3 years ago
8 0

Answer:

Explained

Explanation:

The electrons get rubbed off some items onto others. This causes an excess or electron on one item and deficiency of electron on the other. Therefore an electrostatic force of attraction is produced and hence clothes often cling together after tumbling in a clothes dryer

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Conditional waves contain a crest, trough, wavelength, and amplitude<br> True or False?​
AveGali [126]

Answer:

True

Explanation:

8 0
3 years ago
Light from the sun reaches Earth in 8.3min. The speed of light is 3.00 x 10^8 m/s. How far is the Earth from the sun?
strojnjashka [21]
The answer is this, but i don't know how to simplify it. 3x^100000000<span />
5 0
3 years ago
Read 2 more answers
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
3 years ago
Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?
Komok [63]
Data:

E = 20 V
R_{1} = 65\Omega
R_{2} = 65\Omega
R_{3} = 65\Omega
R_{4} = 65\Omega

<span>Now that we have all the values ​​we need properly identified, simply calculate the equivalent total resistance of the circuit and the intensity of the total electric current using the Ohm's Law:

</span>R_{T} =  R_{1} + R_{2} + R_{3} + R_{4}
R_{T} = 65 + 65 + 65+ 65
R_{T} = 260\Omega

<span>Like this:
</span>
I_{T} =  \frac{E}{ R_{T} }

I_{T} = \frac{20}{ 260 }
I_{T} = 0,076923076...

\boxed{\boxed{I_{T} \approx 0,07A}} 
Answer:
<span>The intensity of the total electric current 
</span>\boxed{\boxed{I_{T} \approx 0,07A}} 

P.S:. Since the association is in series, the current of 0.07A is the same for all resistors.
4 0
3 years ago
Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10 m/s2)
Mars2501 [29]

Answer:

<u>300 J</u>

Explanation:

Given :

  1. Applied force = 30 N
  2. Mass = 2 kg
  3. Height = 10 m
  4. g = 10 m/s²

Work done by the force :

  • Work done = Force x Displacement
  • Work done = mg x h
  • Work done = 30 N x 10 m
  • Work done = <u>300 J</u>

<u></u>

<u><em>Note</em></u> :

  • What you have calculated is the work done by gravitational force on the object (that too, incorrectly)
  • But in the end, it asks for work done by the force of 30N
  • Hence, the given answer ~
3 0
2 years ago
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