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zalisa [80]
3 years ago
5

Why do clothes often cling together after tumbling in a clothes dryer? 1. the dryer heat caused some of the fabrics to melt toge

ther. 2. gravitational force 3. air pressure 4. the clothes are electrically neutral. 5. electrical force?
Physics
2 answers:
nikdorinn [45]3 years ago
5 0
5 as one is positive due to lost of electrons from its surface and another negative due to gain of electrons from the other clothes
as such there is electrostatic forces of attraction
Oduvanchick [21]3 years ago
4 0
I think it is 3 or 2 hope i helped
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Which of the following units would need to be converted before being used for a calculation
Andreyy89

The units which would need to be converted before being used for a calculation are cm and 0 ks.

<h3>What is Unit?</h3>

This is referred to a standard which is used to make comparisons in the aspect of measurement.

The units cm and 0 ks aren't in their standard form which is m and s respectively.

Read more about Unit here brainly.com/question/4895463

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6 0
1 year ago
A thin, metallic spherical shell of radius 0.227 m has a total charge of 6.03 × 10 − 6 C placed on it.
KATRIN_1 [288]

Answer:

Explanation:

Given

radius r=0.227 m

Charge on surface Q=6.03\times 10^{-6} C

Point Charge inside sphere q=1.15\times 10^{-6} C

Electric Field at r=0.735 m

Treating Surface charge as Point charge and applying Gauss law

E_{total}A=\frac{q_{enclosed}}{\epsilon _0}

where A=surface area up to distance r

E_{total}=\frac{Q+q}{4\pi r^2}

E_{total}=\frac{6.03\times 10^{-6}+1.15\times 10^{-6}}{4\pi (0.735)^2\times 8.85\times 10^{-12}}

E_{total}=1.194\times 10^{5} N/C

3 0
3 years ago
The graph of an object's position over time is a horizontal line and y is not equal to 0. What must be true abou
frozen [14]

Answer:D: the velocity is zero

Explanation:

7 0
3 years ago
A ball bounces on the ground. How do the ball and the ground act on each other?(1 point)
liubo4ka [24]

Answer: A is your best answer.

Explanation:

It should be A because the when the ball bounces on the ground the ground will give it force to bounce again but also it wont go as high as it first did. Hope this helps:))

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3 years ago
Read 2 more answers
The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
sattari [20]

Answer:

 k = 11,564 N / m,   w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

 let's apply the equilibrium condition at this point

                 

Axis y

          W_{y} - Fr = 0

          Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

             sin 46 = W_{y} / W

             W_{y} = W sin 46

     

 we substitute

           mg sin 46 = k y

           k = mg / y sin 46

If the length of the bar is L

          sin 46 = y / L

           y = L sin46

 

we substitute

           k = mg / L sin 46 sin 46

           k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

           k = 1.18 9.8 / 1

           k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

          Em₀ = U = mgy

end point. Point at 30º

         Em_{f} = K -Ke = ½ I w² - ½ k y²

          em₀ = Em_{f}

          mgy = ½ I w² - ½ k y²

          w = √ (mgy + ½ ky²) 2 / I

the height by 30º

           sin 30 = y / L

           y = L sin 30

           y = 0.5 m

the moment of inertia of a bar that rotates at one end is

          I = ⅓ mL 2

          I = ½ 1.18 12

          I = 0.3933 kg m²

let's calculate

          w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

          w = 6.06 rad / s

7 0
3 years ago
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