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Luden [163]
3 years ago
15

Why does a balloon that is rubbed on someoneâs shirt stick to a wall at a party?

Physics
1 answer:
alukav5142 [94]3 years ago
8 0
When we rub balloon on a shirt the balloon will steal electrons from the shirt and the shirt will become positively charged and balloon will negatively charged.<span>The reason that the balloon will stick to the wall is because the negative charges in the balloon will make the electrons in the wall move to the other side of their atoms and this leaves the surface of the wall positively charged.</span>
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An isolated system consists of a 1.5 kg mass moving in the presence of the following potential energy function: U open parenthes
pickupchik [31]

Answer: Period T = 4.44secs

Explanation: see attachment

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What state of matter has a define volume but no definite shape ?
mariarad [96]

Liquid has no definite volume or shape.

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3 years ago
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A lunar module weighs 12 metric tons on the surface of the Earth. How much work is done in propelling the module from the surfac
zhuklara [117]

Answer:

W=76.55 miles.metric tons

Explanation:

Given that

Weight on the earth = 12 tons

So weight on the moon =12/6 = 2 tons

 ( because at moon g will become g/6)

As we know that

F=\dfrac{K}{x^2}

Here x= 1100 miles

F 2 tons

2=\dfrac{K}{1100^2}

So

K=2.4\times 10^6

We know that

Work = F. dx

W=\int_{x_1}^{x_2}F.dx

W=\int_{1100}^{1140}\dfrac{2.4\times 10^6}{x^2}.dx

W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}

W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]

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6 0
3 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
A ball at the end of a string is swinging in a horizontal circle of radius 0.85 m. The ball makes exactly 3 revolutions per seco
Anettt [7]

Answer:

It's centripetal acceleration is 301.7 m/s²

Explanation:

The formula to be used here is that of the centripetal acceleration which is

ac = rω²

where ac is the centripetal acceleration = ?

ω is the angular velocity = 3 revolutions per second is to be converted to radian per second: 3 × 2π  = 3 × 2 × 3.14 = 18.84 rad/s

r is the radius = 0.85 m

ac = 0.85 × 18.84²

ac = 301.7 m/s²

It's centripetal acceleration is 301.7 m/s²

8 0
3 years ago
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