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Eva8 [605]
2 years ago
10

A balloon that has been rubbed on wool has a negative charge. When the balloon is held over—but not touching—some bits of paper,

some of the bits jump up to the balloon. Why does the paper move?
Physics
1 answer:
alexandr402 [8]2 years ago
5 0

Answer:

This process is known as static electricity

Explanation:

When balloon is rubbed on the wool , a negative charge is usually formed. When the balloon is then held over—but not touching—some bits of paper, some of the bits jump up to the balloon. The bits which jump up to the balloon are positive charged and we know that unlike charges attract.

This process explains static electricity which is when the charges on a material are not in a balanced state.

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Is this snow or water
fredd [130]
I think its snow. Yup.
4 0
2 years ago
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Certain neutron stars (extremely dense stars) are believed to be rotating at about 1000 rev/s. If such a star has a radius of 14
Sever21 [200]

Answer:

minimum mass of the neutron star = 1.624 × 10^30 kg

Explanation:

For  a material to remain on the surface of a rapidly rotating neuron star, the magnitude oĺf the gravitational acceleration on the material must be equal to the magnitude of the centripetal acceleration of the rotating neuron star.

This can be represented by the explanations in the attached document.

minimum mass of the neutron star = 1.624 × 10^30 kg

8 0
3 years ago
Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same
marissa [1.9K]

Answer:

maximum static frictional force of the wall on the book (Increasing)

normal force of the wall on the book (Decreasing)

weight of the book (Not changing)

Explanation:

Now according to Newton's third law of motion

"Every action has equal but opposite reaction"

By the data given in question, Albert was pressing the book against the wall.Now, Albert started to reduce his force up against the wall.

First we have to consider all the forces applied on book in this scenario.

1. Weight of book acting downwards (y-axis)

2. Friction between book and wall acting upward (y-axis)

3. Albert's force on book against wall (x-axis)

4. Normal reaction of wall against Albert's force (x-axis)

Now, when Albert reduced his force, new scenario will be

1. Weight will be remain constant as it is W = mg

Neither mass nor acceleration due to gravity changed, so weight acting upon the book will remain same.

2. When Albert reduced force, normal reaction of wall reduced against it according to Newton's third law of motion

3. Now notice that friction is a force which acts in accordance with the applied force. For example if a box is placed at floor, no friction is applied, but when you drag the box, friction starts to act and increases until its limit comes. So, when Albert reduced his force, weight will try to pull the book and maximum static friction will increase to hinder the movement of book downwards.

Notice that maximum static friction will hinder the book from movement, since Albert reduced his force, but wight didn't pull the book, which means that maximum static friction increased to hinder downward motion.

7 0
2 years ago
Magnet A has twice the magnetic field strength of magnet B and pulls on magnet B with a force of 100 N. The amount of force that
son4ous [18]

The force exerted by the magnetic in terms of the magnetic field is,

F\propto B

Where B is the magnetic fied strength and F is the force.

Thus, if the magnetic A has twice magnetic field strength than the magnet B,

Then,

B_A=2B_B

Thus, the force exerted by the magnet B is,

\begin{gathered} F_B\propto B_B \\ F_B\propto\frac{B_A}{2} \\ F_B=\frac{F_A}{2} \\ F_B=\frac{100}{2} \\ F_B=50\text{ N} \end{gathered}

Thus, the force exerted by the magnet B on magnet A is 50 N.

The force exerted by the magnet A exerts on the magnet B is exactly 100 N as given.

Hence, the option B is the correct answer.

3 0
8 months ago
A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the
const2013 [10]

Answer:

The equation for the object's displacement is u(t)=0.583cos11.35t

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in

The angular speed is:

w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s

The damping force is:

F_{D} =cu

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in

The critical damping is equal:

c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in

Like cc>c the system is undamped

The equilibrium expression is:

u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t

3 0
2 years ago
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