Answer:
The process by which the balloon is attracted and possibly sticks to the wall is known as static electricity which is the attraction or repulsion between electric charges which are not free to move.
The wall is an insulator.
Explanation:
When a balloon is blown and tied off, and then the balloon is rubbed on the woolly object once in one direction, and the side that was rubbed against the wool is brought near a wall and then released, it is observed that the balloon is attracted to and sticks to the wall. The above observation is due to static electricity.
Static electricity refers to electric charges that are not free to move or that are static. One of the means of generating such charges is by friction. When the balloon is rubbed on the woollen material, electrons are given away to the balloon's surface. Since the balloon is an insulator (materials which do not allow electricity to pass through them easily), the electrons are not free to move. When the balloon is brought near to a wall, there is a rearrangement of the charges present on the wall. Negative charges on the wall move farther away while the positive charges on the wall are attracted to the electrons on the balloon's surface. Because the wall is also an insulator, the charges are not discharged immediately. Therefore, this attraction between opposite charges as well as the static nature of the charges results in the balloon sticking to the wall.
Answer:
She run for, t = 0.92 s
Explanation:
Given data,
The velocity of the runner, v = 10 km/h
The distance covered by the runner, d = 9.2 km
The relationship between the velocity, displacement and time is given by the formula,
t = d / v
Substituting the given values in the above equation,
t = 9.2 / 10
= 0.92 s
Hence, she ran for, t = 0.92 s
The energy of a light wave is calculated using the formula
E = hc/λ
h is the Planck's constant
c is the speed of light
λ is the wavelength
For the ir-c, the range is
<span>6.63 x 10^-34 (3x10^8) / 3000 = 6.63 x 10 ^-29 J
</span>6.63 x 10^-34 (3x10^8) / 1000000 = 1.99 x 10^-31 J
For the ir-a, the range is
6.63 x 10^-34 (3x10^8) / 700 = 2.84 x 10^-28 J
6.63 x 10^-34 (3x10^8) / 1400 = 1.42 x 10^-28 J
Answer:
2.2 meters
Explanation:
Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:
is Coulomb's constant.
#The electric field,
at radius r is expressed as:
From i and ii, we have:
#Substitute actual values in our equation:
Hence, the distance between the charge and the source of the electric field is 2.2 meters
Answer:
: It Decreases.
As the spacecraft gets farther and farther from Earth, the gravitational
forces between the spacecraft and the Earth decrease.
Explanation:
