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.
The work done is 400 J
Explanation:
The work done by you in pushing the box along the slope is given by

where
F is the magnitude of the force applied
d is the distance covered by the box along the slope
Here we have the following:
F = 200 N is the magnitude of the force applied
is the distance covered, where
h = 1 m is the vertical rise
is the slope of the plane
Substituting and solving, we find

Learn more about work:
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Answer:
mercury and alcohol
ii) used to test temperatures
By v = u - at
<span>=>8 = 12 - a x 0.25 </span>
<span>=>a = 4/0.25 km/hr/sec </span>
<span>=>a = 16km/hr/sec
I hope this helped!</span>
Answer:
Imp = 25 [kg*m/s]
v₂= 20 [m/s]
Explanation:
In order to solve these problems, we must use the principle of conservation of linear momentum or momentum.
1)

where:
m₁ = mass of the object = 5 [kg]
v₁ = initial velocity = 0 (initially at rest)
F = force = 5 [N]
t = time = 5 [s]
v₂ = velocity after the momentum [m/s]
![(5*0) +(5*5) = (m_{1}*v_{2}) = Imp\\Imp = 25 [kg*m/s]](https://tex.z-dn.net/?f=%285%2A0%29%20%2B%285%2A5%29%20%3D%20%28m_%7B1%7D%2Av_%7B2%7D%29%20%3D%20Imp%5C%5CImp%20%3D%2025%20%5Bkg%2Am%2Fs%5D)
2)
![(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})\\(0.075*0)+(30*0.05)=(0.075*v_{2})\\v_{2}=20 [m/s]](https://tex.z-dn.net/?f=%28m_%7B1%7D%2Av_%7B1%7D%29%2B%28F%2At%29%3D%28m_%7B1%7D%2Av_%7B2%7D%29%5C%5C%280.075%2A0%29%2B%2830%2A0.05%29%3D%280.075%2Av_%7B2%7D%29%5C%5Cv_%7B2%7D%3D20%20%5Bm%2Fs%5D)