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Lelu [443]
3 years ago
11

Rubbing two objects together may cause large number of electrons to be transferred from one object to the other. True or False

Physics
1 answer:
ahrayia [7]3 years ago
8 0

Answer:

Rubbing two objects together may cause large number of electrons to be transferred from one object to the other. True or False.

Answer is True.

Explanation:

When we Rub two different objects, the contact area increases or we can say friction increases. In response of rubbing two materials there is a transfer of electrons from one material to the other material. This causes one subject to become positively charged ( Electron loser) and the other object become negatively charged ( Electron gainer )  This is also called Triboelectric Charging.

Now If contact area of two different rubbing objects will be larger, Friction will be larger , the larger number of electrons would be transferred from one object to other.  

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Is gravity air resistance?
kolezko [41]

The engineering principles that this project is based upon are gravity and air resistance. Gravity and air resistance are fundamental concepts of many forms of engineering. After we demonstrate the physical principles of gravity and air resistance through various experiments, the children are to pair off into groups of 2 to apply these concepts to an activity. The demonstrations will include dropping a parachute, a small and large feather, and a maple seed and explaining how air resistance affects their movement.


Using only paper towels, paper, scotch tape, scissors and string as construction materials, the children are to brainstorm, design and create a contraption that will make a pencil or a piece of pasta, depending on their age, fall as slowly as possible. This activity is designed for children ages 6-16. The level of difficulty can be adapted with different materials and different falling objects. The older children will be given heavier objects and a limited amount of material.


By participating in this activity, the children will gain a better understanding of how gravity and air resistance affects falling objects. They will have to work cooperatively with team members in order to reach the goal. While each group tests their innovation, the others will compare their ideas to the one that is being tested and learn the strengths and weaknesses of different designs.


As part of a teachers demonstration our group constructed three parachutes with different characteristics out of nylon. One of the parachutes has one large hole in the center of it. Another parachute has several holes throughout. The third parachute has no holes in it. The idea is that the parachute with the most holes will fall the fastest, where as the one without any hole will fall the slowest, and the one with only one hole will fall between the other two. This will demonstrate how the surface area of a parachute causes the air resistance and makes objects fall slowly.


Our activity meets all of the specified requirements. The children are to solve a problem using physical principals that they have learned, thus engaging in an engineering activity. This activity appeals to all types of children; everyone loves to drop things. Components to the kit are safe, durable and do not take up much room. The activity description is very open-ended and so the children are free to design an original contraption anyway they please. The materials are all cheap and easily accessible at home and at school. Given the time it will take for children to brainstorm ideas, design, construct and test a device, it will take the full length of a class period. The whole kit will fit in a plastic storage tub easily because none of the objects in the kit are longer than one foot.

<span>Discovering Gravity & Air ResistanceBy Jason, Allison, and Dave</span><span> <span>If two objects, an elephant and a feather, are dropped off of the side of a building, which would experience a greater force of air resistance during the fall? Which would hit the ground first?<span>  Note: The above graphic requires Flash.</span>If two balls, one made of foam, and one made of metal, are dropped simultaneously from a tall building, which will fall the fastest?</span></span> <span> <span>What's included in the kit:(1) Teachers Manual(3) Demo Parachutes(1) Roll of Paper Towels(1) Pack of Pencils (24)(1) Spool of string(2) Rolls of Scotch Tape(10) Pairs of Safety Scissors(100) Sheets of Paper(1) Box of Paper Clips(1) Box of Pasta(2) Large Feathers(2) Small Feathers(2) Toy Parachutes   </span></span>

7 0
4 years ago
A pair of slits in a double slit experiment are illuminated with monochromatic light of wavelength 480 nm. The slits are separat
Romashka-Z-Leto [24]

Using the appropriate approximations:

dx/L = mλ

d = slit separation

x = fringe spacing

L = distance between slits and screen

m = some integer, used to determine the distance from the central bright fringe to another bright fringe


We don't really need a value for m because we're calculating the distance between any pair of consecutive fringes. Let's just set m = 1


Given values:

d = 1.0mm

L = 2.0m

λ = 480nm


Substitute the terms in the equation with our given values and solve for x:

1.0*10⁻³*x/2 = 480*10⁻9

<h3>x = 0.96mm</h3>
8 0
3 years ago
Help please an observer at the North Pole sees the moon phase shown below. What Moon phase will be observed approximately one we
ziro4ka [17]

Answer:

it should be a new moon

Explanation:

4 0
3 years ago
A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a
lina2011 [118]

Answer:

hinit = 17.5 m

Explanation:

  • Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.
  • At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:

       E_{o} = U_{o} = m*g*h_{init}  (1)

  • When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.
  • We can express this final energy as follows:

       E_{f} = U_{f} + K_{f}  = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2}  (2)

  • In order to find hinit, we need to make (1) equal to (2), and solve for it.
  • In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.
  • Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).
  • We know also that there is a force that keeps the car along the circular track, which is the centripetal force.
  • This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.
  • If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.
  • So, the centripetal force must be equal to the weight.
  • The centripetal force can be expressed as follows:

       F_{c} = m*\frac{v_{f} ^{2}}{R}  (3)

  • We have just said that (3) must be equal to the weight:

       F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)

  • Simplifying, and rearranging, we can solve for vf², as follows:

       v_{f}^{2} = R*g  (5)  

  • Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:

      h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)

7 0
3 years ago
Consider that a clay model of a tiger has a mass of 0.195 kg and glides on ice at a speed of 0.75 m/s. It hits another clay mode
juin [17]

Answer:

The final velocity after the collision is 0.27 m/s.    

Explanation:

Given that,

Mass of tiger, m = 0.195 kg

Initial speed of tiger model, v = 0.75 m/s

Mass of another clay model, m' = 0.335 kg

Initially, second model is at rest, v' = 0

We need to find the final velocity after the collision. It is a case of inelastic collision. Using the conservation of linear momentum as :

mv+m'v'=(m+m')V\\\\V=\dfrac{mv+m'v'}{(m+m')}\\\\V=\dfrac{0.195\times 0.75}{(0.195 +0.335 )}\\\\V=0.27\ m/s

So, the final velocity after the collision is 0.27 m/s.    

6 0
4 years ago
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