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Pavlova-9 [17]
3 years ago
15

Witch of the following is an example of a pull

Physics
1 answer:
joja [24]3 years ago
8 0

To pull is defined as to make something move toward something else by tugging or dragging. An example of pull is hitching a trailer to a car and moving it down the street. An example of pull is someone bringing a door toward themselves to open it.

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You venture out on a cold winter morning to warm up your vehicle. You have layers of cotton/polyester blend clothes on and from
maxonik [38]

Answer:

A. There is a localization of positive charge near the door handle.

Explanation:

  • When on a cold morning a person wearing cotton/ polyester cloth walking on the carpet moves toward his car then due to friction between the feet and the carpet there are transfer of electrons from the carpet to our feet, and since our body is a good conductor of electricity the charges spread throughout on the surface of or body.
  • When the person brings his hands close to the neutral conducting door of the car it gets induced with equal intensity of opposite charge to our hands thus having a concentration of positive charges near to the hand on the car's door is developed as a result of polarization within the conductor.
3 0
3 years ago
How does work affect energy between objects so it can cause a change in the form of energy?
ivolga24 [154]
Im pretty sure the answer would be A
6 0
3 years ago
if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball
grigory [225]
It would exert the same back right?
3 0
3 years ago
A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6
Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6  = V/0.0001

V = 2.0 × 10^6  * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q =  1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

5 0
3 years ago
PLEASE PLEASE HELP!!!!!!
Whitepunk [10]

Answer:

the answer is B

Explanation:

The atomic mass of an atom is the sum of the protons plus neutrons it has.

7 0
3 years ago
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