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satela [25.4K]
3 years ago
7

A girl is swinging on a rope. Does the rope supporting the swing act on the girl at the very top of her swing?

Physics
2 answers:
gladu [14]3 years ago
4 0
No, I don’t think so because is a bad thing
kondaur [170]3 years ago
3 0
Yes yes yes yes yes yes yes yes
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What are two processes that result in rocks being broken down into smaller pieces ?
meriva
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1) A parked car starts to move with a constant velocity of 3 m/s, after moving
makkiz [27]

Answer:

velocity (v) =distance(s) ÷ time (t)

Explanation:

v=s÷t

s=v×t

s=3×3.4

s= 10.2m

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3 years ago
8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?
Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

q_1=q_2=+1C are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N

8 0
3 years ago
Name and state the rule to find out the direction of magnetic field inside and around the loop.
goldfiish [28.3K]

Answer:

The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of F.

Explanation:

5 0
3 years ago
Rank the wavelengths of the following quantum particles from the largest to the smallest. If any have equal wavelengths, display
goldfiish [28.3K]

The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)

  • De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
  • From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.

De Broglie's relationship is given by  \lambda=\frac{h}{mv} .

This can be written as \lambda=\frac{h}{p}  .....(1) where λ  is known as de Broglie wavelength and p is momentum , h = Plank’s constant .

As we know that mass of proton is greater than electron and photon .

(a)For photon , the momentum is given by p=\frac{E}{c}     ...(2)  where c is the speed is the speed of light .

Putting E = 3eV in equation (2) , we get

              p=\frac{3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^7Js/m

Putting this value of p in equation (1) , we get

  \lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-27}}\\\lambda=4.13\times10^{-7}

(b) As we know that  kinetic energy is given by

         K.E=\frac{1}{2} mv^2\\\\2K.E = mv^2\\2K.E\times m = m^2v^2\\2K.E\times m = (mv)^2\\2K.E\times m = p^2\\\\\sqrt{2K.E\times m }=p      ...(3)

Where mass of electron is 9.1\times10^-^3^1 kg .

Putting K.E = 3eV in equation (3) , we get

    \sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 9.31\times10^{-31} }\\p=\sqrt{89.376\times10^{-50}} \\p=9.45\times10^{-25}Js/m

Putting this value of p in equation (1) , we get

\lambda=\frac{6.62\times10^{-34}}{9.45\times10^{-25}}\\\lambda=0.7005\times10^{-9}

(c) Putting m=1.67\times10^{-27}kg and K.E = 3eV in equation (3) , we get

     \sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 1.67\times10^{-27} }\\p=\sqrt{16.032\times10^{-46}} \\p=4.003\times10^{-23}Js/m

Putting this value of p in equation (1) , we get

\lambda=\frac{6.62\times10^{-34}}{4.003\times10^{-23}}\\\lambda=1.65\times10^{-11}

(d) Putting E = 0.3eV in equation (2) , we get

   p=\frac{0.3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^8Js/m

Putting this value of p in equation (1) , we get

\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-28}}\\\lambda=4.13\times10^{-6}

(e) Putting p=3eV/c in equation (1) , we get

    \lambda=\frac{6.62\times10^{-34}\\\times3\times10^8}{3\times1.6\times10^{-19}}\\\lambda=4.13\times10^{-7}

On comparing the wavelength order should be (d) > (a) = (e) > (b) > (c) .

Learn more about de brogile here :

brainly.com/question/28165547

#SPJ4

6 0
2 years ago
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