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Helen [10]
3 years ago
8

How to ask a person out in a different way? LIke insted of do you want to go? Like i like this guy and i gave him a ig teddy bea

r for valenties day and he enjoyed and how do i ask him out by not lie embereessed way like a good way
Physics
2 answers:
almond37 [142]3 years ago
8 0
Well you can start by being friends and little by little start getting into things he's into, liking the stuff he's into.
Don't be shy around him.
Express yourself around him, let him get to know you without him realizing it.
Eventually he will feel "safe" to say things around you.
Then when time has passed say "I'm really into you, want to watch a movie sometime?"
If he takes too long to answer which is like 10 sec. forget it and say "If you don't want to, we don't have to. mabey some other time, i guess."
Next time if HE asks YOU out try not to seem too availble, tell him "Let me think about it."
Dont make yourself seem tooo desperate.
Hope this helps:)
Go Get Him!
@SamSamySamantha
Greeley [361]3 years ago
6 0
You should come off like would you like to hang out sometime 
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3 years ago
In which direction does a bag at rest move when a force of 20 newton's is applied from the right?
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8 0
3 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
Andreas93 [3]

Answer:

v=1.08\times 10^7\ m/s

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, \dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2

Separation between the plates, d=1.75\times 10^{-2}\ m

An electron is released from rest, u = 0

Using equation of kinematics,

v^2-u^2=2ad..........(1)

Acceleration of the electron in electric field, a=\dfrac{qE}{m}............(2)

Electric field, E=\dfrac{\sigma}{\epsilon_o}............(3)

From equation (1), (2) and (3) :

v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}

v = 10840393.1799 m/s

or

v=1.08\times 10^7\ m/s

So, the electron is moving with a speed of 1.08\times 10^7\ m/s before it reaches the positive plate. Hence, this is the required solution.

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Explanation:

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