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slava [35]
3 years ago
11

Who would like to be my girlfiend ?

Physics
2 answers:
Sav [38]3 years ago
8 0

nobody this is a app to help others

Elan Coil [88]3 years ago
3 0

Answer:

I bet my great grandmother would

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Where could you hear a sonic boom
ExtremeBDS [4]
Answer

A sonic boom is a continuous effect that occurs while the object is travelling at supersonic speeds.
7 0
4 years ago
Could you use 6.75 when measuring the<br> pencil?<br> no<br> yes
valentinak56 [21]

Answer: yes u can use a 6.75 when measuring the pencil

3 0
3 years ago
Read 2 more answers
What is the magnitude of the momentum of a 33 g sparrow flying with a speed of 8.7 m/s?
WARRIOR [948]

Answer:

0.2871 kg m/s

Explanation:

p = mv

convert 33g into kg (0.033)

mulitply byt 8.7 to get 0.2871

4 0
3 years ago
A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used
zloy xaker [14]

Answer:

The compression is \sqrt{2} \  d.

Explanation:

A Hooke's law spring compressed has a potential energy

E_{potential} = \frac{1}{2} k (\Delta x)^2

where k is the spring constant and \Delta x the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

E_{kinetic} = \frac{1}{2} m v^2.

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity v_1. Knowing that the energy is constant.

\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2

But, in the left side we can use the previous equation to obtain:

2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2

D^2 =  \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k}

D^2 =  2 \  d^2

D =  \sqrt{2 \  d^2}

D =  \sqrt{2} \  d

And this is the compression we are looking for

3 0
4 years ago
Read 2 more answers
A proton of mass is released from rest just above the lower plate and reaches the top plate with speed . An electron of mass is
xenn [34]

Answer:

  v = √ 2e (V₂-V₁) / m

Explanation:

For this exercise we can use the conservation of the energy of the electron

At the highest point. Resting on the top plate

         Em₀ = U = -e V₁

At the lowest point. Just before touching the bottom plate

        Emf = K + U = ½ m v² - e V₂

Energy is conserved

         Em₀ = Emf

          -eV₁ = ½ m v² - e V₂

           v = √ 2e (V₂-V₁) / m

Where e is the charge of the electron, V₂-V₁ is the potential difference applied to the capacitor and m is the mass of the electron

3 0
4 years ago
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