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My name is Ann [436]
3 years ago
6

??????????????????????????????

Physics
2 answers:
Svet_ta [14]3 years ago
7 0

Answer:

........

Explanation:

Tanzania [10]3 years ago
4 0

Answer:

A) option charging by induction

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Is question II correct?
d1i1m1o1n [39]

Answer:

it most likly right I'm not 100% sure

7 0
3 years ago
Need help transforming one formula to another!
Sindrei [870]

Hello! I don't know if this will help you any but I worked it out, got nothing on it. So, I just had my twin help me and he had nothing either. But, this is just how to find the transformation. Seems like this is already complete to me.

To find the transformation, compare the equation to the parent function and check to see if there is a horizontal or vertical shift, reflection about the x-axis or y-axis, and if there is a vertical stretch.

Parent function : y=√a

Horizontal shift : None

Vertical shift : none

Reflection about the x-axis : none

Vertical stretch : Stretched.

I am so sorry if this doesn't help you but to me, in both my eyes and my brother's eyes, this already looks complete. I hope this helps you out. Again, so sorry if it doesn't.

-Karleif ☺

5 0
4 years ago
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma
andreyandreev [35.5K]

Answer:

Magnitude of the third piece of the vessel after collision is (v√2) m/s

Explanation:

According To Newton's 2nd law, in such a system, momentum is conserved

Momentum before explosion = Momentum after explosion

Total mass of the vessel before explosion = m + m + 3m = 5m

Velocity = (0î + 0j) m/s since it was at rest.

Momentum = 5m (0î + 0j)

Momentum of the first piece after the explosion = m(-vî)

Momentum of the 2nd piece after explosion = m(-vj)

For the momentum of the 3rd piece,

Mass = 3m, let the velocity be (Vₓî + Vᵧj) m/s

Momentum of 3rd piece after collision = 3m (Vₓî + Vᵧj)

Momentum before explosion = Momentum after explosion

5m (0î + 0j) = m(-vî) + m(-vj) + 3m (Vₓî + Vᵧj)

The term m can be factories out,

Grouping the x and y components together,

0î + 0j = (-vî + Vₓî) + (-vj + Vᵧj)

-v + Vₓ = 0, Vₓ = v

-v + Vᵧ = 0, Vᵧ = v

So, the velocity of the 3rd fragment,

(Vₓî + Vᵧj) = (vî + vj) m/s

Magnitude = √(v² + v²) = √(2v²) = v√2 m/s

6 0
4 years ago
Ay which labeled points would the skateboarder have most kinetic energy?
jeka94

Answer: wheres the picture and it could be where they move the fastest

Explanation:

8 0
2 years ago
Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds Vi. Particle m is
nignag [31]

Answer:

v1 = \sqrt{(2)}Vi

v2 = \sqrt{(2/3)}Vi

ANGLE is 35.3 degree celcius

Explanation:

Given data:

mass m and 3m

initial speed Vi

particle with mass m is moving toward left while particle with mass 3m is moving toward right

By using conservation of momentum :

mVi + 3m(-Vi) = mv1 +3mv2

-2mVi = m(v1 + 3v2)

-2Vi = v1 + 3v2

conservation of energy :

m(Vi^2) + 3m(-Vi^2) = mv1^2 + 3mv2^2

4mVi^2 = m(v1^2+3v2^2)

4Vi^2 = v1^2+3v2^2

After collision, particle with mass m moves at right angles, thus by considering conservation of momentum in x & y direction,

x direction : -2mVi = 3m.v2i

-2Vi = 3v2i

y direction : 0 = m(v1)j+3m(v2)j

-v1j = 3v2j

subsitute these value in energy conservation

v1 = \sqrt{(2)}Vi

v2 = \sqrt{(2/3)}Vi

angle = tan^{-1}(\frac{\sqrt{(2)}}{2}) =35.3 degree from x-axis

8 0
3 years ago
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