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motikmotik
2 years ago
6

Plss help with my science

Physics
1 answer:
Neko [114]2 years ago
6 0

Answer:

rotates, axis,not sure,day, not sure, night, revolves, not sure, hemisphere, towards, summer, away, winter

Explanation:

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An uncharged, nonconducting, hollow sphere of radius 10.0cm surrounds a 10.0-μC charge located at the origin of a Cartesian coor
MrMuchimi

The electric flux through the hole is 56.45\ webber .

  • Electric flux is the number of electric field lines cutting through the surface and is measured as surface intregal of electric field over that surface
  • Mathematically it is given by \phi_E=E.A \ Nm^2/C where E is the electric field and A is the area.
  • Gauss's law states that electric flux through closed surface is equal to the 1 / ε₀ times the charge enclosed by that surface which is given by  Ф = q / ε₀ where q is the central charge and ε₀ is the permittivity of the medium.

It is given , hollow sphere of radius 10.0cm surrounds a 10.0-μC charge.

The whole surface of hollow sphere = 4\pi r^2

                                                            = 4\times 3.14\times  (10 \times  10^{-2})^2 \\\\= 12.56\times 10^{-2} m^2

Area of the hole ( both side ) = 2\times \pi  r^2

                                               = 2\times 3.14 \times  (10^-^3)^2\\= 6.28 \times 10^-^6 m^2

According to Gauss's theorem, the flow from a particular charge in the center is given by

 \phi=  \frac{10\times10^-^6}{8.85\times 10^-^1^2}\\\\\phi=1.13\times10^6

This flux flows through the surface of the sphere, so the flux  per unit area which is given by

= \frac{ 1.13\times 10^6 }{ 12.56\times 10^-^2} \\\\= 8.99 \times 10^6 \  weber / m^2

Flux through area of hole is given by :

=  8.99\times10^6 \times6.28 \times 10^-^6\\ = 56.45 \ weber

Learn about more electric flux here :

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8 0
1 year ago
An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of th
Art [367]
The answer is :  We’ll see the bell move, but we won’t hear it ring.  This is    because light can travel through vacuum but sound cannot.  Sound waves are vibrations of particles in any media, so sound requires a medium to travel, and it cannot travel in a vacuum as there is no particles to vibrate.  
5 0
3 years ago
Read 2 more answers
Two cars collide and come to a complete stop. where did all of their energy go?
MA_775_DIABLO [31]
Most of the energy will be absorbed by the materials that make up the cars, causing them to deform. The energy will also be converted into sound energy, causing a loud bang upon collision. Also, some energy will be converted to thermal energy, which will cause the cars to heat up slightly.
3 0
3 years ago
it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J

Therefore, the additional work needed is

\Delta E=E'-E=810-90=720 J

Learn more about work:

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7 0
3 years ago
A spring with force constant of 59 N/m is compressed by 1.3 cm in a hockey game machine. The compressed spring is used to accele
Furkat [3]

Answer:

The puck moves a vertical height of 2.6 cm before stopping

Explanation:

As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.

Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So

1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².

Substituting the kinetic energy of the puck for the potential energy of the spring, we have

1/2kx² = mgh

h = kx²/2mg

= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)

= 0.009971 Nm/0.38416 N

= 0.0259 m

= 2.59 cm

≅ 2.6 cm

So the puck moves a vertical height of 2.6 cm before stopping

3 0
3 years ago
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