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Leviafan [203]
3 years ago
6

A thunderstorm loses energy in the _______ stage.

Physics
2 answers:
RoseWind [281]3 years ago
5 0
A thunderstorm loses energy in the dissipation stage. 
Irina-Kira [14]3 years ago
3 0
 dissipation is the answer ;(
You might be interested in
Which transformation of energy occurs in a hydroelectric power plant?
zysi [14]

D. Kinetic energy to electrical energy


8 0
3 years ago
A long wire carrying a 5.8 A current perpendicular to the xy-plane intersects the x-axis at x=−2.3cm. A second, parallel wire ca
adoni [48]

Answer:

a) v    r = 0.7318 cm , b)  r = 7.23 cm

Explanation:

The magnetic field generated by a wire carrying a current can be found with Ampere's law

       ∫ B. ds = μ₀ I

the length of a surface circulates around the wire is

    s = 2π r

where r is the point of interest of the calculation of the magnetic field

         B = μ₀ I / 2π r

In this exercise we have two wires, write the equation of the magnetic field of each one

wire 1     I = 5.8 A

         B₁ = μ₀ 5,8 / 2π r₁

wire 2    I = 3.0 A

         B₂ = μ₀ 3/2π r₂

the direction of the field is given by the rule of the right hand, the thumb indicates the direction of the current and the other fingers the direction of the magnetic field

Let's apply these expressions to our case

a) the two streams go in the same direction

     using the right hand rule for each wire we see that between the two wires the magnetic fields have opposite directions so there is some point where the total value is zero

          B₁ - B₂ = 0

           B₁ = B₂

         μ₀ 5,8 / 2π r₁ = μ₀ 3 / 2π r₂

          5.8 / r₁ = 3 / r₂

          5.8 r₂ = 3r₁

the value of r is measured from each wire, therefore

        r₁ = 2.3 + r

        r₂ = 2.3 -r

we substitute

          5.8 (2.3 - r) = 3 (2.3 + r)

           r (3 + 5.8) = 2.3 (5.8 - 3)

           r = 2.3 2.8 / 8.8

           r = 0.7318 cm

b) the two currents have directional opposite

with the right hand rule in the field you have opposite directions outside the wires

suppose it is zero on the right side where the wire with the lowest current is

         B₁ = B₂

        5.8 / r₁ = 3 / r₂

        5.8 r₂ = 3 r₁

         r₁ = 2.3 + r

         r₂ = r - 2.3

        5.8 (r - 2.3) = 3 (2.3 + r)

        r (5.8 -3) = 2.3 (3 + 5.8)

        r = 2.3 8.8 / 2.8

        r = 7.23 cm

8 0
3 years ago
11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle
Alja [10]

Answer:

The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

Explanation:

To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.

For M vector you obtain:

M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}

For N vector:

N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}

The resultant vector is the sum of the components of M and N:

F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}

The magnitude of the resultant vector is:

|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm

And the direction of the vector is:

\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°

hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

4 0
3 years ago
24. Every magnet, regardless of its shape, has twoO magnetic charges.O magnetic poles.magnetic fields.O magnetic domains.
Irina18 [472]
I would have to say 2 Magnetic Poles.
7 0
3 years ago
During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of
olasank [31]

Answer:

3.416 m/s

Explanation:

Given that:

mass of cannonball m_A = 72.0 kg

mass of performer m_B = 65.0 kg

The horizontal component of the ball initially \mu_{xA} = 6.50 m/s

the final velocity of the combined system v = ????

By applying the linear momentum of conservation:

m_A \mu_{xA}+m_B \mu_{xB} = (m_A+m_B) v

72.0 \ kg \times 6.50 \ m/s+65.0 \ kg \times 0 = (72.0 \ kg+65.0 \ kg) v

468 kg m/s + 0 = (137 kg)v

v = \dfrac{468\  kg m/s }{137 \ kg}

v = 3.416 m/s

8 0
3 years ago
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