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algol [13]
3 years ago
8

. Friction is a rubbing force that ___________ a spinning yo-yo.

Physics
2 answers:
katovenus [111]3 years ago
5 0

Answer:

i think add energy

Explanation:

Advocard [28]3 years ago
3 0
The yo-yo speeds up when you rub it
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A boy pushes forward a cart of groceries with a total mass of 40.0 kg. What is the acceleration of the cart if the net force on
melisa1 [442]

1.5m/s2

F=ma

a=F/m

a=60/40

a=1.5m/s2

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4 years ago
What is the Ramsar convention?
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The Ramsar Convention on Wetlands of International Importance Especially as Waterfowl Habitat is an international treaty for the conservation and sustainable use of wetlands.
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3 years ago
Which of the following is a component of staying positive while competing ​
Natalija [7]

Answer: B - complimenting others on good plays

Explanation: Reading the first words sort of give it away when staying positive you compliment, not criticize, confront angrily, or refuse.

4 0
3 years ago
Read 2 more answers
A train A travelled a distance of 150 km in 3 hours, whereas, train 'B"
ozzi

Answer:

Therefore, Train A is faster with 50km/h and a 5km/h difference

Explanation:

train A

150km/3h = 50km/h

train B

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8 0
3 years ago
At what distance on the axis of a current loop is the magneticfield half the strength of the field at the center of the loop? Gi
olchik [2.2K]

Answer:

x = 1.26 R

Explanation:

For this exercise let's find the magnetic field using the Biot-Savart law

            B = μ₀ I/4π ∫ ds x r^ / r²

In the case of a loop or loop, the quantity ds is perpendicular to the distance r, therefore the vector product reduces to the algebraic product and the direction of the field is perpendicular to the current loop

suppose that the spiral eta in the yz plane, therefore the axis is in the x axis

         B = μ₀ I/4π ∫ ds / (R² + x²)

     

The total magnetic field has two components, one parallel to the x axis and another perpendicular, this component is annual when integrating the entire loop, so the total field is

            B = Bₓ i^

using trigonometry

            Bₓ = B cos θ

we substitute

            Bₓ = μ₀ I/4π ∫ ds cos θ / (x² + R²)

the cosine function is

            cos θ = R /√(x² + R²)

The differential is

            ds = R dθ

we substitute

             Bₓ = μ₀ I/4π ∫ (R dθ)  R /√( (x² + R²)³ )

we integrate from 0 to 2π

              Bₓ =μ₀ I/4π R² / √(x² + R²)³   2pi

therefore the final expression is

            B = μ₀ I R²/ 2√(x² + R²)³   i^

In our case the distance is requested where B is half of B in the center of the bone loop x = 0

Spire center field   x=0

              B₀ = μ₀ I/2R

Field at the desired point (x)

              B = B₀ / 2

               

we substitute

              R² /√(x² + R²)³ = ½  1 /R

              2R³ =√(x² + R²)³

              (x² + R²)³ = 4 (R²)³

              (x²/R² + 1)³ = 4

               

The exact result is the solution of this equation, but it is quite laborious, we can find an approximate result assuming that the distance x is much greater than R (x »R)

           B = μ₀ I/2x³ 

we substitute

            R² / x³ = 1/2   1 / R

            2R³ = x³

          x = ∛2  R

            x = 1.2599 R

3 0
3 years ago
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