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Oksana_A [137]
3 years ago
10

Resisting force is

Physics
1 answer:
Alina [70]3 years ago
6 0

Answer:

D

Explanation:

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The liquid that will evaporate the quickest would be water. 
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Suppose the carts collided on a surface that had a slight incline to it. Would you expect the momentum to be conserved
lesantik [10]

Momentum is conserved when carts are collided on a slanting plane.

To find the answer, we need to know about the conversation of momentum.

<h3>What's the conversation of momentum?</h3>
  • Conservation of linear momentum says the total momentum before the collision and after the collision remains the same.
  • Mathematically, m1u1+m2u2 = m1v1+m2v2
<h3>How is the momentum conserved when collision occurs on a slanting plane?</h3>
  • On a slanting plane, the velocity has two components,
  1. horizontal component
  2. horizontal component Vertical component
  • So, its momentum has also similar two components.
  • The momentum is conserved along horizontal direction and vertical direction separately.

Thus, we can conclude that the momentum is conserved when carts are collided on a slanting plane.

Learn more about the conversation of momentum here:

brainly.com/question/7538238

#SPJ4

7 0
2 years ago
Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the
astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

B = \frac{\mu_oI}{2\pi R}

Where \mu_o is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

B_1 = \frac{\mu_oI_1}{2\pi R}

B_2 = \frac{\mu_oI_2}{2\pi R}

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  \frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_2}{2\pi R}

or

4.0×10⁻⁶T =  \frac{\mu_o}{2\pi R}\times (I_1-I_2 )

also

\frac{I_1}{I_2} = \frac{3}{1}

⇒I_1 = 3\times I_2

substituting the values in the above equation we get

4.0×10⁻⁶T =  \frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)

⇒I_2 = 1A

also

I_1 = 3\times I_2

⇒I_1 = 3\times 1A

⇒I_1 = 3A

Hence, the larger of the two currents is 3A

3 0
3 years ago
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