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gizmo_the_mogwai [7]
3 years ago
9

a 62.0 kg curler runs into a stationary 78.1 kg curler and they hold on to each other. Together they move away at 1.29 m/s, west

. What was the original velocity of the 62.0 kg curler?
Physics
1 answer:
Lady_Fox [76]3 years ago
7 0
Momentum P is given by:
P = mv

Given: m₁ = 62 kg, m₂ = 78.1 kg, v₁ = ?, v₂ = 0 m/s, v₁₊₂ = 1.29 m/s

before:
P = m_1 v_1 * m_2 v_2 = m_1 v_1

after:
P = (m_1 + m_2)v_{1+2}

Since momentum is conserved, the momentum before the collision must be equal to the momentum after the collision.

P = m_1 v_1 = (m_1 + m_2)v_{1+2} \\ v_1 =  \frac{(m_1 + m_2)v_{1+2}}{m_1}

Solve for v₁.


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Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa
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Answer:

  • No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field \vec{E} is related to the gradient of the electric potential V with :

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})

This means that for constant electric potential the electric field must be zero:

V(\vec{r}) = k

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k

\vec{E} (\vec{r}) = -  (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k

\vec{E} (\vec{r}) = -  (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})

\vec{E} (\vec{r}) = -  (0,0,0)

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4

give an electric field of zero at point (0,0,0)

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v=3.66,h-3.66

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