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2 years ago

The physical quantity represented as rate of change of change in position in a

Physics

1 answer:

Oduvanchick [21]2 years ago

3 0

Answer:

rate of change of its position with respect to a frame of reference, and is a function of time.

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Two tugs are towing a small ship. The tugs pull with forces of 600 N and 800 N respectively and there is an angle of 45 degrees

bija089 [108]

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3 years ago

A point source emits 25.9 W of sound isotropically. A small microphone intercepts the sound in an area of 0.242 cm2, 590 m from

AnnZ [28]

The solution is in the attachment

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3 years ago

If both mass and speed are doubled, what happens to its momentum?

puteri [66]

The general formula is: Momentum = (mass) x (speed)

I never like to just write a bunch of algebra without explaining it.
But in this particular case, there's really not much to say, and
I think the algebra will pretty well explain itself. I hope so:

Original momentum = (original mass) x (original speed)

New momentum = (2 x original mass) x (2 x original speed)

   = (2) x (original mass) x (2) x (original speed)

   = (2) x (2) x (original mass) x (original speed)

   = (4) x (original mass) x (original speed)

   = (4) x (original momentum).

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3 years ago

The image of an object formed by plane mirror is

m_a_m_a [10]

Answer of your question is in this photo

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2 years ago

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

lyudmila [28]

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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3 years ago