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3 years ago

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A balloon is filled with 1500 mL of air at a pressure of 5 atm. What is the pressure of the balloon if the volume is changed to

2500 mL?

1 answer:

Eva8 [605]3 years ago

5 0

Answer:

xcfdbrtbfcgctr dsvrd

Explanation:

ds vfg g cd fg

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Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.

Aleonysh [2.5K]

De broglie wavelength, $\lambda = \frac{h}{mv}$ , where h is the Planck's constant, m is the mass and v is the velocity.

$h = 6.63*10^{-34}$

Mass of hydrogen atom, $m = 1.67*10^{-27}kg$

v = 440 m/s

Substituting

Wavelength $\lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m$

$1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm$

So the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

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2 years ago

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How do climate differences affect the movement at the Mariana Trench

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It pushes the currents to opposite sides

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3 years ago

What does a negative value for the focal length tell you?

Llana [10]

the focal length <span> is much more decent for a concave, and also worse</span><span> for a convex mirror. When the image that is given, distance is good and decent, images are always on the same area of the mirror as the object given , and it is not fake. images distance is </span>never positive <span>, the image is on the oppisite side of the mirror, so the image must be virtual.</span>

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3 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