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3 years ago

How can meteorologists use the jet stream to predict the weather ?

1 answer:

5 0

<span>Jet streams act as an invisible director of the atmosphere and are largely responsible for changes in the weather across the globe.
Hope this helps</span>

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The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t

Vika [28.1K]

Answer:

The hydro static force on the back of the dam is $1.96\times10^{11}\ N$

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of average pressure

$P_{avg}=\rho\times g\times d_{avg}$

Put the value into the formula

$P_{avg}=1000\times9.8\times100$

$P_{avg}=980000\ Pa$

We need to calculate the hydro static force on the back of the dam

Using formula of force

$F = P_{avg}\times A$

Put the value into the formula

$F = 980000\times1000\times200$

$F=1.96\times10^{11}\ N$

Hence, The hydro static force on the back of the dam is $1.96\times10^{11}\ N$

7 0

3 years ago

A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the

jasenka [17]

Answer:

1.8 cm

Explanation:

$m$ = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

$q$ = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

$\Delta V$ =Potential difference through which the ion is accelerated = 215 V

$v$ = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

$(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}$

$r$ = Radius of the path followed by ion

$B$ = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

$qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm$