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Natasha_Volkova [10]

3 years ago

15

A 3.00 kg block moving 2.09 m/s

1 answer:

Talja [164]3 years ago

4 0

Answer:11.64kgm/s

Explanation:

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faltersainse [42]

Seven stars make it up

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3 years ago

There may be installations where grounded conductors of different systems are installed within the same raceway, cable, or box.

Brrunno [24]

It is imperative that the system-grounded conductors remain buried and is denoted as option D.

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2 years ago

1. When the two cars have the same mass, what is true about their<br> velocities?

garri49 [273]

Answer:

I mean they have the same amount of velocity

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2 years ago

What would the speed of each particle be if it had the same wavelength as a photon of violet light ( λ = 400.0 nm ) ?

Crazy boy [7]

To develop this problem we will use the DeBroglie relationship for which the wavelength is considered as

$\lambda = \frac{h}{mv}$

Where,

h = Planck's constant

m = mass

v = Velocity

$\lambda$ = Wavelength

Rearranging the equation we have that the speed would be

$v = \frac{h}{m\lambda}$

Our given values are considered

$\lambda = 400nm = 4*10^{-7}m$

$h = 6.626*10^{-34} J\cdot s$

$m = 1.673*10^{-24}g = 1.673*10^{-27}kg$

The value of the mass varies, therefore its speed would be given as:

Proton $(m=1.673*10^{-27}kg)$

$v = \frac{6.626*10^{-34}}{(1.673*10^{-27})(4*10^{-7})}$

$v = 0.99m/s$

Neutron $(m=1.675*10^{-27}kg)$

$v = \frac{6.626*10^{-34}}{(1.675*10^{-27})(4*10^{-7})}$

$v = 0.988m/s$

Electron $(m=9.109*10^{-31}kg)$

$v = \frac{6.626*10^{-34}}{(9.109*10^{-31})(4*10^{-7})}$

$v = 1818.53m/s$

Alpha particle $(m=6.645*10^{-27} kg)$

$v = \frac{6.626*10^{-34}}{(6.645*10^{-27} )(4*10^{-7})}$

$v = 0.249m/s$

4 0

3 years ago

How much charge does a 9.0Vbattery transfer from the negative to the positive terminal while doing45Jof work?

sveticcg [70]

Answer:

q = 5 coulombs.

Explanation:

Given :

Potential of battery , V = 9 V.

Work done in transferring charge from negative terminal to the positive terminal , W = 45 J.

We know, Work done by charge is product of its charge by its potential.

$W=q\times V$

Therefore, $q=\dfrac{W}{V}$

Putting values of W and V in above equation.

We get , $q=\dfrac{45}{9}=5\ coulombs.$

Hence, this is the required solution.

3 0

3 years ago