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2 years ago

Which of the following can be correct units for acceleration?

Physics

1 answer:

SVETLANKA909090 [29]2 years ago

6 0

Answer:

B. Km/s/hr

Explanation:

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What is the radius of a circle thats the diameter of 480?

Luda [366]

The answer is 12.36. hoped this helped!

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2 years ago

Unscramble the bolded letters to guess the 6 letter word code. Did you get the poses or exercises correct? MOUNTAIN POSE TRIANGL

nikitadnepr [17]

Explanation:

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2 years ago

Calculate the magnitude of the total impulse applied to the car to bring it to rest

n200080 [17]

Answer:

Total impulse = $mv$ = Initial momentum of the car

Explanation:

Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.

The final velocity of the car is 0 m/s as it is brought to rest.

Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.

The magnitude of impulse is the absolute value of the change in momentum.

$|J|=|p_f-p_i|$

Momentum of an object is equal to the product of its mass and velocity.

So, the initial momentum of the car is given as:

$p_i=mv$

The final momentum of the car is given as:

$p_f=m(0)=0$

Therefore, the impulse is given as:

$|J|=|p_f-p_i|=|0-mv|=|-mv|=mv$

Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.

5 0

3 years ago

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

6 0

3 years ago

During the fission reaction shown, how did the target nucleus change ?

Zarrin [17]

Answer:

A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

Explanation:

5 0

2 years ago