You drop a ball from the top of a building, how long does it take to reach the ground

You kick a soccer ball with a mass of 2 kg. The ball leaves your foot with a speed of 30 m/s. How much kinetic energy does the b

Lyrx [107]

Answer:

KE= 900 (I think)

Explanation:

KE=½mv²

KE= ½(2)(30)²

KE=½(60)²

KE=30²

KE=900

Hope this helps!

7 0

2 years ago

Read 2 more answers

Dimension of radius of sphere

Readme [11.4K]

Answer:

The dimension is L

Explanation:

Dimension analysis is a method of representing quantities majorly with respect to some fundamental quantities of mass (M), length (L), time (T).

A sphere has a definite volume which relates to its radius by:

V = $\frac{4}{3}$ $\pi$ $r^{3}$

In this equation $\pi$ is a dimensionless quantity, and the unit of v is $m^{3}$ .

But, metre is a measure of length, thus it has a dimension of L.

So that,

$m^{3}$ ≅ $L^{3}$

Then,

$L^{3}$ = $r^{3}$

Find the cube root of both sides to have,

r = L

Therefore, the dimension of the radius of a sphere is L.

5 0

3 years ago

A soccer ball with a mass of 0.45 kg is kicked and is moving at 8.9 m/s. Find the kinetic

strojnjashka [21]

Answer:

17.82J

Explanation:

Kinetic energy = 1/2 mv^2

Given

Mass M = 0.45kg

Velocity v = 8.9m/s

Therefore,

K.E. = 1/2 x 0.45 x (8.9)^2

= 1/2 x 0.45 x (8.9 x 8.9)

= 1/2 x 0.45 x 79.21

Multiply through

= 35.6445/2

= 17.82J

The kinetic energy of the ball is 17.82J

3 0

3 years ago

Basalt is pushed into the crust by subduction. It will most likely become _____.

suter [353]

Subduction is, "<span>the sideways and downward movement of the edge of a plate of the earth's crust into the mantle beneath another plate." The basalt would most likely be swallowed up into the ground.
Hope this is what you were looking for! :)

</span>

6 0

3 years ago

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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago