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3 years ago

14

The Earth is constantly spinning on its axis, like you might spin a basketball on your finger. It is this spinning of the Earth

that causes

1 answer:

mash [69]3 years ago

5 0

<u>Answer:</u>

The spinning of the earth around its own axis causes day and night.

<u>Explanation:</u>

Earth has two types of motions. It spins around its own axis causing day and night every 12 hours and completes a rotation in 23.93 hours that make a full day. The part of the earth that faces sunlight during spinning experiences day and the other part has night. It also rotates around the sun and completes one rotation in 365 days that makes a year.

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Preston tossed a red ball upward and it reaches a maximum height of 3.0. What is the final velocity when it returns to prestons

Leokris [45]

That will depend on the units of the 3.0. We need to know if it's 3 feet, 3 yards, 3 meters, or 3 miles. Each one will have a different answer.

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3 years ago

Which of the following statements is correct A. College degree isn't as important as it was in the past b. Over time, I can earn

julia-pushkina [17]

Answer:

I'd say C is the answer they want, though my pedantic side wants to argue for B being true as well.

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3 years ago

I need help plz and thank you this is due

xeze [42]

Answer:

Grow up man, this is completely based on your curriculum, we would need your book to answer, and this has to be done by you.

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3 years ago

Read 2 more answers

The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a

Vera_Pavlovna [14]

Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

$r=\frac{kze^{2} }{mpV^{2} }$ (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

energy of the particle = 32 MeV = 5.12x10⁻¹²J

At the potential energy is zero, all the energy will be kinetic energy:

$E_{k} =\frac{1}{2} m V^{2}$

Where

m = 4 mp = mass of proton

$5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}$

Replacing in equation 1

$r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2} }{2.56x10^{-12} } =7.1x10^{-15} m$

8 0

3 years ago

3. A football is kicked with a speed of 35 m/s at an angle of 40°.

jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

b) 26.8 m/s

The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_x = (35)(cos 40)=26.8 m/s$

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

$s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m$

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

$t=2(2.30 s)=4.60 s$

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

$v_x = 26.8 m/s$

Therefore, the horizontal distance travelled during the whole motion is

$d=v_x t = (26.8)(4.60)=123.3 m$

So, the ball lands 123.3 m far from the initial point.

4 0

3 years ago