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2 years ago

Soccer can be traced back to the early Greeks and Romans. O True O False

Physics

1 answer:

Natalija [7]2 years ago

8 0

Can Soccer be traced back to the early Greeks and Romans

Answer:

O False,

Explanation:

Although the game of soccer has been around for more than 2,000 years, soccer as we know it today is traced back to England. The game was once played in ancient China, Greece, Rome, and Japan but with different rules and variations.

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Are the two atoms shown below in the image from the same element?

kvasek [131]

The answer is yes

They are isotopes.

7 0

3 years ago

A 0.45kg baseball is pitched towards home plate at 20 m/s. The ball is hit back towards the pitcher with a speed of 30 m/s. What

Inessa05 [86]

Answer:

4.5kgm/s

Explanation:

Change in momentum is expressed as

Change in momentum = m(v-u)

M is the mass

V is the final velocity

u is the initial velocity

Given

m=0.45kg

v = 30m/s

u = 20m/s

Substitute

Change in momentum = 0.45(30-20)

Change in momentum = 0.45×10

Change in momentum = 4.5kgm/s

3 0

2 years ago

The Faraday disk was an early example of what?

Evgen [1.6K]

Electrical generator which operates using a magnetic field. It is the beginning of modern dynamos.

8 0

3 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$