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2 years ago

To win the game, a place kicker must kick a

Physics

1 answer:

jonny [76]2 years ago

3 0

The answer to the question is that the vertical distance the ball clear the cross bar is 1.465 metres.

We have horizontal distance as 28 m, and velocity as 18 m/s and the angle of projection 46.8 degree.

so, we can say that

In horizontal direction,

28 = (18 cos 46°) t

t = 2.2723 seconds

Now, in vertical direction,

Let be the height of the ball

h = (18 sin 46°)t - (1/2)gt²

h = (18 sin 46°)t - 4.9t²

h = ( (13.1214) × (2.2723) ) - ( (4.9) × (2.2723)² )

h = 4.5154 metre

Vertical distance the ball clear the cross bar = h - 3.05

Vertical distance the ball clear the cross bar = 4.5154 - 3.05

Vertical distance the ball clear the cross bar = 1.465

Thus, we can conclude that after solving and applying the concepts of projectile motion we find out the vertical distance the ball will clear the cross bar is 1.465 metres.

Learn more about Projectile Motion here:

brainly.com/question/10343223

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2 Points How did J. J. Thomson change the atomic model based on his experimental results?

tester [92]

Explanation:

In 1897, J.J. ... Thomson realized that the accepted model of an atom did not account for negatively or positively charged particles. Therefore, he proposed a model of the atom which he likened to plum pudding. The negative electrons represented the raisins in the pudding and the dough contained the positive charge.

6 0

3 years ago

Identical objects, Object X and Object Y, are tied together by a string and placed at rest on an incline, as shown in the figure

goblinko [34]

The slope of the velocity time graph of an object moving with constant acceleration is constant

It will take approximately 3 seconds for the center of mass of Object X to reach point J near the bottom of the incline

The reason why the above time value is correct is given as follows:

Known parameters:

Initial velocity of the objects, u = 0

The graph in the question is a straight line graph with data points

(0, 0), (0.5, 1.0), (1.0, 2), (3.0, 6), and (3.5, 7)

Given that the slope of the velocity-time graph is constant, we have that the acceleration is constant and is given as follows;

$a = \dfrac{\Delta v}{\Delta t } = \dfrac{v_2 - v_1}{t_2 - t_1}$

Therefore;

$a = \dfrac{6 - 2}{3.0 - 1.0} = 2$

The acceleration, a ≈ 2 m/s²

The distance from the center of mass of the Object X to the point J near the bottom = 9 m

The equation for distance travelled is given as follows;

$s = u\cdot t + \dfrac{1}{2} \cdot a \cdot t^2$

Which gives;

$9 = 0\times t + \dfrac{1}{2} \times 2 \times t^2 = t^2$

t = √9 = 3

The time it will take the center of mass of Object X to reach point J near the bottom of the incline is t = 3 seconds

Learn more about motion under constant acceleration here:

brainly.com/question/16391598

6 0

3 years ago

A hydrogen atom has a radius of 2.5 x 10-11 m Determine the radius of a magnesium atom.

baherus [9]

Answer:

$R = 1.5* 10^{-10}m$

Explanation:

Given

$r = 2.5 * 10^{-11}m$ -- radius of hydrogen atom

See attachment

Required

Determine the radius of magnesium atom (R)

From the attachment, the ratio of a hydrogen atom to a magnesium atom is:

$Ratio = 6mm : 36mm$

Simplify

$Ratio =1 : 6$

Represent the radius as ratio:

$Ratio = r : R$

Substitute $r = 2.5 * 10^{-11}m$

$Ratio = 2.5 * 10^{-11}m : R$

Equate both ratios

$2.5 * 10^{-11}m : R = 1 : 6$

Express as fraction

$\frac{2.5 * 10^{-11}m}{R} = \frac{1}{6}$

Cross Multiply

$R * 1 = 2.5 * 10^{-11}m * 6$

$R * 1 = 2.5 * 6* 10^{-11}m$

$R * 1 = 15* 10^{-11}m$

$R = 15* 10^{-11}m$

$R = 1.5*10* 10^{-11}m$

$R = 1.5* 10^{1-11}m$

$R = 1.5* 10^{-10}m$

Hence, the radius of the magnesium atom is: $1.5* 10^{-10}m$

6 0

3 years ago

Five basket ball teams are playing in a tournament. The teams will play with each other only one time. How many games will be pl

77julia77 [94]

Ten games will be played.

The first team can be any one of 5 . For each of those, the opponent can be any one of the other 4 .

Number of ways to pair them up = (5 x 4) = 20 ways .

BUT ... Whether the Reds play the Blues, or the Blues play the Reds, it's the SAME GAME. Each possible game shows up twice in the list of 20 ways. So there are really only 10 different pairs ==> 10 games .

3 0

3 years ago

The theory that light exists in quantized "chunks" was proposed in response

Dafna1 [17]

Its D because of the emission spectra

8 0

3 years ago

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