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3 years ago

If a cannonball were launched with a speed of 37m/s at a 60° angle from the height of Om, what would be the cannonball's range?

Physics

1 answer:

JulijaS [17]3 years ago

8 0

Answer:

Range = 120.98 m, (or rounded to the nearest whole number =121 meters)

Explanation:

We can use the formula for range in projectile motion at ground level:

$Range = \frac{v^2\,\,sin(2\theta)}{g} \\Range= \frac{37^2\,\,sin(120)}{9.8} \\Range\approx 120.98\,\,m$

Then the cannon's range would be approximately 120.98 m, which may be rounded to the nearest whole number as 121 meters

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How many electrons can possess this set of quantum numbers: principal quantum number n = 4, magnetic quantum number mℓ = −1?

LekaFEV [45]

Answer:

6 electrons

Explanation:

To solve this question lets determine the possible quantum numbers for the principal quantum number n = 4.

For the quantum number l which describes the shape of the orbital, we have the possible values : 0 to n-1.

Thus, for n = 4 the l can assume the values 0, 1, 2, 3 ( 4 possible shapes )

For the angular quantum number, ml, which tell us the orientation in space,we have the values - l to + l.

So lets determine the number of orbials which can have the values -l for n=4

l = 0 ml = 0

l= 1 ml = -1,0,1

l= 2 ml = -2, -1, 0, 1, 2

l= 3 ml = -3,-2,-1,0,1,2,3

So we have three orbital with ml = - 1 and from Pauli´s exclusion principle we can have up two electrons in each orbital. Thus for n= 4 we can have up to 6 electrons with ml = -1

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3 years ago

Which of the following is not part of costa rica’s geography?

Romashka-Z-Leto [24]

costa rica does not have deserts hoped this helped and have a great day:)

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3 years ago

Read 2 more answers

All of the following are ways in which sedimentary rocks form EXCEPT

Anna71 [15]

Except A. Lava cooling

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3 years ago

Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r

Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

$v=u-gt$

$u=gt$

$t=\dfrac{u}{g}$

$t=\dfrac{13.7}{9.8}$

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = -g

$h=ut-\dfrac{1}{2}gt^2$

$h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2$

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

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3 years ago

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Answer:

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