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3 years ago

How many shots does one get when he fouls a shooter? (basketball)

Physics

1 answer:

Tamiku [17]3 years ago

4 0

Answer:

Depends on where he was fouled

If it is beyond the 3 pt line, then it is 3

If it is within the 2 pt line, then it is 2

I hope that helps :)

Explanation:

Plz mark B R A I N L I E S T

You might be interested in

A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average

frez [133]

(u) = 20 m/s
(v) = 0 m/s
 (t) = 4 s

0 = 20 + a(4)

4 x a = -20

so, the answer is -5 m/s^2. or -5 meter per second

8 0

3 years ago

Read 2 more answers

As the time required to run up the stairs increases, the power developed by that person

weeeeeb [17]

Yes. Power will decrease.

'cause Power = Work / time
So, power is indirectly proportional to time so, when one increases other would decrease

Hope this helps!

8 0

3 years ago

For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along t

Mariulka [41]

Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

$y=2+\frac{1}{x}$

$y=2+x^{-1}$

$y'=-x^{-2}$

$y'=-\frac{1}{x^{2}}$

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(1)^{2}}$

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-3=-1(x-1})$

$y-3=-1x+1$

$y=-x+1+3$

$y=-x+4$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-x+4=0$

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2)^{2}}$

$m=y'=-\frac{1}{4}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.5=-\frac{1}{4}(x-2})$

$y-2.5=-\frac{1}{4}x+\frac{1}{2}$

$y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}$

$y=-\frac{1}{4}x+3$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{4}x+3=0$

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2.5)^{2}}$

$m=y'=-\frac{4}{25}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.4=-\frac{4}{25}(x-2.5})$

$y-2.4=-\frac{4}{25}x+\frac{2}{5}$

$y=-\frac{4}{25}x+\frac{2}{5}+2.4$

$y=-\frac{4}{25}x+\frac{14}{5}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{4}{25}x+\frac{14}{5}=0$

and solve for x

$x=\frac{35}{20}$

so, when fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(4)^{2}}$

$m=y'=-\frac{1}{16}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.25=-\frac{1}{16}(x-4})$

$y-2.25=-\frac{1}{16}x+\frac{1}{4}$

$y=-\frac{1}{16}x+\frac{1}{4}+2.25$

$y=-\frac{1}{16}x+\frac{5}{2}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{16}x+\frac{5}{2}=0$

and solve for x

$x=40$

so, when fired from (4,2.25) the rocket will hit the target at (40,0)

I uploaded a graph that represents each case.

8 0

3 years ago

Which fossil fuel is produced as a by-product that occurs when bacteria decompose organic material under anaerobic conditions?

Mumz [18]

Bio-gas is the naturally produced fossil fuel, a by-product when bacteria decompose organic material under anaerobic conditions.

<h3>Explanation:</h3>

Organic matter particularly waste material is broken down by bacteria through fermentation in an environmental condition without any presence of oxygen. This process of decomposition leads to formation of bio-gas with "carbon dioxide and methane" in a 2:3 ratio.

The above biological process is termed as bio-digestion or anaerobic digestion. Methane is flammable and thus bio-gas can be used as "energy source", a waste-to-energy transformation. The remaining decomposed matter is ideal as manure for plants due to its rich nutrient level.

5 0

3 years ago

In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220-m-long cannon with final speed of 10.

Sidana [21]

Answer:

The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².

How many times stronger than gravity is this force? 2.79 x 10⁴ g.

Explanation:

given information:

s = 220 m

final speed, vf = 10.97 km/s = 10970 m/s

g = 9.8 m/s²

he unrealistically large acceleration experienced by the space travelers during their launch

vf² = v₀²+2as, v₀ = 0

vf² = 2as

a =vf²/2s

= (10970)²/(2x220)

= 2.7 x 10⁵ m/s²

Compare your answer with the free-fall acceleration

a/g = 2.7 x 10⁵/9.8

a/g = 2.79 x 10⁴

a = 2.79 x 10⁴ g

7 0

3 years ago