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4 years ago

14

A basketball referee tosses the ball straight up for the starting tipoff. at what velocity must a basketball player leave the gr

ound to rise 1.25 m above the floor in an attempt to get the ball?

1 answer:

sladkih [1.3K]4 years ago

5 0

Mgh=mv²/2

v=√2gh=√2·9.8·1.25=4.95m/s

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Maria makes a graphic organizer to compare open, closed, and short circuits. She adds the labels shown.

Lina20 [59]

The Answer is Label 1:Y Label 2: W

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3 years ago

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You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.

san4es73 [151]

Answer: The mean value <u>= 9.85m/s².</u>

Explanation:

Mean = $\dfrac{\text{Sum of n observations}}{n}$

The given measurements the acceleration of gravity (units of m/s²): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90.

Number of measurements =9

Sum of measurements = 88.69

Mean = $\dfrac{88.69}{9}=9.85444444\approx9.85$

Hence, the mean value<u> = 9.85m/s².</u>

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3 years ago

Can someone solve this problem and explain to me how you got it

Zarrin [17]

1) The electric force changes by a factor of 25

2) The electric force changes by a factor of 16/9

Explanation:

1)

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, let's call F the initial force between the two charges when they are at a distance of r.

Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is

r' = 5r

Therefore, the new force between the charges is:

$F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}$

So, the force has changed by a factor of 25.

2)

The original force between the two charges is

$F=k\frac{q_1 q_2}{r^2}$

In this problem, we have:

- The distance between the charges is changed by a factor of 6:

r' = 6r

- The charges are both changed by a factor of 8:

$q_1' = 8q_1$

$q_2' = 8q_2$

Substituting into the equation, we find the new force:

$F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F$

So, the force has changed by a factor of 16/9.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago

What is the upper temperature range for stars?

padilas [110]

The upper temperature range for stars is : c. 40,000 K

Every object has their own upper and lower temperature
This indicate the maximum possible energy that a star could release to its surrounding if its somehow exploded.

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A seaplane flies horizontally over the ocean at 50 meters/second. It releases a buoy, which lands after 21 seconds. What's the v

pochemuha

The motion of the buoy is a composition of two independent motions:
- a uniform motion on the horizontal axis, with constant speed vx=50 m/s
- an uniformly accelerated motion on the vertical axis, with constant acceleration $g=9.81 m/s^2$

Since we want to find the vertical displacement, we are only interested in the vertical motion.
The law of motion on the vertical direction is given by:
$y(t)=h+v_{0y} t+ \frac{1}{2}gt^2$
where
h is the initial height of the buoy
$v_{0y}$ is the initial vertical velocity of the buoy, which is zero
t is the time

We know that the buoy lands after t=21 seconds, this means that the vertical position at t=21 s is y(21 s)=0. If we substitute these data into the equation, we can find the value of h, the initial height of the buoy:
$0=h+ \frac{1}{2}gt^2$
$h= -\frac{1}{2}gt^2= -\frac{1}{2}(9.81 m/s^2)(21 s)^2=-2163 m$
And this corresponds to the vertical displacement of the buoy.

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4 years ago