What is the frequency of a wave that has a wavelength of 20 cm and a speed of 10 m/s

Marysya12 [62]

28.65m x 15.24m

Explanation:

The size of a basketball court in the NBA is usually 28.65m x 15.24m giving a total court area of 436.62m².

The NBA is the acronym for National Basketball Association. It comprises of several teams from across United states and Canada in basketball league.

The game is played by hand with a target basket.
For each successive throw into the basket, 1 or 2 or 3 points can be awarded.
In feets the dimensions are 94 by 50 feet
From the dimension we can accurately say that a basketball court is rectangular in shape.
James Naismith is credited with the invention of the game of basketball

learn more:

Basketball jump brainly.com/question/11222255

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6 0

3 years ago

Roger pushes a box on a 30° incline. If he applies a force of 60 newtons parallel to the incline and displaces the box 10 meters

Dmitry [639]

Work = Force * distance

He's going up an incline but pushing parallel to the ground. Draw out a diagram to show what's going on. )

Work = force* distance*Cos(incline angle)
60 N force * 10 meters * cos(30)
= 519.61 J

7 0

4 years ago

A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2

Rudiy27

Answer:

Approximately $7.66\; \rm m$ .

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of $u = 1.75\; \rm m \cdot s^{-1}$ and came to a stop when her speed became $v = 0\; \rm m \cdot s^{-1}$ . How much time would that take if her acceleration is $a = -0.20\; \rm m \cdot s^{-1}$ ?

$\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}$ .

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

$\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m$ .

In other words, the skateboarder travelled $15.3\; \rm m$ up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

$\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a}$ ,

where:

$u$ and $v$ are the initial and final velocity of the object,
$a$ is the constant acceleration that changed the velocity of this object from $u$ to $v$ , and
$x$ is the distance that this object travelled while its velocity changed from $u$ to $v$ .

For the skateboarder in this question:

$\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}$ .