How can light change matter

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0

2 years ago

How did the South American plate and African plate move?

garri49 [273]

Answer:

How did the South American Plate and African Plate move? Earth's plates move on top of a soft, solid layer of rock called the mantle. ... The South American and African Plates moved apart as a divergent boundary formed between them and an ocean basin formed and spread.

Explanation:

pls make me brainliest

4 0

2 years ago

Read 2 more answers

For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ

slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

157 lb = $157 lb \times \frac{1 kg}{2.2046 lb}$

= 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

$180 \frac{mg}{kg} \times 71.215 kg$

= 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

0.65% (w/w) = $\frac{0.65 g}{100 g}$

= $\frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}$

= $\frac{0.65 g}{100 ml}$

Therefore, calculate the volume which contains the amount of caffeine as follows.

12818.7 mg = 12.8187 g = $\frac{12.8187 g}{\frac{0.65 g}{100 ml}}$

= 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0

2 years ago

2. a) A disc rotates about its axis at speed 25 revolutions per minute and takes 15 s to stop. Calculate the

ASHA 777 [7]

The statement shows a case of rotational motion, in which the disc <em>decelerates</em> at <em>constant</em> rate.

i) The angular acceleration of the disc ( $\alpha$ ), in revolutions per square second, is found by the following kinematic formula:

$\alpha = \frac{\omega_{f}-\omega_{o}}{t}$ (1)

Where:

$\omega_{o}$ - Initial angular speed, in revolutions per second.
$\omega_{f}$ - Final angular speed, in revolutions per second.
$t$ - Time, in seconds.

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $t = 15\,s$ , then the angular acceleration of the disc is:

$\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}$

$\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$

The angular acceleration of the disc is $\frac{1}{36}$ radians per square second.

ii) The number of rotations that the disk makes before it stops ( $\Delta \theta$ ), in revolutions, is determined by the following formula:

$\Delta \theta = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha}$ (2)

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$ , then the number of rotations done by the disc is:

$\Delta \theta = 3.125\,rev$

The disc makes 3.125 revolutions before it stops.

We kindly invite to check this question on rotational motion: brainly.com/question/23933120

6 0

2 years ago

Do you think the amount of positive charge on the glass after rubbing it with silk cloth will be equal to the amount of negative

IrinaK [193]

This is not a answer but can’t you just search that up on the internet and get the answer?

5 0

2 years ago