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2 years ago

Why is the rotation system used in volleyball? Physics was the closest thing to PE

Physics

2 answers:

iris [78.8K]2 years ago

5 0

Eso es por que.. lo que la persona dijo

tankabanditka [31]2 years ago

3 0

Answer:

A rotation occurs after every side out, which is when the receiving team gains the right to serve by winning a rally. ... The new serving team will rotate clockwise one spot. The purpose of this is to rotate all the players through the serving position. If you continue winning points, you stay in position.

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A student walks 3 north and 4 m west. The magnitude of the resultant displacement for the student is

evablogger [386]

Answer:

Explanation:

Using Pythagoras theorem,

a^2+ b^2=c^2

3^2+4^2=c^2

25=c^2

√(25)=c

5m=c

6 0

2 years ago

Which two factors does the power of a machine depend on? А. work and distance B.. force and distance C. work and time D. time an

Ivahew [28]

$Hello$ $There!$

$AnimeVines$ $is$ $here!$

The answer is...

C. Work and time.

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6 0

3 years ago

A 13.6 kg block is tied at the top of an incline to a tree. If the incline is 35.5 degrees and the coefficient of friction betwe

Gre4nikov [31]

Answer:

Explanation:

ASSUMING that block = sled AND that the rope is parallel to the slope.

The force acting parallel due to the weight is

13.6(9.81)sin35.5 = 77.475 N

The maximum friction force is

(0.45)13.6(9.81)cos35.5 = 48.877 N

If rope tension is T

77.475 - 48.877 < T < 77.475 + 48.877

28.6 N < T < 126 N

28.6 N will occur if the block is on the verge of sliding downhill

126 N will occur if the block is on the verge of sliding uphill

Could be any value between them.

5 0

2 years ago

A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

6 0

3 years ago

Light with a wavelength of about 490 nm is made to pass through a diffraction grating. the angle formed between the path of the

polet [3.4K]

The number of lines per mm in the diffraction grating is 326.

<h3>What is diffraction grating?</h3>

A diffraction grating is a type of optical instrument obtained with a continuous pattern. The pattern of the diffracted light by a grating depends on the structure and number of elements present.

The given data in the problem is

$\rm \theta$ is the angle formed between the path of the incident light and the diffracted light = 9. 2°

λ is the wavelength of the light=490nm=4.9

N is the number of lines per mm in the diffraction grating=?

n is ordered = 1

The formula for the diffraction grating is;

$n \lambda = d sin\theta \\\\ d = \frac{n \lambda}{sin \theta } \\\\ d = \frac{1 \times 4.90 \times 10^{-7}}{sin 9.2^0 } \\\\ d=3.06 \times 10^{-6} \\\\ d=3.06 \times 10^{-3} \ mm$

The number of lines per mm is found as;

$\rm N= \frac{1}{d} \\\\ N= \frac{1}{3.06 \TIMES 10^{-3}} \\\\ N=326.8 /mm$

Hence the number of lines per mm in the diffraction grating is 326.

To learn more about diffraction grating refer to the link;

brainly.com/question/1812927

5 0

2 years ago