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2 years ago

9

Draw a standard VOLLEYBALL COURT to show the correct player’s positioning and rotation by writing the numbers of each player.

1 answer:

miss Akunina [59]2 years ago

8 0

Answer:

this is the picture okkkkkkkkkkkkkkkk

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In which of the following is positive work done by a person on a suitcase

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pendajo sMh so stupid

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3 years ago

When air resistance equals the weight of an object, the object has reached

MissTica

Answer:

When air resistance equals the weight of an object, the object has reached free fall.

Explanation:

When an object has only force acting on it as gravity then, it experiences free fall.
During free fall all the forces except gravity is balanced by one another.
In the question, object's weight is balanced by air resistance so it is in the state of free fall.
At the null point of free fall, object experiences weightlessness i.e. it feels like object is not attracted by any force.

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3 years ago

Two farmers are trying to move a large rock that sits in a field. They use a long stick as a lever and a brick as the fulcrum, a

kumpel [21]

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3 years ago

Read 2 more answers

our lab partner wears a new pair of sneakers to lab and, rather than performing the required experiments, you decide to measure

Dafna1 [17]

Answer:

The coefficient of static friction between your partner and the floor is 0.55

Explanation:

Given:

Mass $m = 59$ Kg

Frictional force $F_{s} = 318.3$ N

From the formula of frictional force,

$F_{s} = \mu_{s} mg$

Where $\mu _{s} =$ coefficient of static friction, $g = 9.8 \frac{m}{s^{2} }$

Put the above values and find the coefficient of static friction.

$318.3 = \mu_{s} \times 59 \times 9.8$

$\mu_{s} = 0.55$

Therefore, the coefficient of static friction between your partner and the floor is 0.55

4 0

3 years ago

We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu

liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

$\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}$

$\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}$

f = 30 cm

using lens formula

$\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})$

$R_1 = R\ and\ R_2 = -R$

$\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})$

$R = (n -1)\ f$

$R = 2(1.5 -1)\ 30$

R = 30 cm

hence, the radii of curvature is 30 cm.

3 0

3 years ago