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3 years ago

The area of the court in which the server must serve from is called the...

Physics

1 answer:

SSSSS [86.1K]3 years ago

5 0

Answer:

The Correct Answer is Left Court

Explanation:

The server serves from the left court.

If the serving side wins a rally, the serving side scores a point and the same server serves again from the alternate service court.

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The grocery store is 20 miles away

adoni [48]

Answer:

40mph

Explanation:

20miles .5 miles away if you go 20 mph you get there in a hour soo if you double to 40 mpg you get there in 0.5

5 0

3 years ago

Read 2 more answers

A wheel is rotating freely at angular speed 660 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi

sweet-ann [11.9K]

Answer:

(a) 110 rev/ min

(b) 5/6

Explanation:

As per the conservation of linear momentum,

L ( initial ) = L ( final )

I' ω' = ( I' + I'' ) ωf

I' is the rotational inertia of first wheel and I'' is the rotational inertia of second wheel which is at rest.

(a)

So, ωf = I' ω' / ( I' + I'' )

As I'' = 5I'

ωf = I' ω' / ( I' + 5I' )

ωf = ω'/ 6

now we know ω' = 660 rev / min

therefore ωf = 660/6

= 110 rev/ min

(b)

Initial kinetic energy will be K'

K' = I'ω'² / 2

and final K.E. will be K'' = ( I' + I'' )ωf² / 2

K'' = ( I' + 5I' ) (ω'/ 6)²/ 2

K'' = 6I' ω'²/72

K'' = I' ω'²/ 12

therefore the fraction lost is

ΔK/K' = ( K' - K'' ) / K'

= {( I'ω'² / 2) - (I' ω'²/ 12)} / ( I'ω'² / 2)

= 5/6

6 0

4 years ago

THE LARGEST HORIZONTAL PLATES IS SEPARATED BY 4mm .The plate is at the potencial of -6V . What potencial should be applied to th

Serhud [2]

Answer:

V₂ = -22 V

Explanation:

Electric potential and field are related

ΔV = - E d

where ΔV is the potential difference between the plates, E the electric field and d the separation between the plates

In this exercise we are given the parcionero d = 4 mm = 0.004 m, the potential of one of the plates V1 = -6V and the value of the electric field E = 4000 V / m

V₂- V₁ = - E d

V₂ = - Ed + V₁

V₂ = - 4000 0.004 + (- 6)

V₂ = -16 - 6

V₂ = -22 V

6 0

3 years ago

A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate

Mariana [72]

According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass $m=1.0 \cdot 10^{-4} kg$ . The acceleration of the insect is $a=1.0 \cdot 10^2 m/s^2$ , therefore we can calculate the force exerted by the car on the insect:
$F=ma=(1.0 \cdot 10^{-4} kg)(1.0 \cdot 10^2 m/s^2)=0.01 N$

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.

6 0

3 years ago

Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the

svetoff [14.1K]

Answer:

$\mu_k=0.18$

Explanation:

First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

$x: T+F-f_k=0\\\\y:N-mg=0$

Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.

Since $f_k=\mu_k N$ and $N=mg$ , we can rewrite the first equation as:

$T+F-\mu_k mg=0$

Now, we solve for $\mu_k$ and calculate it:

$\mu_k=\frac{T+F}{mg}\\ \\\mu_k =\frac{220N+390N}{(350kg)(9.8m/s^{2})} =0.18$

This means that the crate's coefficient of kinetic friction on the floor is 0.18.

6 0

4 years ago