All categories

3 years ago

Explain whether a tennis ball dropped from a high distance experiences an elastic collision or inelastic collision

Physics

1 answer:

hodyreva [135]3 years ago

4 0

I need help on that too poop head

You might be interested in

It's important to do cardiovascular exercise on a regular basis because it _____.

n200080 [17]

Answer:

I believe it improves your hearts stroke volume

If im wrong plz let me know

4 0

3 years ago

How long has eggs been around

Paraphin [41]

Answer:

About six million years

Explanation:

3 0

3 years ago

Read 2 more answers

What is the angular speed of the tip of the minute hand on a clock

julsineya [31]

Answer:

Its angular speed is 0.105 rad /s. The minute hand of a clock takes 60 minutes = 60 x 60 seconds to complete one rotation.

Hope this Helps!!

Have a nice day :)

Explanation:

4 0

3 years ago

A plane travels 200 meters in 20 seconds. What is its velocity?

ki77a [65]

10 is the correct answer. Velocity Formula is s= d/t (s=speed, d= distance, t= time)

3 0

3 years ago

Two 2.4 cm -diameter disks face each other, 1.0 mm apart. They are charged to ±11nC. What is the electric field strength between

stiv31 [10]

Answer:

$\rm 1.374\times 10^6\ N/C.$

Explanation:

<u>Given:</u>

Diameter of each disc, D = 2.4 cm.
Distance between the discs, d = 1.0 mm.
Charges on the discs, q = ±11 nC = $\rm \pm 11\times 10^{-9}\ C.$

The surface area of each of the disc is given by

$\rm A=\pi \times Radius^2.\\\\Radius = \dfrac D2=\dfrac{2.4\ cm}{2} = 1.2\ cm = 1.2\times 10^{-2}\ m.\\A = \pi \times (1.2\times 10^{-2} )^2=4.524\times 10^{-4}\ m^2.$

For the case, when d<<D, the strength of the electric field at a point due to a charged sheet is given by

$\rm E=\dfrac{|\sigma|}{2\epsilon_o}$

<em>where</em>,

$\sigma$ = surface charge density of the disc = $\rm \dfrac qA$ .

$\epsilon_o$ = electrical permittivity of free space = [tex\rm ]9\tiimes 10^9\ Nm^2/C^2[/tex].

The electric field strength between the discs due to negative disc is given by

$\rm E_1 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.$

Since, the electric field is directed from positive charge to negative charge, therefore, the direction of this electric field is towards the negative disc.

The electric field strength between the discs due to positive disc is given by

$\rm E_2 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.$

The direction of this electric field is away from the positive disc, i.e., towards negative disc.

The electric field between the discs due to both the disc is towards the negative disc, therefore the total electric field strength between the discs is given by

$\rm E=E_1+E_2\\=\dfrac{|q|}{2A\epsilon_o}+\dfrac{|q|}{2A\epsilon_o}\\=\dfrac{|q|}{A\epsilon_o}\\=\dfrac{11\times 10^{-9}}{2\times (4.524\times 10^{-4})\times (8.85\times 10^{-12})}\\=1.374\times 10^6\ N/C.$

8 0

3 years ago