What would happen if friction suddenly disappears while you are driving a car? (full explanation)

You are walking at 3.75 km per hour across a frozen lake in the snow. You do not realize that with each step you turn 0.350 degr

VladimirAG [237]

Answer:

242.27929622673 meters

5 0

3 years ago

When you push on a wall can you push harder on the wall than it can on you?

IgorLugansk [536]

Every action have an equal and opposite reaction.

When you push with a force it will push on you with the same force

5 0

3 years ago

Three different liquids are mixed together in a calorimeter. The masses of the liquids are m1 = 520 g, m2 = 715 g, m3 = 370 g. T

DochEvi [55]

....................................

6 0

3 years ago

7. A photon of violet light has a frequency of 7.57 x 10' Hz. How much

hichkok12 [17]

Answer:

3.13 eV

Explanation:

E = hf

f =7.57 x 10' Hz.

h = 6.62× 10-34Js

E = 6.62× 10-34 × 7.57 x 10' Hz

E = 5.011× 10^ -30 J

1eV = 1.6×10^-19J

Hence ,

5.011× 10^ -30 J ÷1.6×10^-19J

= 3.132×10^-49

8 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

2 years ago