Ask question

Ask question

All categories

I am Lyosha [343]

3 years ago

13

When you are moving up at a constant speed in an elevator, there are two forces acting on you: the floor pushing up on you (F1)

and gravity pulling down (F2).
What's the relation of the magnitude of F1 and F2?

1 answer:

WARRIOR [948]3 years ago

6 0

Answer:

F₁ = F₂

Explanation:

Given that

floor pushing up on you by force F₁

gravity pulling down by force F₂

From second law of Newton's

F₁ - F₂ = m a

But here given that ,elevator is moving with constant speed .It means that acceleration of the elevator is zero.

a= 0 m/s²

F₁ - F₂ = m x 0

F₁ = F₂

You might be interested in

Answer the question please

kirill [66]

Answer:

i think is third

Explanation:

7 0

4 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

4 years ago

Two parallel square metal plates that are 1.5 cm and 22 cm on each side carry equal but opposite charges uniformly spread out ov

Mumz [18]

Answer:

The number of excess electrons are on the negative surface is $4.80\times10^{10}\ electrons$

Explanation:

Given that,

Distance =1.5 cm

Side = 22 cm

Electric field = 18000 N/C

We need to calculate the capacitance in the metal plates

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}$

$C=0.285\times10^{-10}\ F$

We need to calculate the potential

Using formula of potential

$V=Ed$

Put the value into the formula

$V=18000\times1.5\times10^{-2}\ V$

$V=270\ V$

We need to calculate the charge

Using formula of charge

$Q=CV$

Put the value into the formula

$Q=0.285\times10^{-10}\times270$

$Q=76.95\times10^{-10}\ C$

Here, the charge on both the positive and negative plates

$Q=+76.95\times10^{-10}\ C$

$Q=-76.95\times10^{-10}\ C$

We need to calculate the number of excess electrons are on the negative surface

Using formula of number of electrons

$n=\dfrac{q}{e}$

Put the value into the formula

$n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}$

$n=4.80\times10^{10}\ electrons$

Hence, The number of excess electrons are on the negative surface is $4.80\times10^{10}\ electrons$

8 0

3 years ago

A material having an index of refraction of 1.35 is used as an antireflective coating on a piece of glass (n = 1.50). What shoul

exis [7]

Answer:

103.70 nm

Explanation:

The minimum thickness of the film is given by $t=\frac{\lambda }{4n}$ where $\lambda$ is the wavelength and n is the index of the refraction

We have n =1.35

Wavelength $\lambda =560nm$

So the minimum thickness of the film $t=\frac{\lambda }{4n}=\frac{560}{4\times 1.35}=103.70\ nm$

3 0

3 years ago

Plzz helpp mee!!!! plzzz

Gnesinka [82]

If you add 2 miles from west then 2 miles east then it would 4 miles all together.

7 0

3 years ago