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3 years ago

A book that weighs 20 N sits on a table. How big and in what direction does the force of gravity from the Earth act on the book?

Physics

2 answers:

taurus [48]3 years ago

7 0

Answer:

A. 20 N down

Explanation:

It's asking how much force of gravity is acting on it. The book weighs 20 Newtons so that's how much gravity is being applied. Hope this helps

Anna71 [15]3 years ago

5 0

I think the answer is A. 20 N down

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An electron initially 3.00 m from a nonconducting infinite sheet of uniformly distributed charge is fired toward the sheet. The

Cloud [144]

Answer:

$4.4443704375\times 10^{-18}\ C/m^2$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\Delta l$ = Distance charge traveled = 2 m

v = Velocity of electron = 420 m/s

E = Electric field

$m_e$ = Mass of electron = $9.11\times 10^{-31}\ kg$

$q_e$ = Charge of electron = $1.6\times 10^{-19}\ C$

As the energy of the system is conserved we have

$q_eE\Delta l=\dfrac{1}{2}m_ev^2\\\Rightarrow E=\dfrac{1}{2}\dfrac{m_e}{q_e}\times \dfrac{v^2}{\Delta l}\\\Rightarrow E=\dfrac{1}{2}\dfrac{9.11\times 10^{-31}}{1.6\times 10^{-19}}\times \dfrac{420^2}{2}\\\Rightarrow E=2.51094375\times 10^{-7}\ N/C$

For an infinite non conducting sheet electric field is given by

$E=\dfrac{\sigma}{2\epsilon}\\\Rightarrow \sigma=2E\epsilon\\\Rightarrow \sigma=2\times 2.51094375\times 10^{-7}\times 8.85\times 10^{-12}\\\Rightarrow \sigma=4.4443704375\times 10^{-18}\ C/m^2$

The surface charge density is $4.4443704375\times 10^{-18}\ C/m^2$

5 0

3 years ago

A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a batte

vovikov84 [41]

<h2>Answer:</h2>

(e) halved

<h2>Explanation:</h2>

The electrical enery (E) stored in a capacitor is related to its capacitance (C) and potential difference (V) as follows;

E = $\frac{1}{2}$ x C x $V^{2}$ ------------------------(i)

Also, the capacitance (C) of a capacitor consisting of parallel plates is related to the area (A) of the plates and distance (d) between the plates as follows;

C = A x ε₀ / d ------------------------(ii)

Where;

ε₀ is the permittivity of free space.

Substituting equation (ii) into equation (i) gives;

E = $\frac{1}{2}$ x A x ε₀ / d x $V^{2}$ --------------------(iii)

From equation(iii)

When the potential difference (V) is constant, then the electrical energy (E) stored is <em>inversely </em>proportional to the distance between the plates. i.e

E = k / d ----------------(iv)

Where;

k = proportionality constant = $\frac{1}{2}$ x A x ε₀ x $V^{2}$ (which is the product of all constants)

Therefore from equation (iv);

=> E₁ x d₁ = E₂ x d₂ ---------------------------(v)

Where;

E₁ and E₂ are the initial and final values of the electrical energy stored.

d₁ and d₂ are the initial and final values of the distance between the plates.

<em>So, when the distance is doubled, i.e.</em>

d₂ = 2 x d₁

<em>Substitute the value of d₂ into equation (v) to give;</em>

=> E₁ x d₁ = 2 x d₁ x E₂

<em>Divide through by d₁ to give;</em>

=> E₁ = 2 x E₂

<em>Make E₂ subject of the formula</em>

=> E₂ = $\frac{1}{2}$ x E₁

Therefore, the electrical energy stored in the capacitor will be halved.

6 0

4 years ago

A light beam strikes a piece of glass at a 65.00 ∘ incident angle. The beam contains two wavelengths, 450.0 nm and 700.0 nm, for

Leokris [45]

Answer:

Explanation:

Lower the refractive index , higher the wave length

Refractive index becoming high reduces the velocity and hence wavelength as frequency remains unchanged.

So refractive index 1.4831 will correspond to wavelength of 450 nm.

For refractive index is 1.4831 , angle of incidence i , angle of refraction r .

Sin i / Sinr = 1.4831

sin65 / sinr = 1.4831

sir r = sin65 / 1.4831

= .9063 / 1.4831

= .6111

r = 37.67 degree

For refractive index is 1.4754 , angle of incidence i

Sin i / Sinr = 1.4754

sin65 / sinr = 1.4754

sir r = sin65 / 1.4754

= .9063 / 1.4754

= .61427

r = 37.9 degree

angle between two refracted ray

37.9 -37.67

= 0. 23 degree

6 0

3 years ago

Read 2 more answers

A small metal bead, labeled A, has a charge of 26 nC . It is touched to metal bead B, initially neutral, so that the two beads s

adell [148]

Answer:

$q_A=25.953\ \rm nC$

$q_B=0.047\ \rm nC$

Explanation:

Given:

Total initial charge on bead A=26 nC
The distance between them=5 cm
Magnitude o the force between them $=4.4\times10^{-4}$

Using coulombs law the force between two charged particle is $\dfrac{q_Aq_B}{4\pi \epsilon_0 r^2}$

where r is the radial distance between them

According to question we have

$4.4\times10^{-4}=\dfrac{q_Aq_B\times9\times10^9}{0.05^2}\\\\4.4\times10^{-4}=\dfrac{q_A(q_A-26\times10^{-9})\times9\times10^9}{0.05^2}\\q_A=25.953\ \rm nC\\q_B=0.047\ \rm nC$

Hence the charge on two metal beads is calculated.

5 0

3 years ago

11. If a man is standing more than one focal length away from the focal point of a concave mirror, how will his image form in th

Vladimir79 [104]

Upside down if its further than 1 focal point you have seen this with a spoon and enlarged right side up if closer than 1 focal point

5 0

3 years ago