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3 years ago

How do you post a picture

1 answer:

4 0

Click the ask a question button or answer button then at the bottom above the keyboard there are two little icons. A camera lens and a Mountain View. If you click the camera lens you can take a picture to post and if you click the Mountain View it will give you the option to post a picture from your photos.

You might be interested in

Which of the following expressions gives the ratio of the energy density of the magnetic field to that of the electric field jus

miss Akunina [59]

Answer:

$(d) \ \ \frac{\mu_o}{\epsilon_o} (\frac{L}{2\pi r*R} )^2$

Explanation:

Energy density in magnetic field is given as;

$U_B = \frac{1}{2 \mu_o} B^2$

where;

B is the magnetic field strength

Energy density of electric field

$U_E = \frac{1}{2}\epsilon E^2$

where;

E is electric field strength

Take the ratio of the two fields energy density

$\frac{U_B}{U_E} = \frac{1}{2\mu_o} B^2 / \frac{1}{2}\epsilon E^2\\\\\frac{U_B}{U_E} = \frac{B^2}{2\mu_o} *\frac{2}{\epsilon E^2} \\\\\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B^2}{E^2})$

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B}{E})^2$

But, Electric field potential, V = E x L = IR (I is current and R is resistance)

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{E*L})^2$

Now replace E x L with IR

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{IR})^2$

Also, B = μ₀I / 2πr, substitute this value in the above equation

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{\mu_oI*L}{2\pi r* IR})^2$

cancel out the current "I" and factor out μ₀

$\frac{U_B}{U_E} = \frac{\mu_o^2}{\mu_o \epsilon} (\frac{L}{2\pi r* R})^2$

Finally, the equation becomes;

$\frac{U_B}{U_E} = \frac{\mu_o}{\epsilon} (\frac{L}{2\pi r*R })^2$

Therefore, the correct option is (d) μ₀/ϵ₀ (L /R 2πr)²

3 0

3 years ago

A hamster of mass 0.139 kg gets onto his 20.8−cm-diameter exercise wheel and runs along inside the wheel for 0.823 s until its f

Anna11 [10]

Explanation:

Given that,

Mass of the hamster, m = 0.139 kg

Diameter of the wheel, d = 20.8 cm

Radius, r = 10.4 m

Frequency of the wheel, f = 1 Hz

Time, t = 0.823 s

(a) There exist a relationship between the tangential and angular acceleration of the wheel. It is given by :

$a=\alpha \times r$

Since, $\alpha =\dfrac{\omega}{t}$

$a=\dfrac{\omega}{t} \times r$

$a=\dfrac{2\pi f}{t} \times r$

$a=\dfrac{2\pi \times 1}{ 0.823} \times 10.4\times 10^{-2}$

$a=0.793\ m/s^2$

(b) In radial direction, applying the second law of motion as :

$N-mg=ma$

a is the radial acceleration, $a=\dfrac{v^2}{r}$

$N=mg+ma$

$N=mg+m(\dfrac{v^2}{r})$

$N=mg+m(\dfrac{(r\omega)^2}{r})$

$N=mg+m\omega^2 r$

$N=m(g+\omega^2 r)$

$N=m(g+(2\pi f)^2 r)$

$N=0.139\times (9.8+(2\pi 1)^2\times 10.4\times 10^{-2})$

$N=1.93\ N$

Hence, this is the required solution.

8 0

4 years ago

What is the acceleration of a ball thrown vertically upward in the presence of Earth’s

adelina 88 [10]

Answer:

-9.8 m/s/s

Explanation:

8 0

3 years ago

Consider the two-body situation at the right. A 3.50x103-kg crate (m1) rests on an inclined plane and is connected by a cable to

4vir4ik [10]

Answer: $a=2.42 m/s^2$

Explanation:

Given

mass $m_1=3.50\times 10^{3} kg$

$m_2=1.00\times 10^{3} kg$

$\theta$ (inclination)= $30^{\circ}$

$\mu =0.21$

Let T be the tension in the rope

From Diagram

$m_1g\sin \theta -T-f_r=m_1a$ -----------------1

where $f_r=friction\ force$

$f_r=\mu m_g\cos \theta$

For block $m_2$

$T=m_2a$ -----------2

From 1 & 2

$m_1g\sin \theta -m_2a-\mu m_1g\cos \theta =m_1a$

$m_1g(\sin \theta -\mu \cos \theta )=(m_1+m_2)a$

$\frac{9.8\times 3.5\times 10^3}{4.5\times 10^3}(0.5-\0.21\cos 30)=4.5a$

$a=2.42 m/s^2$

4 0

4 years ago

While watching some tv you see a circus show in which a woman drives a motorcycle around the inside of a vertical ring. you dete

Mila [183]

Since the biker is moving at constant speed

so we will have

$v = \frac{distance}{time}$

here we know that

$distance = 2\pi R$

time = 4 second

now we have

$v = \frac{2\pi R}{4} = \frac{\pi R}{2}$

now at the top point it feels free fall

so it is

$mg = \frac{mv^2}{R}$

$v^2 = Rg$

$(\frac{\pi R}{2})^2 = Rg$

$R = \frac{4g}{\pi^2}$

so height of the bike at the top is same as diameter

$h = 2R$

$h = \frac{8g}{\pi^2} = 7.95 meter$

Yes it will hurt on falling from this height

5 0

4 years ago