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3 years ago

HELP ME ASAP!!!!100 points!!!!!How do you get your fingers in the Labia?

Physics

2 answers:

Korvikt [17]3 years ago

5 0

genitals. And p,ease put appropiate questions.

Nina [5.8K]3 years ago

4 0

You literally just put your fingers in your genitals? is this for a sex ed course...?

You might be interested in

Electrons in excited hydrogen atoms are in the

Sidana [21]

I can see three different transitions here:

3 --> 1

3 --> 2
followed by
2 --> 1 .

So we should expect to see three different 'colors'
being emitted from this excited mob.

4 0

3 years ago

It is interesting to speculate on the properties of a universe with different values for the fundamental constants.

qwelly [4]

Answer:

Part a)

$\lambda = 0.345 m$

Part b)

$\Delta x = 0.274 m$

Part c)

$r = 2.8 \times 10^{11} m$

Explanation:

Part a)

De broglie wavelength is given as

$\lambda = \frac{h}{mv}$

$\lambda = \frac{1}{(0.145)(20)}$

$\lambda = 0.345 m$

Part b)

By principle of uncertainty we know that

$\Delta x \times \Delta P = \frac{h}{4\pi}$

$\Delta x \times (0.145)(21 - 19) = \frac{1}{4\pi}$

$\Delta x = 0.274 m$

Part c)

As we know that

$\frac{kq_1q_2}{r^2} = \frac{mv^2}{r}$

also we know

$mvr = \frac{nh}{2\pi}$

$v = \frac{h}{2\pi mr}$

now we have

$\frac{ke^2}{r} = \frac{mh^2}{4\pi^2m^2 r^2}$

$r = \frac{h^2}{4\pi^2mke^2}$

$r = 2.8 \times 10^{11} m$

5 0

3 years ago

How do you calculate the total pressure?

neonofarm [45]

Answer:

You could put a pressure stick against the pressure and see the pressure or estimate it from the power its coming out.

Explanation:

4 0

3 years ago

A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

telo118 [61]

Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

$\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}$ (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

$\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}$ (2)

dB/dt = -0.280T/s (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

$E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}$

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

6 0

4 years ago

suppose that the man pictured on the front side is orbiting the earth (mass=5.98 x 10^24kg at a distance of 310 miles above the

wlad13 [49]

Answer:

a = 9.81[m/s^2]; v = 18683.5[m/s]

Explanation:

The stament of the problem is:

Suppose that the man pictured on the front side is orbiting the earth (mass = 5.98 x 1024kg) at a distance of 310 miles (1600 meters = 1 mile) above the surface of the earth (radius = 4000 miles).

a. What acceleration does he experience due to the earth's pull?

b. What tangential velocity must he possess in order that he orbit safely (in m/s)?

First we need to convert all the initial data to units of the SI

Rs = 310 [mil] = 498897 [m]

RT= 400 [mil] = 643738 [m]

r = Rs + RT = 1142635 [m] "distance from the center of the earth to the man"

$F=G*\frac{M*m}{r^{2} } \\where:\\$

G = universal gravitational constant

M = mass of the earth [kg]

m = mass of the man [kg]

r = distance from the center of the earth to the man [m]

The acceleration he is experimenting is the same acceleration given by the gravity, therefore:

a = g = 9.81[m/s^2]

To find the tangential velocity, we must determinate the force exerted by the earth.

Now we will find the force exerted by the gravity when the man is orbiting the earth at distance r.

$G = 6.67*10^{-11} [\frac{N*m^{2} }{kg^{2} } ]\\M=5.98*10^{24}[kg]\\ m=100 [kg]\\Replacing:\\F = G*\frac{M*m}{r^{2} }$

$F = 6.67*10^{-11}*\frac{5.98*10^{24}*100 }{1142635^{2} } \\ F= 30550 [N]$

And this force will be equal to the following expression:

$F = m*\frac{v^{2} }{r} \\where:\\v= tangential velocity [m/s]\\v=\sqrt{\frac{F*r}{m} } \\v=\sqrt{\frac{30550*1142635}{100} } \\v=18683.5[m/s]$

8 0

3 years ago