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3 years ago

When are the two days of equinox

Physics

2 answers:

galina1969 [7]3 years ago

5 0

March 20 and September 22

oee [108]3 years ago

4 0

- around March 21
- around September 22

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A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

$\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s$

$\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM$

Therefore the speed in RPM will be 62.64 RPM.

5 0

2 years ago

Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of

mart [117]

Answer:

The value is $B = 3.33 *10^{-6} \ T$

Explanation:

From the question we are told that

The distance of separation is $d = 0.6 \ m$

The current on the one wire is $I_1 = 9 \ A$

The current on the second wire is $I_2 = 4 \ A$

Generally the magnitude of the field exerted between the current carrying wire is

$B = B_1 - B_2$

Here $B_1$ is the magnetic field due to the first wire which is mathematically represented as

$B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}$

Here $d_1$ is the distance to the half way point of the separation and the value is

$d_1 = 0.3 \ m$

$B_2$ is the magnetic field due to the first wire which is mathematically represented as

$B_2 = \frac{\mu_o * I_2 }{2 \pi * d_2}$

Here $d_2$ is the distance to the half way point of the separation and the value is

$d_2 = 0.3 \ m$

This means that $d_1 = d_2 = a = 0.3$

$B = \frac{\mu_o * I_1 }{2 \pi * d_1} - \frac{\mu_o * I_2 }{2 \pi * d_2}$

=> $B = \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }$

=> $B = \frac{ 4\pi * 10^{-7} * (9- 4)}{2 * 3.142 *0.3 }$

=> $B = 3.33 *10^{-6} \ T$

5 0

2 years ago

When finding the upper bound of the density, you put what number in the denominator?

lilavasa [31]

Answer: The result of "the upper bound of the density" does not go on the denominator.
So simplified, no. The answer is no.

3 0

3 years ago

Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin

Sergio [31]

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

= - GMm/ (R+h)

Potential difference

= GMm/ R - GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.

Extra energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

8 0

3 years ago

What’s the answer to 12

Schach [20]

Speed is v = d/t

Or speed is distance over time

So...

40min / 60min = 0.6667 or 2/3 --> Finding what proportion 40 minutes is to an hour or 60 minutes as we need the units of hours to match up

45km/h = d/0.6667h

d = (45)(0.667)

d = 30.0015 or 30km

4 0

3 years ago

Read 2 more answers