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3 years ago

During the Water Wheel simulation you were able to change the flow of water

Physics

1 answer:

Law Incorporation [45]3 years ago

4 0

Answer:

Explanation Home Page, THE FLOW OF ENERGY OUT OF THE SUN, Please sign our mailing list ... install itself when you run it on your computer ... To show how spectral lines are formed by random processes of absorption and re-emission in its ... One set of simulations deals with the scattering of photons in the interior of a star.n:

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What was significant about the discovery of gallium

Kryger [21]

It confirmed medeleeve's hypothesis (prediction) and showed the use of his table

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3 years ago

Read 2 more answers

Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.

Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = $\frac{mv}{qB}$

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r = $\frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T}$ = 0.0328 m, or 3.28 cm.

8 0

3 years ago

A 12 volt car battery is connected to a 3 ohm brake light. What is the current carrying energy to the lights?

zloy xaker [14]

Answer:

4 A

Explanation:

V = IR, where V=voltage, I=current, R=resistance. This is Ohm's Law. (remember that for units V = volts, Ω = ohms, A = amperes.)

V = IR

12 V = I * 3 Ω

12/3 = I

8 0

3 years ago

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it i

andreyandreev [35.5K]

The apparent weight of a 1.1 g drop of water is 4.24084 N.

<h3>What is Apparent Weight?</h3>

According to physics, an object's perceived weight is a characteristic that describes how heavy it is. When the force of gravity acting on an object is not counterbalanced by a force of equal but opposite normality, the apparent weight of the object will differ from the actual weight of the thing.
By definition, an object's weight is equal to the strength of the gravitational force pulling on it. It follows that even a "weightless" astronaut in low Earth orbit, with an apparent weight of zero, has almost the same weight that he would have if he were standing on the ground; this is because the gravitational pull of low Earth orbit and the ground are nearly equal.

Solution:

N = Speed of rotation = 1250 rpm

D = Diameter = 45 cm

r = Radius = 22.5 cm

M = Mass of drop = 1.1 g

Angular speed of the water = $\omega = \frac{2\pi N}{60}$

$\omega = \frac{2\pi \times 1250}{60}$

$\omega = 130.89 rad/s$

Apparent weight is given by

$W _a = M\omega^{2}R$

$W_a = 1.1 \times 10^-^3\times (130.89)^2\times 0.225$

$W_a$ = 4.24084 N

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Question:

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?

7 0

2 years ago

Calculate the magnitude of the gravitational force exerted by the Moon on a 79 kg human standing on the surface of the Moon. (Th

Luden [163]

Answer:

F= 134.92 N

Explanation:

Given that

The mass of the moon ,M = 7.4 x 10²² kg

The mass of the man ,m = 79 kg

The radius ,R= 1.7 x 10⁶ m

The force exerted by moon is given as

$F=G\dfrac{Mm}{R^2}$

Now by putting the values in the above equation we get

$F=6.67\times 10^{-11}\times \dfrac{79\times 7.4\times 10^{22}}{(1.7\times 10^6)^2}\ N\\F=134.92 N$

Therefore the force will be 134.92 N.

F= 134.92 N

8 0

4 years ago