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-Dominant- [34]
3 years ago
11

What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make

a larger set of waves in the same direction as that of the smaller waves. A label Wave motion is above the series of waves and an arrow next to the label points right. The particles will move up and down over large areas. The particles will move up and down over small areas. The particles will move side to side over small areas. The particles will move side to side over large areas.
Physics
1 answer:
Sonja [21]3 years ago
4 0

Answer:

The transverse wave

Explanation:

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A cyclist turns a corner with a radius of 50m at a speed of 10m/s. What is the cyclist's acceleration?
garri49 [273]

Answer:

2 m/s^2

Explanation:

a = v^2/r

a = (10m/s)^2 / 50m

a = 2 m/s^2

Leave a like and mark brainliest if this helped

Leave a like and mark brainliest if this helped

5 0
3 years ago
Draw the following patterns as they would appear when viewed through a compound microscope.:
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(marking brainliest) pease help asap! both of the questions are in the pdf, and please let me know which is question one and whi
REY [17]
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<em>1) The relationship in between the electrical energy carriesd by the transmission wires and the amount of the heat loss in it is due to the reason that when the electricity is flown through the wires there are some resistance found in these wires which creates a disturbance in the efficient flow of electricty.Also we know that current have an heating effect when it is in motion as due to if a large amount or magnitude of electricity is flown through the transmission wires it will carry a larger heat effected and also due to the resistance is provided by the wires and so the process of heat loss takes place.</em>

<em>2)It is important to minimize current in transmission wires due to minimize the heat loss and resistance on flowing electric current to make the system more efficient </em>

<em><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u></em><em> 3)Given Resistance = 250 ohms </em>

<em>Electric potential = 150 volts </em>

<em>so we know Power = </em>

<em>volt^2/Resistance = </em>

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<em>4)Heat energy (H) = Power(P)×Time(t)</em>

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4 0
3 years ago
An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1
Ksivusya [100]
<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

                 F=\frac{GMm}{r^2}

           Where G =  6.67 x 10⁻¹¹ N m²/kg²

                       M = Mass of body 1

                       M = Mass of body 2

                       r = Distance between them

Here we have

                 M = Mass of Sun = 1.99×10³⁰ kg

                 m = Mass of asteroid = 4.00×10¹⁶ kg

                 F = 3.14×10¹³ N

Substituting

                   F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m

The asteroid is 4.11 x 10¹¹ m far from Sun

3 0
3 years ago
If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa
shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0
3 years ago
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