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nekit [7.7K]
3 years ago
6

Which of the following is a component of staying positive while competing ​

Physics
2 answers:
vredina [299]3 years ago
5 0

Answer:

B

Explanation:

Edge 2021

Natalija [7]3 years ago
4 0

Answer: B - complimenting others on good plays

Explanation: Reading the first words sort of give it away when staying positive you compliment, not criticize, confront angrily, or refuse.

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Lelu [443]

Answer:

okay then

Explanation:

6 0
3 years ago
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What do we mean when we say that energy levels are quantized in atoms?
jek_recluse [69]

Answer:

Electrons are located in specific orbit corresponding to discrete energy levels

Explanation:

In Bohr's model of the atom, electron orbit the nucleus in specific levels, each of them corresponding to a specific energy. The electrons cannot be located in the space between two levels: this means that only some values of energy are possible for the electrons, so the energy levels are quantized.

A confirmation of Bohr's model is found in the spectrum of emission of gases. In fact, when an electron jumps from a higher energy level to a lower energy level, it emits a photon whose energy is exactly equal to the difference in energy between the two levels: since the energy levels are discrete, this means that the emitted photons cannot have any value of wavelength, but also their wavelength will appear as a discrete spectrum. This is exactly what it is observed in the spectrum of emission of gases.

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
a cyclist coasting down a 5.0 ◦ incline at a constant speed of 6.0 km/h because of air resistance. If the total mass of the bicy
Dvinal [7]

Answer:

F_{net}= 85.41\ N

Explanation:

mass of the bicycle + cyclist = 50 kg

constant speed = 6 km/h

a cyclist coasting down a 5.0° incline

the downward velocity is constant, so net acceleration must be zero

the air drag must be equal to gravitational force downward along the ramp

F_a = mg sin \theta  

now for upward motion

F_{net} = mg sin \theta + air\ drag

F_{net} = mg sin \theta + mg sin \theta

F_{net} = 2 mg sin \theta

F_{net} = 2\times 50 \times 9.8 sin 5^0

F_{net}= 85.41\ N

3 0
4 years ago
At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field
vesna_86 [32]

Answer:

9.7\times 10^{-5} T

Explanation:

Velocity =5\times 10^4i-2\times 10^4j

r=0.03i+0.05j

r=\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058

v=\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}

We know that

B=\frac{mv}{qr}

Where q=1.6\times 10^{-19} C

Mass of proton=1.67\times 10^{-27} kg

Using the formula

B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}

B=9.7\times 10^{-5} T

3 0
3 years ago
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