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Vladimir79 [104]
3 years ago
8

PLEASE HELP Scientists have observed that the average sunspots have increased in the last 50 years. Which of these statements be

st describes an effect of
this increase in sunspots?
A Earth's temperature has become cooler.
B Earth's temperature has not changed over time.
C Earth's temperature has become warmer
D Earth's temperature has reduced by fifty times.
Physics
1 answer:
Serga [27]3 years ago
6 0

Answer:

option c is correct, earth temp increase

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Question C) needs to be answered, please help (physics)
Zarrin [17]

(a) Differentiate the position vector to get the velocity vector:

<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>

<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>

<em></em>

(b) The velocity at <em>t</em> = 2.00 s is

<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>

<em></em>

(c) Compute the electron's position at <em>t</em> = 2.00 s:

<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>

The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:

||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that

tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s)   ==>   <em>θ</em> ≈ -79.4º

or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.

3 0
3 years ago
An object of mass 2 kg starts from rest and is allowed to slide down a
WARRIOR [948]

Answer: B

Explanation:

Given that an object of mass 2 kg starts from rest and is allowed to slide down a frictionless incline so that its height changes by 20 m. 

The parameters given from the question are:

Mass M = 2kg

Height h = 20m

Let g = 9.8m/s^2

At the bottom of the incline plane, the object will experience maximum kinetic energy.

From conservative of energy, maximum K.K.E = maximum P.E

Maximum P.E = mgh

Maximum P.E = 2 × 9.8 × 20 = 392 J

But

K.E = 1/2mv^2

Substitute the values of energy and mass into the formula

392 = 1/2 × 2 × V^2

V^2 = 392

V = sqrt( 392 )

V = 19.8 m/s

V = 20 m/s approximately

7 0
3 years ago
What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa
dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}

E=3.315\times10^{-17}\ J

(b).  We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}

E=6.709\times10^{-21}\ J

(c). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}

E=3.315\times10^{-11}\ J

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}

E=6.709\times10^{-9}\ J

Hence, This is the required solution.

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I think it's just 20/20
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There is a threshold frequency for each metal, and only light of a frequency higher than this threshold causes electrons to be emitted from the metal surface.
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