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postnew [5]
3 years ago
15

Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of app

roximately 4.0 cm/yr. Assuming this to be a constant rate, how many years will pass before the radius of the Moon's orbit increases by 3.84x 10^7 m (10%)?
Physics
1 answer:
melomori [17]3 years ago
7 0

Answer:

967500000 years

Explanation:

The Speed at which the radius of the orbit of the Moon is increasing is 4 cm/yr

Converting to m

1 m = 100 cm

1\ cm=\frac{1}{100}\ m

4\ cm\y=\frac{4}{100}=0.04\ m/yr

The distance by which the radius increases is 3.84×10⁷ m

Time = Distance / Speed

\text{Time}=\frac{3.87\times 10^7}{0.04}\\\Rightarrow \text{Time}=967500000\ yr

967500000 years will pass before the radius of the orbit increases by 10%.

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When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
At a location near the equator, the earth’s magnetic field is horizontal and points north. An electron is moving vertically upwa
ivann1987 [24]

Answer:

(b) EAST

Explanation:

you can assume that the magnetic field points rightward, that is, in the positive x direction (NORTH). Furthermore, you can assume that the direction of the motion of the electron is in the positive y direction. Hence, you have:

\vec{B}=B_o\hat{i}\\\\\vec{v}=v_o\hat{j}

You use the Lorentz formula to known which is the direction of the magnetic force over the electron:

F=qv\ X\ B

which implies the cross product between the unitary vecors j and i, that is

\hat{i} \ X\ \hat{j} = -\hat{k}  (WEST)

However, the minus sign of the charge of the electron changes the direction 180°. Hence, the direction is k. That is, to the EAST

3 0
3 years ago
Read 2 more answers
The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an em
Vilka [71]

Answer:

\rho=995.50\ kg.m^{-3}

\bar w=9765.887\ N.m^{-3}

s=0.9955

Explanation:

Given:

  • volume of liquid content in the can, v_l=0.355\ L=3.55\times 10^{-4}\ L
  • mass of filled can, m_f=0.369\ kg
  • weight of empty can, w_c=0.153\ N

<u>So, mass of the empty can:</u>

m_c=\frac{w_c}{g}

m_c=\frac{0.153}{9.81}

m_c=0.015596\ kg

<u>Hence the mass of liquid(soda):</u>

m_l=m_f-m_c

m_l=0.369-0.015596

m_l=0.3534\ kg

<u>Therefore the density of liquid soda:</u>

\rho=\frac{m_l}{v_l} (as density is given as mass per unit volume of the substance)

\rho=\frac{0.3534}{3.55\times 10^{-4}}

\rho=995.50\ kg.m^{-3}

<u>Specific weight of the liquid soda:</u>

\bar w=\frac{m_l.g}{v_l}=\rho.g

\bar w=995.5\times 9.81

\bar w=9765.887\ N.m^{-3}

Specific gravity is the density of the substance to the density of water:

s=\frac{\rho}{\rho_w}

where:

\rho_w= density of water

s=\frac{995.5}{1000}

s=0.9955

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an immigrant

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