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3 years ago

How does heat move through the atmosphere?

Physics

1 answer:

hodyreva [135]3 years ago

6 0

Answer:

conduction radiation and convection helps it be moved

Explanation:

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A 1650 kg car accelerates at a rate of 4.0 m/s^2. How much force is the car's engine producing?

Westkost [7]

F = M A

Force = (mass) x (acceleration)

= (1,650 kg) (4 m/s²) = 6,600 kg-m/s² = <em>6,600 Newtons</em>

7 0

3 years ago

Read 2 more answers

Blood is 92% water. Blood is

Pepsi [2]

Answer:

Plasma, which constitutes 55% of blood fluid, is mostly water (92% by volume), and contains proteins, glucose, mineral ions, hormones, carbon dioxide (plasma being the main medium for excretory product transportation), and blood cells themselves.

Explanation:

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2 years ago

Landslides are most common in

nikitadnepr [17]

shorelines of the southeast U.S.

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3 years ago

What is the magnitude of the electric field strength between them, if the potential 7.95 cm from the zero volt plate (and 2.05 c

Dmitriy789 [7]

Answer:

ΔVab = Ed

ΔVab = Va-Vb = Va-V0 = Va

E = Va/ d

= 413V / 0.0795 m

= 5194.97 V/M

Explanation:

the potential difference between two uniform plates is calculated by the formula of electric field.

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2 years ago

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