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2 years ago

An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50

Physics

1 answer:

Setler79 [48]2 years ago

6 0

Answer:

very hard others will answer it

Explanation:

hard

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When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted p

jarptica [38.1K]

Answer:

1.76 eV

Explanation:

Maximum kinetic energy of the emitted photo electrons = energy of the photon- work function of the metal

K.E' = (hc/λ)-∅.................. Equation 1

Where K.E' = maximum kinetic energy of the emitted photo electrons, h = Planck's constant, c = speed, λ = wave length, ∅ = work function of the metal.

make ∅ the subject of the equation

∅ = (hc/λ)-K.E'.................. Equation 2

Given: h = 6.63×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 400 nm = 400×10⁻⁹ m, K.E' = 1.1 eV = 1.1(1.602×10⁻¹⁹) J = 1.7622×10⁻¹⁹ J

Substitute into equation 2

∅ = [6.63×10⁻³⁴(3×10⁸)/400×10⁻⁹ ]-1.7622×10⁻¹⁹

∅ = (4.973×10⁻¹⁹)-(1.7622×10⁻¹⁹)

∅ = 3.21×10⁻¹⁹ J.

The maximum kinetic energy of the photo electrons when the wave length is 330 nm is

K.E' = [6.63×10⁻³⁴( 3×10⁸ )/330×10⁻⁹]-(3.21×10⁻¹⁹)

K.E' = (6.03×10⁻¹⁹)-(3.21×10⁻¹⁹)

K.E' = 2.82×10⁻¹⁹ J

K.E' = 2.82×10⁻¹⁹/1.602×10⁻¹⁹

K.E' = 1.76 eV

7 0

3 years ago

At which point in the diagram above is the kinetic energy the greatest

Mars2501 [29]

Answer:

its a or more probably e

Explanation:

the ball goes back and forward in kinetic energy.

3 0

3 years ago

Do not have definite size and always take the shape of their contanier

sladkih [1.3K]

Answer:

water or h2o

Explanation:

7 0

3 years ago

A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha

Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

$PQ = \sqrt{4^{2}+3^{2}}=5 m$

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

$E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C$

Electric field at P due to the charge q is

$E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C$

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

$E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C$

3 0

3 years ago

Number 2<br> Is the answer is right?

garik1379 [7]

In question 2
The second line of equation would be 54 = 108 + 10a
get the rest from that

6 0

3 years ago