Soot particles ("black carbon aerosols" generally cause ________ of earth's atmosphere by _______

Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const

bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

$S = ut + \frac{1}{2}at^{2}$

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

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7 0

1 year ago

HELP QUICK!!!

RoseWind [281]

One is for far distance the other is for close range

7 0

2 years ago

What is the threshold frequency ν0 of cesium? note that 1 ev (electron volt)=1.60×10−19 j. express your answer numerically in he

andreyandreev [35.5K]

The threshold frequency fo of Cesium is 4.83 × 10¹⁴ Hz

<h3>Further explanation</h3>

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:

$\large {\boxed {E = h \times f}}$

E = Energi of A Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Eletromagnetic Wave ( Hz )

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

$\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}$

$\large {\boxed {E = qV + \Phi}}$

E = Energi of A Photon ( Joule )

m = Mass of an Electron ( kg )

v = Electron Release Speed ( m/s )

Ф = Work Function of Metal ( Joule )

q = Charge of an Electron ( Coulomb )

V = Stopping Potential ( Volt )

Let us now tackle the problem!

Given:

Φ = 2.1 eV = 2 × 1.60 × 10⁻¹⁹ Joule = 3.20 × 10⁻¹⁹ Joule

Unknown:

fo = ?

Solution:

$\Phi = h \times fo$

$3.20 \times 10^{-19} = 6.63 \times 10^{-34} \times fo$

$fo = ( 3.20 \times 10^{-19} ) \div ( 6.63 \times 10^{-34} )$

$\large {\boxed{fo = 4.83 \times 10^{14} ~ Hz} }$

<h3>Learn more</h3>

Photoelectric Effect : brainly.com/question/1408276

Statements about the Photoelectric Effect : brainly.com/question/9260704

Rutherford model and Photoelecric Effect : brainly.com/question/1458544
Photoelectric Threshold Wavelength : brainly.com/question/10015690

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt

8 0

2 years ago

Read 2 more answers

imagine that you are on a spaceship that is far from the sun or any planet. Explain why a 500 kg piece of equipment would be dif

Mila [183]

Even though the object is weightless, it would need inertia, I.e, you pushing it or any form of transportation. So you would still have to push that 500kg to just keep it moving in space. Say if it were a planet with less gravitational force, it would be weightless.

8 0

2 years ago

A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and

Troyanec [42]

The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

• net vertical force:

∑ F = N - W = 0

• net horizontal force:

∑ F = Fs = m a

where

N = magnitude of normal force

W = car's weight

Fs = mag. of static friction

m = car's mass

a = v ²/R = mag. of the centripetal acceleration

v = car's speed

R = radius of curve

Now,

• compute the car's weight:

W = m g = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

N = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

Fs = 0.5 N = 7350 N

• solve for the (maximum) acceleration:

7350 N = (1500 kg) a → a = 4.9 m/s²

• solve for the (maximum) speed:

4.9 m/s² = v ²/ (35 m) → v ≈ 13 m/s

4 0

3 years ago

Soot particles ("black carbon aerosols" generally cause ________ of earth's atmosphere by ________ solar energy.

Soot particles ("black carbon aerosols" generally cause of earth's atmosphere by solar energy.