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svetoff [14.1K]
4 years ago
8

If displacement per unit time is tripled, the velocity is by a factor of .

Physics
1 answer:
luda_lava [24]4 years ago
7 0
I think the velocity has tripled also. If the unit time stayed the same.
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Since force is dp/dt, the force due to radiation pressure reflected off of a solar sail can be calculated as 2 times the radiati
QveST [7]

Answer:

8.67×10⁻⁶ N/m²

Explanation:

p = Momentum of a photon

E = Energy of a photon

c = Speed of light

I = Intensity of light

Force = dp/dt

p=\frac{E}{c}

\\\Rightarrow F=\frac{\frac{dE}{dt}}{c}

As given in question

F=2\frac{\frac{dE}{dt}}{c}

Now F/A = Pressure

P=\frac{2I}{c}\\\Rightarrow P=\frac{2\times 1300}{3\times 10^{8}}\\\Rightarrow P=8.67\times 10^{-6}\ N/m^2

∴ Magnitude of the pressure on the sail is 8.67×10⁻⁶ N/m²

4 0
4 years ago
Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a
lilavasa [31]

Answer:

\lambda = 570\ nm

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

tan \theta = \dfrac{y}{L}

tan \theta = \dfrac{4}{4\times 10^3}

\theta = 0.0572

Now.

W sin \theta = m \lambda

m = 1

5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda

\lambda = 569.99 \times 10^{-9}\ m

\lambda = 570\ nm

4 0
4 years ago
The source of energy that keeps the Sun shining today is _________.
Reil [10]

The source of energy that keeps the Sun shining today is _________.

<u>Is nuclear fusion</u>

~Hope this answers your question!~

6 0
4 years ago
Help with speed problems #1-3
-Dominant- [34]
1. Each plot represents the meters traveled by both the Hare and the Tortoise over a certain period of time (minutes).

2. The Tortoise lines show it lines is steadily increasing over a period of time. So as more time elapses the faster the tortoise becomes it travels more meters. The Tortoise line shows steady acceleration.

3. The Hare in the first 5 minutes had a rapid fast advancement up to 40 meters. But for the 5-20 mins. period the Hare did not move at all. Its speed stayed at the same place. But towards the end 20-25 mins. marks the Hare started moving again. At the end the Hare at first had a rapid acceleration but stopped for a long time then it sped up briefly. 
8 0
3 years ago
The rate at which a metal alloy oxidizes in an oxygen-containing atmosphere is a typical example of the practical utility of the
Ray Of Light [21]

Answer:

The activation energy is  Q = 328.31 \ K J/mol

Explanation:

From the question we are told that

      The rate constant is  k

       at the temperature T_1  = 300 =  300 + 273 =  573 \ K

      The value of k is  k_1 = 1.05 *10^{-8} \  kg /m^4 \cdot s

      at temperature T_2 = 400 ^oC =  400 + 273 =  673 \ K

       The value of  k is  k_2 = 2.95 *10^{-4} \ kg /m^4 \cdot s

The rate constant is mathematically represented as

       k  =  Ce^{- \frac{Q}{RT} }

Where Q is the activation energy

         R is the ideal gas constant with a value of  R =  8.314 \ J /mol \cdot K

          C is a constant

           T is the temperature

For the first  rate constant

       k_1 = Ce ^{-\frac{Q}{RT_1} }

For the second   rate constant

       k_2 = Ce ^{-\frac{Q}{RT_2} }

Now the ratio between the two given rate constant is  

      \frac{k_1 }{k_2}  =  e^{(\frac{Q}{R} [\frac{1}{\frac{T_2 - 1}{T_1} } ] )}

  =>    ln [\frac{k_1}{k_2} ] =  \frac{Q}{R}  * [\frac{1}{\frac{T_2 -1}{T_1} } ]

substituting values  

       ln [\frac{1.05 *10^{-8}}{2.95 *10^{-4}} ] =  \frac{Q}{8.314}  * [\frac{1}{\frac{673 -1}{573} } ]

=>     Q = 328.31 \ K J/mol

7 0
3 years ago
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