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nlexa [21]
2 years ago
11

9. If the frequency of a certain light is 3.8 x 1024 Hz, what is the energy of this light?

Physics
1 answer:
german2 years ago
4 0

Answer:

E=hf

Were, h = Planck constant 6.67*10^11

E=3.8*10^24 * 6.67*10^11= 2.508*q0^36j

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How long will your trip take if you travel 4000 m at an average speed of 8m/s
SpyIntel [72]
  • Answer:

<em>500 sec</em>

<em>8 min 20 sec</em>

  • Explanation:

<em>Hi there !</em>

<em />

<em>8 m ................ 1 s </em>

<em>4000 m ........ x s</em>

<em>x = 4000m×1s/8m = 500 sec = 8 min 20 sec</em>

<em />

<em>Good luck ! </em>

8 0
3 years ago
Read 2 more answers
How much soil is in a 5 cm to 9 cm deep hole?
vlabodo [156]
There is no soil in a hole

;)
6 0
3 years ago
Name nine elements that have been around so long that we don't even know when they
lora16 [44]

Answer:

gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon

Explanation:

gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon

-  (These are the most longest elements that have been around very long )

7 0
3 years ago
Read 2 more answers
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

5 0
3 years ago
Question 1 of 10
Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field E at a distance R from a charge Q is given by

E = k\dfrac{Q}{R^2}

where k = 9*10^9Nm/C is the coulomb's constant.

Now, in our case

R = 0.0075m

Q = 0.0035C;

therefore,

E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}

\boxed{E = 5.6*10^{11}N/C.}

which is choice C from the options given<em> (at least it resembles it).</em>

6 0
3 years ago
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