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andrew11 [14]
2 years ago
10

If wavelength and speed of a wave are 4 m and 332 m/s respectively, calculate its frequency​

Physics
1 answer:
Furkat [3]2 years ago
4 0

Explanation:

<em>Given </em>

<em>wavelength </em><em>=</em><em> </em><em>4</em><em> </em><em>m</em>

<em>speed </em><em> </em><em>=</em><em> </em><em>3</em><em>3</em><em>2</em><em> </em><em>m/</em><em>s</em>

<em>frequency </em><em>=</em><em> </em><em>?</em>

<em>We </em><em>know </em><em>we </em><em>have </em><em>the </em><em>formula </em>

<em>wavelength</em><em> </em><em>=</em><em> </em><em>speed </em><em>/</em><em> </em><em>frequency </em>

<em>4</em><em> </em><em>=</em><em> </em><em>3</em><em>3</em><em>2</em><em> </em><em>/</em><em> </em><em>frequency </em>

<em>frequency </em><em>=</em><em> </em><em>3</em><em>3</em><em>2</em><em>/</em><em>4</em>

<em>Therefore </em><em> </em><em>frequency </em><em>is </em><em>8</em><em>3</em><em> </em><em>Hertz </em><em>.</em>

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The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone'
Helen [10]
We can use 3rd equation of motion to find the velocity of the stone just before it strikes the ground.

Height = S = 318 meters
Acceleration = a= 9.8 m/s²
Initial velocity = u = 0
Final velocity = v = ?

According to the 3rd equation of motion:

2aS = v² - u²

2(9.8)(318)=v²

v²=6232.8

⇒

v = 78.95 m/s

So, the velocity rounded of to nearest integer will be 79 meters per second.

Thus, C option is the correct answer
5 0
3 years ago
Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From the frame of reference of car 1, what is the veloci
EleoNora [17]
We subtract the velocity of car 1 from the velocity of car 2:
v=(30\ mph\ North)-(20\ mph\ South)
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4 0
2 years ago
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What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?
EleoNora [17]
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4 0
2 years ago
What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on
Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is  252.52\ m/s^2.

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = \frac{2500}{9}\ m/s

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

\therefore 0=\frac{2500}{9} + a(1.1)

\Rightarrow a=-\frac{2500}{9(1.1)}

\therefore a = - 252.52 \ m/s^2

Hence , the deceleration of the rocket is  252.52\ m/s^2.

To learn more about Attention here:

brainly.com/question/28500124

#SPJ4

6 0
1 year ago
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc
ludmilkaskok [199]

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

  • Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

       \Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2}  (1)  

  • Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       \omega_{f1}  = \omega_{o} + \alpha * \Delta t  (2)

  • Since the machine starts from rest, ω₀ = 0.
  • We know the value of ωf₁ (the operating speed) in rev/min.
  • Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

       3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)

  • Replacing by the givens in (2):

       57.5 rev/sec = 0 + \alpha * 6 s  (4)

  • Solving for α:

       \alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)

  • Replacing (5) and Δt in (1), we get:

       \Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev  (6)

  • in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       \Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev  (7)

  • In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       \Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2}  (8)

  • First of all, we need to find the value of the angular acceleration during the second period.
  • We can use again (2) replacing by the givens:
  • ωf =0 (the machine finally comes to an stop)
  • ω₀ = ωf₁ = 57.5 rev/sec
  • Δt = 32 s

       0 = 57.5 rev/sec + \alpha * 32 s  (9)

  • Solving for α in (9), we get:

       \alpha_{2}  =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)

  • Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

        \Delta \theta_{2}  = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)

  • In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
  • \Delta \theta_{2h}  = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)
7 0
2 years ago
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