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Nana76 [90]
3 years ago
15

Which equation shows the relationship of wave speed to

Physics
1 answer:
BaLLatris [955]3 years ago
5 0

Answer:

Answer is A

Speed = Wavelength × Frequency

Explanation:

Googled it

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A plasma wave moving through a plasma has a frequency of 109 Hz and a speed of 3.0 × 107 m/s. What is the wavelength of this wav
tigry1 [53]

Answer:0.27*10^{6} (m)

Explanation:

with a wave we know that  speed= wavelenght* frecuency   so

wavelenght= 3*10^{7} / 109  = 0.27*10^{6} (m)

8 0
3 years ago
Uma massa de 500 Kg desloca-se com velocidade 58 km por hora. Calcule o módulo de sua quantidade por movimento
Simora [160]

The momentum of the object is 8050 kg m/s

Explanation:

The momentum of an object is defined as

p=mv

where

p is the momentum

m is the mass

v is the velocity of the object

For the object in this problem, we have

m = 500 kg is its mass

v = 58 km/h is its velocity

Converting the velocity into m/s,

v=58 \frac{km}{h}\cdot \frac{1000 m/km}{3600 s/h}=16.1 m/s

Therefore now we can find the momentum of the object:

p=(500)(16.1)=8050 kg m/s

Learn more about momentum:

brainly.com/question/7973509

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7 0
4 years ago
A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,
Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

Explanation:

The free-body  diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

  • the peripheral velocity that is directed downward (-V_y) along the y-axis
  • the linear velocity (V_x) that is directed along the x-axis

Now;

V_x = \frac{d}{dt}(12t^3+2) = 36 t^2

V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s

Also,

-V_y = R* \omega

where \omega(angular velocity) = \frac{d\theta}{dt} = \frac{d}{dt}(8t^4)

-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

3 0
4 years ago
"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attach
Lisa [10]

Answer:

<em>The time interval required to lift the spacecraft to this specified height is 123.94 seconds</em>

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2  = <em>14691.456 J  </em> is lost to friction.

Total useful mechanical energy =  81619.2 J - 14691.456 J = 66927.74<em> J</em>

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

P = \frac{w}{t}

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = <em>123.94 seconds</em>

8 0
4 years ago
Two automobiles of equal mass approach an intersection. One vehicle is traveling with speed 13.0 m/s toward the east, and the ot
puteri [66]

Answer:

Explanation:

We shall apply law of conservation of momentum to know the Speed of northward moving vehicle before collision to check the veracity of driver's statement .

Let v be the velocity of composite mass after collision

Applying law of conservation of momentum in north direction

m v₂ = 2m v sin55.08

Applying law of conservation of momentum in east  direction

m x 13 = 2m v cos55.08

Dividing these two equations

v₂ / 13 = tan55.08

v₂ = 13  tan55.08

= 18.62 m/s

= (18.62 x60 x 60) / 1000

= 67 km/h

= 67 x 5/8 mi/h

= 42 mi/h

So he is lieing.

7 0
4 years ago
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