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Mrrafil [7]
3 years ago
8

A wave emitted from a source has a frequency of 10 Hz and wavelength 2.5 m. How much time will it take to reach a person located

5 m from the source?
Physics
1 answer:
const2013 [10]3 years ago
4 0

Answer:

time taken by the wave to reach the person is 0.2 s

Explanation:

As we know that the speed of the wave is given as

v = \lambda f

here we know that the wavelength of the wave is

\lambda = 2.5 m

f = 10 Hz

now speed of the wave is given as

v = 10(2.5)

v = 25 m/s

Now time taken by the wave to reach 5 m distance is

t = \frac{L}{v}

t = \frac{5}{25}

t = 0.2 s

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Calculate the kinetic energy of a 64.g bullet moving at a speed of 411.ms . Round your answer to 2 significant digits.
Mandarinka [93]

Answer:

The kinetic energy of the bullet is 5.4 × 10³ J

Explanation:

Hi there!

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the bullet.

v = speed of the bullet.

Let´s convert the mass unit to kg so that our result is in Joules:

64 g · ( 1 kg / 1000 g) = 0.064 kg

Then, the kinetic energy will be the following:

KE = 1/2 · 0.064 kg · (411 m/s)²

KE = 5.4 × 10³ J

8 0
3 years ago
Is this right? plzz anwser soon
-Dominant- [34]

Answer:

yes

Explanation:

6 0
2 years ago
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A 70kg man runs at a pace of 4 m/s and a 50g meteor travels at 2 km/s. Which has the most kinetic energy?.
kirill115 [55]

Answer:

the 70kg man

Explanation:

because he has more weight and is moving faster

4 0
3 years ago
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

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3 years ago
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Gnoma [55]

Answer:

Bile helps in the digestion of fats

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3 years ago
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