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Burka [1]
3 years ago
5

Boyle's Law mainly involves _______.

Physics
1 answer:
goblinko [34]3 years ago
6 0
Your answer is B, gases
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"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its
prisoha [69]

Answer:

1.3m/s

Explanation:

Data given,

Mass,m=1.0kg,

Amplitude,A=0.10m,

Frequency,f=2.0Hz.

From the equation of a simple harmonic motion, the displacement of the object at a given time is define as

x=Acos\alpha \\

we can express the velocity by the derivative of the displacement,

Hence

V=-Awsin\alpha \\

at equilibrium, the velocity becomes

V=wA\\w=2\pi f

Hence if we substitute values we arrive at

V=2\pi fA\\V=2\pi *2*0.1\\V=1.3m/s

4 0
3 years ago
Read 2 more answers
A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go
nata0808 [166]

<u>Answer:</u>

Golf ball will go a maximum of 270.36 meter.

<u>Explanation:</u>

  Projectile motion has two types of motion Horizontal and Vertical motion.

 Vertical motion:

          We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

          Considering upward vertical motion of projectile.

          In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

         0 = u sin θ - gt

          t = u sin θ/g

     Total time for vertical motion is two times time taken for upward vertical motion of projectile.

     So total travel time of projectile = 2u sin θ/g

Horizontal motion:

   We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

   In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

  So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

  Now in the given problem

     A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

     u = 51.5 m/s, for maximum range θ = 45⁰

   So maximum distance reached = \frac{51.5^2sin(2*45)}{9.81}=270.36 meter

So it will go a maximum of 270.36 meter.

5 0
3 years ago
What effect do sound waves have on air particles
valkas [14]

Answer:

Sound waves travel at 343 m/s through the air and faster through liquids and solids. The waves transfer energy from the source of the sound, e.g. a drum, to its surroundings. Your ear detects sound waves when vibrating air particles cause your ear drum to vibrate. The bigger the vibrations the louder the sound.

Explanation:

7 0
3 years ago
All simple machines are types of
taurus [48]

Answer: Metal. There are six of them.  inclined plane, the wedge, the screw, the lever, the wheel and axle, and the pulley.

Hope this helps!

~Jarvis

Explanation:

6 0
3 years ago
Prima problema este urgent
Aleonysh [2.5K]

Answerana alyom kint gahda amshi

Explanation:

5 0
2 years ago
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