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Gre4nikov [31]
3 years ago
14

During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of

74.0 kg and its horizontal component of velocity is 7.00 m/s just before the 65.0 kg performer catches it. If the performer is initially motionless on nearly frictionless roller skates, what is his speed immediately after catching the cannon ball
Physics
1 answer:
Ipatiy [6.2K]3 years ago
4 0

Answer:

m v1 = (m + M) v2

v2 = m v1 / (m + M)

v2 = 7 * 74 / (74 + 65)

3.73 m/s

74 kg is too heavy for the cannonball (over 150 lbs)

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Natali [406]
He moved 2 km
I speak arabik ;)

8 0
3 years ago
Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter
kaheart [24]

The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is

T_e=2\pi \sqrt{\frac{L}{g_e}}

where

g_e = 9.8 m/s^2 is the acceleration of gravity on Earth

The period of the pendulum on the Moon is

T_m=2\pi \sqrt{\frac{L}{g_m}}

where

g_m = 1.6 m/s^2 is the acceleration of gravity on the Moon

Calculating the ratio of the period on the Moon to the period on the Earth, we find

\frac{T_m}{T_e}=\frac{g_e}{g_m}=\frac{9.8}{1.6}

Therefore, for every hour interval on Earth, the Moon clock will display a time of

A) (9.8/1.6)h

#LearnwithBrainly

6 0
3 years ago
Lf a dropped hammer's velocity increases from 0.Omis to 15.0 m/s in 4.04s
Slav-nsk [51]

Answer: The acceleration due to gravity on the surface of the Moon is approximately 1.625 m/s2, about 16.6% that on Earth's surface or 0.166 ɡ.

Heres your answer.

6 0
3 years ago
You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be
Alex

Answer:

The amplitude  is  A =  90.2 \ m

Explanation:

From the question we are told that

    The frequency of when sound is approaching observer is   f = 392 Hz

     The frequency as the move away from observer  is  f_ a =  330 \ Hz

    The time between the pitch are t =  10 \ s

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

       T =  2 t

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

          T = \frac{2 \pi}{w}

Where  w is the angular velocity

=>   \frac{2 \pi}{w}  =  2 * 10

=>   w =  0.314 \ rad/sec

Now using Doppler Effect,

   The source of the sound is approaching the observer

The

          f = f_o (\frac{v}{v- wA} )

         392  = f_o (\frac{v}{v- wA} )

Where A is the amplitude

    So when the source is moving away from the observer

         f_a =  f_o (\frac{v}{v+ wA} )  

        330  =  f_o (\frac{v}{v+ wA} )  

Here  f_o is the fundamental frequency

Dividing the both equation  we have

           \frac{392}{330}  =  \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}

           1.1878  = \frac{v+wA}{v-wA}

         1.1878 v -  1.1878 wA = v+wA

        1.1878 v = 2.1878 wA

=>     A =  \frac{(0.1878 * (330))}{(2.1878)* (0.314)}

         A =  90.2 \ m

7 0
3 years ago
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ivanzaharov [21]
The correct answer for the question that is being presented above is this one: "(a)4." Suppose that during any period of 1/4 second there is one instant at which the crests or troughs of component waves are exactly in phase and maximum <span>reinforcement occurs, in 1 second, there will be 4 beats.</span>
8 0
3 years ago
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