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Aloiza [94]
3 years ago
12

A billiard ball moving at 0.5 m/s strikes another identical billiard ball, which is at rest, in an elastic head-on collision on

a level table. Which of the following statements best describes the billiard balls after the collision?
Physics
1 answer:
NikAS [45]3 years ago
4 0

Answer: assuming that the billiard balls are of identical weight the impacted billiard ball will move forward at around 0.5m/s (not considering energy conservation). The ball impacting the 2nd one would stop because most of its Kinetic energy would have been transferred into the not moving ball.

Explanation: hope this helps!

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A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same mome
lozanna [386]

A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same momentum?A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same momentum?

8 0
4 years ago
Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke
umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

      M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

       thrust = 5.26 N

       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

5 0
3 years ago
An eccentric inventor attempts to levitate by first placing a large negative charge on himself and then putting a large positive
GuDViN [60]

Answer:

shirt is little attached to the body, it can come off and fly away

Explanation:

In electrostatics, charges of different signs attract and charges of the same sign repel.

In this case, when a negative charge is placed on it, both the inventor and the shirt are charged, therefore there is a repulsive force, also there is an attraction between the positive charge of the roof attracts the negative charge, such as the shirt. of weak the two forces not greater than the resistance of the walk.

As the shirt is little attached to the body, it can come off and fly away

3 0
3 years ago
A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic fr
Licemer1 [7]

Complete Question

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz.

(a)

What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding

(b) What wavelength is detected by a person on the platform as the train approaches?

 

Answer:

a

  \Delta  f  =  81.93 \ Hz

b

  \lambda_1 =  0.867 \ m

Explanation:

From the question we are told that

      The speed of the train is  v_t  =  39.6 m/s

      The frequency of the train horn is  f_t =  350 \ Hz

Generally the speed of sound has a constant values of  v_s  =  343 m/s

  Now  according to dopplers equation when the train(source) approaches a person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as  

        f_1 =  f *   \frac{v_s}{v_s - v_t}

substituting values

        f_1 =  350 *  \frac{343 }{343-39.6}

       f_1 =  395.7 \ Hz

  Now  according to dopplers equation when the train(source) moves away from  the  person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as  

           f_2 =  f *   \frac{v_s}{v_s +v_t}

substituting values

        f_2 =  350 *   \frac{343}{343  + 39.6}

       f_2 =  313.77 \ Hz

The overall change in frequency is detected by a person on the platform as the train moves from approaching to receding is mathematically evaluated as

        \Delta  f  =  f_1 - f_2

        \Delta  f  =  395.7 - 313.77

        \Delta  f  =  81.93 \ Hz

Generally the wavelength detected by the person as the train approaches  is mathematically represented  as

          \lambda_1 =  \frac{v}{f_1 }

          \lambda_1 =  \frac{343}{395.7 }

         \lambda_1 =  0.867 \ m

4 0
4 years ago
A 2,200-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.40 m before coming into contact
love history [14]

Answer:

The magnitude of Force is 8.58×10⁵N and direction is upwards

Explanation:

The work beam does on the pile driver is given by

W=(FCos180°)Δx= -F(0.088m)

From work energy theorem

W_{nc}=(KE_{f}-KE_{i})+(PE_{f}-PE_{i})\\W_{nc}=1/2m(vf^{2}-vi^{2})+mg(yf-yi)

Choosing y=0 at the the level where the driver first contacts the beam and vi=0 at yi=+3.40m and comes to rest again vf=0 at yf= -0.088m

So

-F(0.088m)=1/2m(0-0)+2,200kg(9.81m/s^{2} )(-0.088m-3.40m)\\-F(0.088)=-75508.224\\F=75508.224/0.088\\F=8.58*10^{5} N

The magnitude of Force is 8.58×10⁵N and direction is upwards

6 0
4 years ago
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