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valkas [14]
2 years ago
8

Football player A has a mass of 210 pounds and is running at a rate of 5.0 mi/hr. He collides with player B. Player B has a mass

of 190 pounds and is running in the same direction at 3.0 mi/hr. Which of the following statements is true?
-The momentum of player A is equal to that of player B.
-The momentum of player A is greater than that of player B.
-The momentum of player A is less than that of player B.
Physics
2 answers:
Naddika [18.5K]2 years ago
6 0

The equation for momentum is p = mv where p is the omentum, m is the mass and v is the velocity. Calculating the momentum for each football player, player A will have a momentum of 1050 lb-mi/h and player B will have a momentum of 570 lb-mi/h. Therefore, momentum of player A is greater than that of player B.

IgorLugansk [536]2 years ago
3 0

Answer:

the momentum of player A is greater than player B.

Therefore, Option 2 is correct.

Step-by-Step explanation:

We have been given the velocity of the players and mass of the players.

Player A:

Velocity is 5m/hr

Mass 210 pounds

Player B:

Velocity is 3m/hr

Mass 190 pounds

We have a formula for momentum:

p=mv

p is momentum

m is mass

v is velocity

We will  find momentums for both the players:

Momentum of player A:

on substituting the values we get:

p=210\cdot 5=1050 pounds  

Momentum of player B:

on substituting the values we get:

p=190\cdot 3=570 pounds  

Hence, the momentum of player A is greater than player B.

Therefore, Option 2 is correct.




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Answer:

The distance traveled during its acceleration, d = 214.38 m

Explanation:

Given,

The object's acceleration, a = -6.8 m/s²

The initial speed of the object, u = 54 m/s

The final speed of the object, v = 0

The acceleration of the object is given by the formula,

                                      a = (v - u) / t   m/s²

       ∴                              t = (v - u) / a

                                         = (0 - 54) / (-6.8)

                                         = 7.94 s

The average velocity of the object,

                                       V = (54 + 0)/2

                                           = 27 m/s

The displacement of the object,

                                 d = V x t   meter

                                    = 27 x 7.94

                                    = 214.38 m

Hence, the distance the object traveled during that acceleration is, a = 214.38 m

3 0
2 years ago
A string, 0.28 m long and vibrating in its third harmonic, excites an open pipe that is 0.82 m long into its second overtone res
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Answer: 98.

Explanation: it has been stated in the question that sound wave in the string and in the pipe resonated at a specific frequency, this simply implies that the frequency of sound wave in the string equals frequency of sound wave in pipe.

Fs = Fp.

The length (l) of the string is 0.28m and it is vibrating at it third harmonic.

The length of stationary wave on a string at third harmonic is given below as

l = 3λ/2

Where λ = wavelength of sound wave in pipe (λs)

By substituting l = 0.28m into the equation above, we have that

0. 28 = 3λs/2

3λs = 0.28 * 2

3λs = 0.56, λs= 0.56/ 3

λs = 0.187m

Thus the wavelength of wave in the string is 0.187m.

Sound from the string in the pipe is produced at the second overtone ( which is the third harmonic).

Therefore the length of air in the pipe at second overtone ( third harmonic) is given below as

l = 5λp/ 4, we need to get the wavelength of sound in the pipe.

Thus

λp = 4*l/5

λp = 4 * 0.82 / 5

λp = 0.656m.

The velocity of sound waves produced in the pipe is 345m/s thus the frequency of sound in the pipe is gotten using the formulae below

V = fpλp

V= velocity of sound in pipe, fp = frequency of sound in pipe, λp= wavelength of sound in pipe

345 = f / 0.656

fp = 525.92Hz.

As stated in the question, the frequency of sound in pipe is the same as that in string (fp = fs = f) , thus to get the velocity of sound wave in string we use the same formulae of

v = fλ

Where f = frequency of sound in pipe = frequency of sound in string = 525.92Hz.

λ = wavelength of sound in string = 0.187m

Thus v = 525.92 * 0.187 = 98.34 which is closest to 98.

3 0
2 years ago
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Answer:

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Explanation:

a) By Newton's second law we can find the relation between force and acceleration of the SUV:

F=ma

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.

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A cyclist goes around a circular track once every 2 minutes. If the radius of the circular track is 105 meters calculate his spe
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Answer

given,                                                                      

angle between two polarizing filters = 45°

filter reduce intensity = ?                          

a) I = I₀ Cos² θ                                

here θ = 45⁰                                

I = \dfrac{I_0}{2}                      

intensity of the light is reduced by 0.500

correct answer from the given option D

b) direction of the polarization                    

                        θ = 45°                  

7 0
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