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KIM [24]
3 years ago
7

A white billiard ball with mass mw = 1.43 kg is moving directly to the right with a speed of v = 3.39 m/s and collides elastical

ly with a black billiard ball with the same mass mb = 1.43 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θw = 38° and the black ball ends up moving at an angle below the horizontal of θb = 52°. 1)What is the final speed of the white ball? m/s 2)What is the final speed of the black ball? m/s 3)What is the magnitude of the final total momentum of the system? kg-m/s 4)What is the final total energy of the system?

Physics
1 answer:
Anna71 [15]3 years ago
4 0

Answer: a) VW = 1.28m/s

b) Vb = 3.86m/s

c) p = 5.82kgm/s

d) E = 11.84J

Explanation: To solve this question, we make use of explosion formula in linear momentum concept.

Please find the attached file for the solution

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A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the s
yan [13]

Answer:

The  force is F_1  = 400.8 \  N

Explanation:

From the question we are told that

   The first  diameter is  d_1 =  4.0 \ cm =  0.04 \ m

   The second diameter is  d_2  =  8.0 \ cm  = 0.08 \  m

   

Generally the first area is  

         A_1  =  \pi  * \frac{d^2_1 }{4}

=>      A_1  = 3.142  * \frac{0.04^2}{4}

=>       A_1  =  0.00126 \ m^2

The  second area is  

     A_2 =  \pi  * \frac{d^2_2 }{4}

     A_2  = 3.142  * \frac{0.08^2}{4}

     A_2  =  0.00503 \ m^2

For a hydraulic press the pressure at both end must be equal .

Generally  pressure is mathematically represented as

    P =  \frac{F}{A}

=>  

   \frac{F_1}{A_1 }  =  \frac{F_2}{A_2 }

=>   F_1  =  \frac{1600}{0.00503}  *  0.00126

=>    F_1  = 400.8 \  N

8 0
3 years ago
A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of and performs 4.1 kJ of net work, and rejec
Sati [7]
There are some missing data in the problem. The full text is the following:
"<span>A </span>real<span> (</span>non-Carnot<span>) </span>heat engine<span>, </span>operating between heat reservoirs<span> at </span>temperatures<span> of 710 K and 270 K </span>performs 4.1 kJ<span> of </span>net work<span>, and </span>rejects<span> 9.7 </span>kJ<span> of </span>heat<span>, in a </span>single cycle<span>. The </span>thermal efficiency<span> of a </span>Carnot heat<span> engine, operating between the same </span>heat<span> reservoirs, in percent, is closest to.."

Solution:
The efficiency of a Carnot cycle working between cold temperature </span>T_C and  hot temperature T_H is given by
\eta = 1 - \frac{T_C}{T_H}
and it represents the maximum efficiency that can be reached by a machine operating between these temperatures. If we use the temperatures of the problem, T_C=270 K and T_H=710 K, the efficiency is
\eta = 1 - \frac{270 K}{710 K}=0.62 = 62%

Therefore, the correct answer is D) 62 %.
6 0
3 years ago
The distance that a free falling object falls is directly proportional to the square of the time it falls (before it hits the gr
Harman [31]

If the distance is proportional to the square of the time, then the distance in 9 seconds will be  (9/8)²  as long as the distance in 8 seconds.

(9/8)² = 1.265625

(64 ft) · (1.265625) = <em>81 ft. </em>

4 0
4 years ago
How much force can a 2.5 kg sledge hammer excerpt on a nail if you can swing the hammer at 20 m/s and the hammer contacts the na
AleksAgata [21]

Answer:

625 N

Explanation:

The impulse given to the nail is equal to the change in momentum of the hammer:

I=F \Delta t=m \Delta v

where

F is the force exerted by the hammer

\Delta t=0.08 s is the time of contact

m=2.5 kg is the mass

\Delta v=20 m/s is the change in velocity of the hammer

Substituting the data and re-arranging the equation, we can find the force:

F=\frac{m \Delta v}{\Delta t}=\frac{(2.5 kg)(20 m/s)}{0.08 s}=625 N

4 0
4 years ago
In your own words describe hot spot and the formation of the Hawaiian island
vesna_86 [32]
The Hawaiian Island was formation by volcanic activity
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