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DochEvi [55]
3 years ago
5

Becky has a shower and uses 20,000g of water with a specific heat capacity of 4200J/KG. When the water is supplied with 336,000J

of energy, it heats up to 50^c. What is the startig temperature of the water?
Physics
1 answer:
lara [203]3 years ago
7 0

Answer:

46°C

Explanation:

q = mCΔT

336,000 J = (20 kg) (4200 J/kg/°C) (50°C − T)

T = 46°C

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Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1
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Temperature after ignition=7883.205 K

Explanation:

The number of moles is,

n=PV/RT

=(1.18x10^6)(47.9x10^-6)/8.314(325)

= 0.0209 moles

a) In this process volume is constant

Q=U

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dT= Q/nCv

=1970/(1.5x8.314)(0.0209)

= 7558.205 K

The final temperature is,

= 7558.205+325

= 7883.205 K

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Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and
lawyer [7]

1 kilometre=1000 metre

      1 hour = 3600 second

       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


3 0
3 years ago
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