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Alona [7]
3 years ago
6

Precise measurements of earth indicate that its polar diameter is

Physics
1 answer:
Hatshy [7]3 years ago
3 0

Answer:

12713.6 km (7899.86 mi)

Explanation:

The polar diameter of Earth is slightly shorter that the equitorial diameter of the earth. The polar diameter to be precise is 12713.6 km (7899.86 mi), whereas the equitorial diameter is about 12756 Km (7926 mi). This mean that objects located along the equator are about 21 km further away from centwe of the Earth than objects located on the poles.

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single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev
Goryan [66]

Answer:

The sound level of the 26 geese is  Z_{26}= 96.15 dB

Explanation:

From the question we are told that

    The  sound level is Z_1 =  81.0 \ dB

     The number of geese is N = 26

Generally the intensity level of sound is mathematically represented as

        The intensity of sound level in dB  for one  goose is mathematically represented as

                       Z_1 = 10 log [\frac{I}{I_O} ]

Where I_o is the  threshold level of intensity with value  I_o = 1*10^{-12} \  W/m^2

            I is the intensity for one goose in W/m^2

For 26 geese the intensity would be  

          I_{26} = 26 * I

   Then  the intensity of 26 geese in dB is  

              Z_{26} = 10 log[\frac{26 I }{I_o} ]

               Z_{26} = 10 log (\ \ 26 *  [\frac{ I }{I_o} ]\ \ )

               Z_{26} = 10 log (\ \ 26  \ \ ) *   (\ \  10 log [\frac{ I }{I_o} ]\ \ )

 From the law of logarithm we have that

              Z_{26} = 10 log 26 +  10 log [\frac{I}{I_0} ]

                    = 14.15 + 82

                    Z_{26}= 96.15 dB

               

               

           

4 0
3 years ago
Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
Alexandra [31]

11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

4 0
3 years ago
50 POINTS!!!!!!!!<br><br> In a GUT (Grand Unified Theory), what is "unified"?
melomori [17]

Subatomic particles

Explanation:

Grand unification theory or GUT is a model that tries to describe the universe.

7 0
2 years ago
A train increase its speed steadily from 10ms to 20ms in 1 minutes
maxonik [38]

Answer:

Explanation:  nigerian what

8 0
2 years ago
Free point cuz i cewl like dat‍♀️‍♂️
DedPeter [7]
Yes thank u teehee




.................... x
8 0
2 years ago
Read 2 more answers
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