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sdas [7]
3 years ago
11

What changes occurred when Zoroastrianism became a state religion during the Sasanian Empire?

Physics
1 answer:
statuscvo [17]3 years ago
8 0
Zoroastrianism affected the way the Persians governed their subjects by allowing the territories under their rule to worship their own religion.
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An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e
frozen [14]

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

Tsin\theta= \frac{mv^2}{r}

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

Tcos\theta = mg

Matching both expressions with respect to the tension we will have to

T = \frac{\frac{mv^2}{r}}{sin\theta}

T = \frac{mg}{cos\theta}

Then we have that,

\frac{\frac{mv^2}{r}}{sin\theta} =  \frac{mg}{cos\theta}

\frac{mv^2}{r} = mg tan\theta

Rearranging to find the velocity we have that

v = \sqrt{grTan\theta}

The value of the angle is 14.5°, the acceleration  (g) is 9.8m/s^2 and the radius is

r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}

r = \frac{11.1}{2}+2.41

r = 7.96m

Replacing we have that

v = \sqrt{(9.8)(7.96)tan(14.5\°)}

v = 4.492m/s

Therefore the speed of each seat is 4.492m/s

6 0
4 years ago
Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car
blagie [28]

Describing motion from each frame of reference:

Since observer A is on one of the train's cars, he will find that he is at rest with respect to the train even when it is pulling away because,  he is also moving with respect to the train.

Observer B on the stationary platform will observe that the train is pulling away from the station towards the right of the platform. as is described


Observer C will notice that the train is approaching him in the opposite direction at a speed which is the sum of the speeds at which both the train are traveling


If the conductor applies brakes on the train, since the platform is a stationary frame of reference. The motion will be observed as a simple decelerationg. he will observe that the force due to the brakes will cause the velocity of the train with respect to the train to decrease


Universal law of gravitation effects all objects alike . There will be a constant force of gravity acting on the train that will keep the train on its tracks. The tracks in turn exert a reaction force on the train. There will not be any affect on the train's motion as such( assuming that the train is moving along the tracks and gravitational force is exactly perpendicular)

3 0
3 years ago
cup of hot chocolate has a spoon in it. How does the heat move from the hot chocolate to the metal spoon handle?
serg [7]

The heat moves from the hot chocolate to the handle of the spoon by a process called thermal conduction.

It is the transfer of heat energy from one object to another when they are in contact with eachother.

Hope this answers your question.

5 0
4 years ago
The Earth's magnetic field is approximately 0.5 Gauss. What is the magnetic energy density of this field
ser-zykov [4K]

Answer:

The  magnetic energy density  is  E_B  =  9.95 *10^{-4} \  J/m^3

Explanation:

From the question we are told that  

    The  earths magnetic field is  B  =  0.5 \  Gauss

Now

          1 \ Gauss  \to  1 *10^{-4} \ T

=>        0.5  \ Gauss  \to  x T

So  

          x =  \frac{0.5 *  1*10^{-4}}{1}

          x = B= 5.0 *10^{-5} \  T

Generally the magnetic energy density is  mathematically represented as

        E_B  =  \frac{B^2}{ 2 *  \mu_o  }

  Here \mu_o is the  permeability of  free space with a constant value

          \mu_o  =  4\pi * 10^{-7} N/A^2

substituting values for equation above  

       E_B  =  \frac{(5*10^{-5})^2}{ 2 *   4\pi * 10^{-7}  }

       E_B  =  9.95 *10^{-4} \  J/m^3

5 0
3 years ago
A skater with an initial speed of 7.30 m/s stops propelling himself and begins to coast across the ice, eventually coming to res
Rufina [12.5K]

Answer:

a. a= 1.029 ms⁻²

b. S=25.89 m

Explanation:

given

initial speed is u = 7.30 m/sec

final speed is v = 0 m/sec

mu_k = 0.105

kinetic frictional force is Fk = mu_k×m×g

mu_k×m×g = m×a

m cancels

mu_k×g = a

deccelaration a = mu_k×g = 0.105×9.8 =1.029 ms⁻²

b) using v²-u² = 2aS

0²-7.3² = -2×1.029×S

53.29=2.058×S

S = 53.29/2.058

S=25.89 m

6 0
3 years ago
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