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nekit [7.7K]
2 years ago
5

jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

gine of her train 10 seconds to go from the first pole to the third pole. How long would it take for the engine to go the entire distance around the track?
Physics
1 answer:
d1i1m1o1n [39]2 years ago
5 0

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

\theta = \frac{2\pi}{6}

\theta = \frac{\pi}{3}

so the angular speed of the train is given as

\omega = \frac{\theta}{t}

\omega = \frac{\pi}{30} rad/s

now we have time period of the train given as

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{\frac{\pi}{30}}

T = 60 s

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Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount
Mariulka [41]

Answer:

The value is  \Delta s  = 8.537 *10^{25 } \ J/K

Explanation:

From the we are told that

   The radius of the sphere is r =  6.32 *10^{8} \  m

   The temperature is T_x  =  5350 \  K

    The average temperature of the rest of the universe is  T_r  =  2.73 \  K

Generally the change in entropy of the entire universe per second is mathematically represented as

         \Delta s  =  s_r - s_x

Here s_r is the entropy of the rest of the universe which is mathematically represented as

          s_r =  \frac{Q}{T_r}

Here Q is the quantity of heat radiated by the star which is mathematically represented as

           Q =  4 \pi *  r^2 *  \sigma * T^4_x

Here \sigma is the Stefan-Boltzmann constant with value  

           \sigma =  5.67 * 10^{-8 }W\cdot  m^{-2} \cdot  K^{-4}.

=>         Q =  4 \pi *  (6.32*10^{8})^2 *  5.67 * 10^{-8 }  * 5350 ^4

=>         Q =  2.332 *10^{26} \  J

So

      s_r =  \frac{2.332 *10^{26}}{2.73}

=>   s_r =  8.5415 *10^{25}\  J/K

Here s_x is the entropy of the rest of the universe which is mathematically represented as

      s_x =  \frac{Q}{T_x}

=>   s_x =  \frac{2.332 *10^{26} }{5350}

=>   s_x =  4.359 *10^{22} \  J/K

So

      \Delta s  = 8.5415 *10^{25}  - 4.359 *10^{22}

=>   \Delta s  = 8.537 *10^{25 } \ J/K

7 0
2 years ago
1. Susie wondered if the height of a hole punched in the side of a quart-size milk carton would affect how far from the containe
sdas [7]
Hypothesis: The water will squirt far.
Indep.V.: Height of hole.
Depend.V.: Range of squirt
Constant: Everything that isnt the independant var. such as filled liquid.
Control: None,I believe.
Number of groups: 4
Trials per group: 4
7 0
2 years ago
Read 2 more answers
A muon formed high in the Earth's atmosphere is measured by an observer on the Earth's surface to travel at speed V - 0.983c for
Alex_Xolod [135]

Answer:

The moun lives 2.198*10^-6 s as measured by its own frame of reference

The Earth moved 648 m as measured by the moun's frame of reference

Explanation:

From the point of view of the observer on Earth the muon traveled 3.53 km at 0.983c

0.983 * 3*10^8 = 2.949*10^8 m/s

Δt = d/v = 3530 / 2.949*10^8 = 1.197*10^-5 s

The muon lived 1.197*10^-5 s from the point of view of the observer.

The equation for time dilation is:

\Delta t' = \Delta t * \sqrt{1 - \frac{v^2}{c^2}}

Then:

\Delta t' = 1.197*10^-5 * \sqrt{1 - \frac{(0.983c)^2}{c^2}} = 2.198*10^-6 s

From the point of view of the moun Earth moved at 0.983c (2.949*10^8 m/s) during a time of 2.198*10^-6, so it moved

d = v*t = 2.949*10^8 * 2.198*10^-6 = 648 m

7 0
2 years ago
A cow’s mass is 410 kg and a car’s mass is 565 kg. What is the difference between their weights?
solmaris [256]
B. 1520 is the difference between their weights.
5 0
2 years ago
A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation
LuckyWell [14K]

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given that

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The expression for heat generation is given by

q = πr²Lq'

q = π . 0.1² . L . 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Using energy balance equation,

Energy going in = Energy coming out

Which is = q, which is 754L

From the attachment, we deduce that the temperature between the rod and the sleeve is 71.8° C

At the same time, we find out that the temperature on the outer surface is 51° C

Also, from the second attachment, the temperature at the center of the rod was calculated to be, 191.8° C

6 0
2 years ago
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