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forsale [732]
3 years ago
9

Can you go on glowing coals and if so why ? Please help meee

Physics
1 answer:
castortr0y [4]3 years ago
6 0
The coals or wood embers that are used in fire walking have a low heat capacity. Sweat produced on the bottom of people's feet also helps form a protective water vapor. All of this together makes it possible, if moving quickly enough, to walk across hot coals without getting burned.
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To practice tactics box 13.1 hydrostatics. in problems about liquids in hydrostatic equilibrium, you often need to find the pres
OLEGan [10]

Answer:

A. The pressure denoted as Pa and Pb at the surfaces of A and B in the tube is

PA= Pgas

PB= Patmos

B. The second sketch

C. The gas pressure is

Pgas= Patmos+ rho.g(h2-h1)

= 1atm + rho.g (h2-h1)

Explanation:

5 0
3 years ago
An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper
dimulka [17.4K]

Answer:

The power output of this engine is  P =  17.5 W

The  the maximum (Carnot) efficiency is  \eta_c  = 0.7424

The  actual efficiency of this engine is  \eta _a  = 0.46

Explanation:

From the question we are told that

    The temperature of the hot reservoir is  T_h = 1250 \ K

      The temperature of the cold reservoir  is  T_c  =  322 \ K

     The energy absorbed from the hot reservoir is E_h  = 1.37 *10^{5} \ J

       The energy exhausts into  cold reservoir is  E_c  = 7.4 *10^{4} J

The power output is mathematically represented as

      P  =  \frac{W}{t}

Where t is the time taken which we will assume to be 1 hour =  3600 s  

W is the workdone which is mathematically represented as

      W =  E_h  -E_c

substituting values

       W = 63000 J

So

    P =  \frac{63000}{3600}

    P =  17.5 W

The Carnot efficiency is mathematically represented as

          \eta_c  =  1 - \frac{T_c}{T_h}

         \eta_c  =  1 - \frac{322}{1250}

         \eta_c  = 0.7424

The actual efficiency is mathematically represented as

        \eta _a  =   \frac{W}{E_h}

substituting values

         \eta _a  =  \frac{63000}{1.37*10^{5}}

         \eta _a  = 0.46

     

7 0
3 years ago
The law of suggests that the orbit of planets is not circular but .
SVEN [57.7K]
One of Kepler's laws is that the orbits of planets are elliptical. It's not a suggestion. BTW, circles are ellipses too, but so special that their likelihood is close to zero.
3 0
3 years ago
Read 2 more answers
A stone is dropped from the upper observation deck of a tower, "500" m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dis
Yuri [45]

Answer:

h₍₁₎ = 495,1 meters

h₍₂₎ = 480,4 m

h₍₃₎ = 455,9 m

...

..

Explanation:

The exercise is "free fall". t = \sqrt{\frac{2h}{g} }

Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s

The heights (h) according to his time (t) are found according to the formula:

h(t) = 500 - 1/2 * g * t²

Remplacing "t" with the desired time.

4 0
3 years ago
In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. the dish is on a shelf above the poi
Georgia [21]

components of the speed of the coin is given as

v_x = v cos60

v_x = 6.4 cos60 = 3.2 m/s

v_y = vsin60

v_y = 6.4 sin60 = 5.54 m/s

now the time taken by the coin to reach the plate is given by

t = \frac{\delta x}{v_x}

t = \frac{2.1}{3.2}

t = 0.656 s

now in order to find the height

h = vy * t + \frac{1}{2} at^2

h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2

h = 1.52 m

so it is placed at 1.52 m height

3 0
3 years ago
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