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3 years ago

9

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

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