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tino4ka555 [31]

3 years ago

6

What part of the sweating process promotes cooling?

2 answers:

lora16 [44]3 years ago

8 0

Evaporation of liquid sweat from the skin.

Mandarinka [93]3 years ago

6 0

You actually sweating if the cooling part. you sweat so you can cool down.

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With regard to the pH scale, a solution with a pH

Montano1993 [528]

Close to 14 is considered a strong base.

6 0

3 years ago

A wrecking ball has a mass of 315 kg. If it is moving at a speed of 5.12 m/s, what is its kinetic energy?

zubka84 [21]

The formula for Kinetic energy is:
KE (Kinetic Energy) = 1/2 M (mass) x V (velocity) ^2

Substitute the known values into the equation:
KE = 1/2 (315) x (5.12)^2

Find out the values in sections (split the formula in half)
1/2 x 315 = 157.5
5.12^2 = 26.2144

Now times the answers together to complete the formula.
157.5 x 26.2144 = 4,128.768 km^2/s

5 0

3 years ago

What happens to a light beam that passes through a polarizing crystal? What happens when plane-polarized light passes through a

levacccp [35]

Answer:

it gives rise to vibrating in one plane.when light passes through second polarizing crystal no light passing through.

Explanation:

in the second polarizing crystal only darkness emerges.

5 0

4 years ago

In February 1955, a paratrooper fell 1200 ft from an

-Dominant- [34]

Answer:

$s=1.1107\ m$ is the minimum depth of snow for survivable stopping.

Explanation:

Given:

terminal velocity of fall, $u=56\ m.s^{-1}$

mass of the paratrooper, $m=85\ kg$

force on the paratrooper by the ice to stop him, $F=1.2\times 10^5\ N$

<u>Firstly, we calculate the deceleration caused in the snow:</u>

$a=\frac{F}{m}$

$a=\frac{120000}{85}$

$a=1411.765\ m.s^{-2}$

Now, using equation of motion:

$v^2=u^2+2a.s$ ....................(1)

where:

v = final velocity of the body after stopping

u = initial velocity of the body just before hitting the snow

a = acceleration of the body in the snow

s = distance through in the snow

Putting respective values in eq. (1)

$0^2=56^2+2\times (-1411.765)\times s$

$s=1.1107\ m$

5 0

3 years ago

Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you

Alekssandra [29.7K]

Answer:

(a) F = 6.14 *10⁻⁴ N

(b) P = 6.14* 10⁻¹⁰ Pa

(c) t = 27.2 min

Explanation:

Area of sail A = 1.0 km² = 1.0 * 10⁶m²

Wavelength of light λ = 650 nm = 650 * 10⁻⁹ m

Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s

(a)

Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹

= 1.0192 * 10⁻²⁷ kg.m/s

Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.

Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.

dp/dt = R * h/ λ

This is the force acting on the sail.

F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N

F = 6.14 *10⁻⁴ N

b)

Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail

P = F/A = 6.14 * 10⁻⁴ / 1.0 * 10⁶ = 6.14* 10⁻¹⁰ Pa

P = 6.14* 10⁻¹⁰ Pa

c)

Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²

As the spacecraft starts from rest, initial speed u=0,m/s ,

final speed is u = 1.0 m/s after time t

v = u+at

t = 1.0 - 0/ 6.14 * 10⁻⁴ = 1629s = 27.2 min

t = 27.2 min

4 0

4 years ago