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2 years ago

If you dip hand in cold water after having dipped in warm water wii you feel the water colder than actually is?

Physics

1 answer:

cluponka [151]2 years ago

4 0

Your hand feels colder when it transitions from warm to "colder" (room temperature) water.

<h3>What happens when you put your hand from warm water to cold water?</h3>

Your hand feels colder when you move it from the warm water to the "colder" (room temperature) water. Skin surface temperature will decrease in a constant ambient environment as a result of sympathetic nerve activity increasing and blood flow decreasing as a result of cutaneous receptors triggered by ice-water immersion of one hand in the palm skin of the non-immersed opposite hand.

Learn more about the temperature phenomenon here:

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A 1 mW laser beam is incident onto a detector. Determine the fractional fluctuation in number of photons intercepted by the dete

Ugo [173]

Answer:

(a) $625 photons$

(b) $625\times 10^{6}photons$

Explanation:

Given that , the power of the laser beam is,

$P=1 mW\\P=1\times 10^{-3} W$

and time is given is,

$t=1\mu s\\t=10^{-6}s$

Now the energy formula for the laser beam is,

$E=P\times t$

Now,

$E=10^{-3}\times 10^{-6} \\E=10^{-9}J$

(a) The value of energy is given,

$E_{1}=10MeV\\E_{1}=10\times 10^{6}\times 1.6\times 10^{-19}J\\ E_{1}=16\times 10^{-13}J$

Now the no of photons's fraction fluctuation is,

$n=\frac{E}{E_{1} }\\ n=\frac{10^{-9} }{16\times 10^{-13} }\\ E_{1}=625 photons$

Therefore the no of photons is $625 photons$ .

(b)The value of energy is given,

$E_{z}=10eV\\E_{z}=16\times 10^{-19}J$

Now the no of photons's fraction fluctuation is,

$n=\frac{E}{E_{z} }\\ n=\frac{10^{-9} }{16\times 10^{-19} }\\n=625\times 10^{6}photons$

Therefore the no of photons is $625\times 10^{6}photons$ .

5 0

3 years ago

It is desired to create a particle of mass 7920 MeV/c^2 in a head-on collision between a proton and an antiproton (each having a

Olegator [25]

Answer:

$v=0.9714c$

Explanation:

The kinetic energy possessed by particles will be

$K.E=\frac{1}{2}Mc^2$

where,

M is the mass of the particle (7920938.3 MeV/c²)

c is the speed of the light

Also,

energy of the proton particle = $\frac{m_pc^2}{\sqrt{1-\frac{v^2}{c^2}}}$

where,

v is the velocity

m_p is the mass of the proton (938.3 MeV/c²)

since the energy is equal

thus,

$\frac{m_pc^2}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{2}Mc^2$

$1-\frac{v^2}{c^2}=[\frac{2m_p}{M}]^2$

substituting the values in the above equation, we get

$1-\frac{v^2}{c^2}=[\frac{2\times 938.3 }{7920}]^2$

$v=0.9714c$

Hence,<u> the speed necessary for the specified condition to occur is </u><u>0.9714 times the speed of the light</u>

5 0

3 years ago

The mass of 1 dozen dimes is 27.22 g. the numerical value of the mass of 5 dozen dimes can be obtained by

Jobisdone [24]

Answer:

by obtaining the total mass of the dimes present:

d = 27.22 g / dozen the density of dimes

M = n * d = 5 dozen * 27.22 g / dozen = 126.1 g

4 0

2 years ago

A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

babunello [35]

26.10 N is the vertical component of the force.

Rx represents the Horizontal component of force

Ry represents The Vertical component of force

According to the given diagram

Rx - Tcosθ = 0

Rx = Tcosθ

And,

Ry + Tsinθ = mg

Ry = mg - Tsinθ

The horizontal component of force =The Vertical component of force

Rx = Ry

Tcosθ = mg - Tsinθ

T(cosθ + sinθ) = 29 × 9.8 = 284.2 N

T√2 cosθ = 284.2 N

T × √2 ×0.544 = 284.2 N

T × 0.769 = 284.2 N

T = 370 N (app)

So,

Ry = 284.2 - 370 (sin 57°)

= 284.2 - 310.3 = -26.10 N

Hence, 26.10 N is the vertical component of the force exerted.

Learn more about the horizontal and vertical components here

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7 0

2 years ago

Directions: Complete the concept map using the terms listed below.

Komok [63]

Answer:

this is confusing

Explanation:

6 0

3 years ago