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Nikolay [14]
3 years ago
10

What is an effect of an object getting wet?

Physics
1 answer:
djverab [1.8K]3 years ago
7 0

Answer:

once light hits a wet shirt, that water layer causes less of the blue shirt's blue wavelengths of light to be reflected toward your eyes and more of the blue light to be refracted, or bounce away from you, back into the fabric.

Explanation:

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I had gotten this one wrong please help thank you
Nata [24]
Check the 1st 2nd 3rd and 4th boxes
4 0
4 years ago
The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic
Drupady [299]

a) Density at 100 degrees: 1.34\cdot 10^4 kg/m^3

Explanation:

The density of mercury at 0 degrees is d=1.36\cdot 10^4 kg/m^3

Let's take 1 kg of mercury. Its volume at 0 degrees is

V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3

The formula to calculate the volumetric expansion of the mercury is:

\Delta V= \alpha V \Delta T

where

\alpha=180\cdot 10^{-6} K^{-1} is the cubic expansivity of mercury

V is the initial volume

\Delta T is the increase in temperature

In this part of the problem, \Delta T=100 C-0 C=100 C=100 K

So, the expansion is

\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3

So, the new density is

d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3


b) Density at 22 degrees: 1.355\cdot 10^4 kg/m^3

We can apply the same formula we used before, the only difference here is that the increase in temperature is

\Delta T=22 C-0 C=22 C=22 K

And the volumetric expansion is

\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3

So, the new density is

d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3


8 0
3 years ago
If a machine exerts a force of 250 N on an object and no work is done, what must have occurred?
valina [46]

Answer:

<h2>1) There is no work done by the machine because</h2><h2>B) The object has not moved</h2><h2>2) There is no work done by the prisoner because</h2><h2>D) The prisoner does no work because the wall goes no distance</h2><h2>3) The kinetic energy when it is half the way down is</h2><h2>6.0 J</h2>

Explanation:

1) As we know that the work done is the product of force and displacement

It is given as

W = Fdcos\theta

so if the object is not displaced due to the force exerted by the object then the work done by the object must be ZERO

so correct answer is

B) The object has not moved

2) As we know that the work done is the product of force and displacement

It is given as

W = Fdcos\theta

As we know that the wall is not displaced due to applied force so here work done by the prisoner must be zero

D) The prisoner does no work because the wall goes no distance

3) As we know by work energy theorem that work done by all forces is equal to change in its kinetic energy

So we will have

W_g + W_f = \frac{1}{2} mv^2

so we will have

3(10)(4) + W_f = \frac{1}{2}(3)(3)^2

120 + W_f = 13.5

W_f = -106.5 J

now when cart moves half the distance then again using the same

W_g + W_f = K

K = 3(10)(2) - \frac{106.5}{2}

K = 6.5 J

8 0
3 years ago
When a medium is disturbed by a wave, does it move with the wave?
DerKrebs [107]

Answer:

As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on. Particles of the medium don't actually travel along with the wave. Only the energy of the wave travels through the medium

8 0
3 years ago
What would most likely cause an increase in cloud formation?
podryga [215]
<span>increased evaporation</span>
6 0
3 years ago
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