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2 years ago

How does the color of water affect its evaporation rate

1 answer:

8 0

Well my thinking is that the lighter the slower the water evaporates or the darker the faster becuase dark colors absorb light and allows a lot of heat to be generated or not at all.

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A 256 Hz tuning fork is resonating in a closed tube on a warm day when the speed of sound is 346 m/s. What is the length of the

Ilya [14]

Assuming our "closed tube" is closed at only one end, then
 v = fλ = f*4L/n
 where "n" is the harmonic number. So
 L = nv / 4f = n*346m/s / 4*256Hz = n*0.38 m
 Since the only option in your list that is an integer multiple of 0.38 m is 1.35 m
 I'd say that we're hearing the fourth harmonic.
answer is
A. 1.35 m

5 0

2 years ago

A football player kicks a ball horizontally off a hill with an initial velocity of 42.0 m/s. It travels a horizontal distance of

damaskus [11]

Explanation:

.........................

7 0

2 years ago

This is the last question

koban [17]

Answer:

It's C

Explanation:

3 0

2 years ago

An electromagnetic wave of wavelength

Ivanshal [37]

Answer:

$4.01\cdot 10^{-7} m$

Explanation:

When an electromagnetic wave passes through the interface between two mediums, it undergoes refraction, which means that it bents and its speed and its wavelength change.

In particular, the wavelength of an electromagnetic wave in a certain medium is related to the index of refraction of the medium by:

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0$ is the wavelength in a vacuum (air is a good approximation of vacuum)

n is the refractive index of the medium

In this problem:

$\lambda_0 = 5.89\cdot 10^{-7} m$ is the original wavelength of the wave

n = 1.47 is the index of refraction of corn oil

Therefore, the wavelength of the electromagnetic wave in corn oil is:

$\lambda=\frac{5.89\cdot 10^{-7}}{1.47}=4.01\cdot 10^{-7} m$

8 0

2 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

2 years ago