Which equation below is the correct equation for photosynthesis?

What is the velocity of a wave with a wavelength of 9 meters and a period of 0.006

Ipatiy [6.2K]

Answer: 1,500m/s

Explanation:

Relationship existing between velocity of a wave (v), wavelength(¶) and frequency(f) is

v = f¶... (1)

Since Frequency (f) is the reciprocal of the period (T);

Frequency = 1/Period i.e F = 1/T... (2)

Substituting equation 2 into 1 we have;

v = 1/T × ¶

v = ¶/T

Given wavelength ¶ = 9m

Period T = 0.006s

v = 9/0.006

v = 1,500m/s

The velocity of the wave will be 1,500m/s

8 0

3 years ago

A woman is standing in the ocean, and she notices that after a wave crest passes, five more crests pass in a time of 54.0 s. The

timama [110]

Answer:

a) f=0.1 Hz ; b) T=10s

c)λ= 36m

d)v=3.6m/s

e)amplitude, cannot be determined

Explanation:

Complete question is:

Determine, if possible, the wave's (a) frequency, (b) period, (c) wavelength, (d) speed, and (e) amplitude.

Given:

number of wave crests 'n'= 5

pass in a time't' 54.0s

distance between two successive crests 'd'= 36m

a) Frequency of the waves 'f' can be determined by dividing number of wave crests with time, so we have

f=n/t

f= 5/ 54 => 0.1Hz

b)The time period of wave 'T' is the reciprocal of the frequency

therefore,

T=1/f

T=1/0.1

T=10 sec.

c)wavelength'λ' is the distance between two successive crests i.e 36m

Therefore, λ= 36m

d) speed of the wave 'v' can be determined by the product of frequency and wavelength

v= fλ => 0.1 x 36

v=3.6m/s

e) For amplitude, no data is given in this question. So, it cannot be determined.

5 0

3 years ago

If two solutions have unequal concentrations of a solute, the solution with the lower concentration is called

marta [7]

Obviouskly lower concentrstion

6 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

According to the picture of the Electromagnetic spectrum [EMS] below, What color has a wavelength of 700 nm?

White raven [17]

Answer:

Red

Explanation:

I looked at the picture...where it says red is 700

5 0

3 years ago

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