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kotegsom [21]
3 years ago
14

Explain the process of photosynthesis, se sure to include the reactants and products.

Physics
1 answer:
steposvetlana [31]3 years ago
6 0

Answer:

The process by which green plants and some other organisms use sunlight to synthesize foods from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a byproduct.

Explanation:

That's how plants make food for themselves.

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A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and
cricket20 [7]

Answer:

The frequency f = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = \frac{d}{2} =\frac{44}{2} = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = \frac{d}{2} =\frac{3.0}{2} = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = \frac{d}{2} = \frac{2.0}{2} = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that  the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2

pgy_1=\frac{1}{2}pv^2_2 +pgy_2

v_2=\sqrt{2g(y_1-y_2)}

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = \sqrt{2*9.8*(0.35-0)}

v₂ = \sqrt {6.86}

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

v_3 = \frac{v_2r_2^2}{v_3^2}

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = (2.62  m/s)* (\frac{1.5mm^2}{1.0mm^2} )

v₃ = 0.0589 m/s

∴ velocity  of the fluid in the cylinder =  0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

f=\frac{v_s}{4(h-v_3t)}

where;

v_s = velocity of sound

h = height of the fluid

v₃ = velocity  of the fluid in the cylinder

f=\frac{343}{4(0.40-(0.0589)(0.4)}

f= \frac{343}{0.6576}

f = 521.59 Hz

∴ The frequency f = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})

\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)

\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}

where;

v_s (velocity of sound) = 343 m/s

v₃ (velocity  of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s  when the cylinder has been filling for 4.0 s.

8 0
3 years ago
Solve the inequality 2(n+3) – 4&lt;6. Then graph<br> the solution.
Aloiza [94]
The solution is 22 2(n+3)-4&6
6 0
3 years ago
How does temperature influence a chemical change?
Serga [27]
Your weight #sorryfortheweight
8 0
3 years ago
Based on the data you found, about how many of the 100 aliens would become thin if the temperature were 35 C?
saw5 [17]

Answer:

Option D: More than 90

Explanation:

Thickness probably depends on temperature. The higher the temperature, the more aliens we expect to be thin.

5 0
3 years ago
Read 2 more answers
A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l
mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be \who_w = 1000 kg/m^3

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

F_s + F_b = W

Where F_s = kxN is the spring force, F_b = W_w = m_wg = \rho_w V_s g is the buoyancy force, which equals to the weight W_w of the water displaced by the submerged portion of the cylinder, which is the product of water density \rho_w, submerged volume V_s and gravitational constant g. W = mg is the weight of the metal cylinder.

kx + \rho_w V_s g = mg

The submerged volume would be the product of cross-section area and the submerged length x

V_s = Ax = \pi(d/2)^2x

Plug that into our force equation and we have

kx + \rho_w \pi(d/2)^2x g = mg

x(k + \rho_w g \pi d^2/4) = mg

x = \frac{m}{(k/g) + (\rho_w\pi d^2/4)} = \frac{1.5}{(33/9.8) + (100*\pi * 0.048^2/4)} = 0.423 m

6 0
3 years ago
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