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3 years ago

Why would it be hard to find the ideal light intensity if the temperure were very hot or cold?

Physics

1 answer:

grandymaker [24]3 years ago

6 0

The correct answer for this question is this one:

Light intensity is not dependent on temperature, with that said, plants (assuming that is what were talking about) would dehydrate faster at a higher temperature (unless they had mechanisms to cope e.g. C4 and CAM plants) with the same light intensity.

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1. Place the following in logical order by writing the numbers 1 through 6 in the spaces provided.

grin007 [14]

Answer:

ecdbaf

Explanation:

4 0

3 years ago

An object of volume 0.0004m^3 and density 6000kg/m^3 is immersed inside a fluid of density 5000kg/m^3. The force exerted by the

FromTheMoon [43]

We have that the density of the fluid is

$\rho_f=178.57 kg/m3$

From the question we are told that

object of volume 0.0004m^3

density 6000kg/m^3

fluid of density 5000kg/m^3.

Force F=0.7N

Generally the equation for the Force is mathematically given as

$F= v*\rho_f*g\\\\0.7= 0.0004*\rho_f*9.8\\\\\rho_f=\frac{0.7}{0.0004*9.8}$

$\rho_f=178.57 kg/m3$

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8 0

3 years ago

Calculate the force of friction for a 5kg aluminum block being pulled with constant velocity (uniform motion) across a horizonta

Stolb23 [73]

The purpose of this lab is to determine whether the surface of an area would affect the coefficient of Friction. My classmates and I have learned a lot in this lab and that there could have been some errors in our lab because the strength of how a person pulls it might be a slight different than the normal force. I learned from this lab that the <span>surface area would have no effect on the coefficient of friction. </span>

4 0

3 years ago

Two charges that are 1 meter apart repel each other with a force of 2 N. If the distance between the charges is increased to 2 m

Savatey [412]

Answer:

b) 0.5 N

Explanation:

From coulomb's law,

F = kq'q/r².................... Equation 1

Where F =force of repulsion between the charges, q' = first charge, q = second charge, r = distance between the charges, k = proportionality constant.

q'q = Fr²/k........................... Equation 2

Given: F = 2 N, r = 1 m, k = 9.0×10⁹ Nm²/C²

Substituting into equation 2

q'q = 2(1)²/(9.0×10⁹)

q'q = 2/9.0×10⁹ C².

If the distance between the charges is increased to 2 meters,

r = 2 m, q'q = 2/9.0×10⁹ C².

Substitute into equation 1

F = 9.0×10⁹(2/9.0×10⁹)/2²

F = 2/4

F = 1/2 = 0.5 N.

The right option is b) 0.5 N

5 0

4 years ago

A 9 m3 container is filled with 300 kg of r-134a at 10°c. what is the specific enthalpy (kj/kg) of the r-134a in the container?

andreev551 [17]

Given the mass of R-134a m = 300kg; Volume of the container V = 9 cu. meter; Temperature of R-134a T = 10 degrees Celsius;
Formula of specific volume : v = V / m = 9 / 300 = 0.03 cu. m / kg.
At T = 10 degrees Celsius from saturated R-134a tables, vf = 0.0007930 cu. m /kg; vg = 0.049403 cu. m/kg. We know v = vf + x (vg - vf), so 0.03 = 0.0007930 + x (0.049403 - 0.0007930), which makes x = 0.601.
Specific enthalpy of R-134a in the container is h = hf + x*hfg = 65.43 + (0.601 * 190.73). Answer is 180.0587 kJ/kg

8 0

3 years ago