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kow [346]
3 years ago
11

Which of the following is true about plants and cellular energy?

Physics
1 answer:
kkurt [141]3 years ago
3 0

Answer:

D

Explanation:

6CO² + 6H²O > sunlight, chlorophyll, enzymes > C⁶H¹²O⁶ + 6O²

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A river flowing steadily at a rate of 175 m3/s is considered for hydroelectric power generation. It is determined that a dam can
Alex_Xolod [135]

Answer:

137200000 watts or 137200 kilowatts

Explanation:

The formula for power is P= dhrg

Where P = Power in watts

d = density of water (1000 kg/m^3)

h = height in meters

r = flow rate in cubic meters per second,

g = acceleration due to gravity of 9.8 m/s^2,

Plugging in the known values,

we get

P = 1000 kg/m^3 * 80 m * 175 m^3/s * 9.8 m/s^2

P = 80000 kg/m^2 * 175 m^3/s * 9.8 m/s^2

P = 14000000 kg m/s * 9.8 m/s^2

P = 137200000 kg m^2/s^3

P = 137200000 watts or 137200 kilowatts

The above figure assumes 100% efficiency which is impossible. A good efficiency would be 90% so the actual power available would be close to 0.90 * 137200 = 123480 kilowatts

6 0
3 years ago
The visible spectrum refers to the tiny portion of the electromagnetic spectrum that we ________.
marusya05 [52]

Answer:

are able to see/observe

Explanation:

Humans are not able to see most wavelengths in the universe--but there is a select range that is visible to our eyes. This (which is usually shown on an electromagnetic spectrum diagram/chart/depiction by a small portion of rainbow) is the visible light spectrum

3 0
2 years ago
g Doppler Radar gathers information about precipitation by sending out pulses of ______ energy that is reflected back by the pre
allsm [11]

Answer:

Doppler Radar gathers information about precipitation by sending out pulses of ___Radio wave___ energy

7 0
3 years ago
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
The focal length of a lens is inversely proportional to the quantity (n-1), where n is the index of refraction of the lens of th
Ainat [17]

Answer:

46.22 cm

Explanation:

The focal refraction, fr is given by

fr = \frac {c}{(1.572 -1)}  = \frac {c}{0 .572}  

The focal red light is given by

fv = \frac {c}{(1.605 - 1)} = \frac {c}{0.605}

\frac {fv}{fr} = \frac {0.572}{0 .605} = 0.945455

\frac {1}{fr} = \frac{1}{image} + \frac {1}{object} and making fr the subject we obtain

fr = \frac {image * object}{(image + object)} = \frac {24.00 * 55} {(24.0 + 55)} = 16.70886 cm

fv = 0.945455* 16.70886 cm = 15.79747 cm

image = \frac {object * f} {(object - f)} = \frac {15.79747 * 24.0}{(24.0 - 15.79747)} = 46.22222 cm

Therefore, violet image is approximately 46.22 cm

5 0
3 years ago
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