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2 years ago

When sunlight shines on a leaf the leaf looks green why does the leaf look green

Physics

2 answers:

KonstantinChe [14]2 years ago

7 0

The correct answer is the third, It reflects the green light waves and absorbs most of the rest.

Sonja [21]2 years ago

3 0

Answer:

Green light is not absorbed but reflected, making the plant appear green. Chlorophyll is found in the chloroplasts of plants. There are various types of chlorophyll structures, but plants contain chlorophyll a and b. These two types of chlorophyll differ only slightly, in the composition of a single side chain.

Explanation:

Hope that helps.

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an object of mass 4 kg is placed on an inclined plane at an angle of 60. if the object slides down the plane with an acceleratio

balandron [24]

The forces on the y axis are:

N-mgcos(60)=0 , wich becomes

N=mgcos(60)

Rember that the friction force is always contrary to the motion of an object and its formula is f=μ * N

The forces in the x axis are:

-f + mgsin(60)= m * a

-μ*mgcos(60) + mgsin(60)=m*a ,

μ = ( m*a - mgf=μ[sin(60) )/ ( mgcos(90) )

6 0

3 years ago

Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses? A. specific heat B. conductiv

KengaRu [80]

Answer:

Efficiency

Explanation:

The amount of mechanical work an engine can do per unit of heat energy it uses is called its efficiency.
It is also defined as the output divided by the total electrical power consumed.
In terms of heat, efficiency of engine is given by :

$\eta=1-\dfrac{Q_o}{Q_i}$

$Q_o\ and\ Q_i$ are output heat and input heat respectively.

Hence, the correct option is (d) "efficiency"

4 0

3 years ago

A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t

Ganezh [65]

Answer:

The value is $\eta = 54.4 \%$

Explanation:

From the question we are told that

The work input is $W = ( 245 + A ) \ J$

The heat delivered is $Q = (142 + B) \ J$

The value of A is A = 14

The value of B is B = 72

Generally the efficiency of the heat engine is mathematically represented as

$\eta = \frac{W}{Q_t}$

Here $Q_t$ is the total out energy produce by the heat engine and this is mathematically represented as

$Q_t= Q + W$

=> $Q_t= 245 + A + 142 + B$

=> $Q_t= 390 + A+B$

$\eta = \frac{245 + A }{390 + A+ B}$

=> $\eta = 0.544$

=> $\eta = 0.544 *100$

=> $\eta = 54.4 \%$

5 0

3 years ago

A student wants to determine the impulse delivered to the lab cart when it runs into the wall. The student measures the mass of

forsale [732]

Impulse = Force * times and also Impulse = change in momentum.

Given that the mass does not change, change if momentum = mass * (final velocity - initial velocity)

Given that you know mass and initial velocity (which is the velicity before the cart hits the wall) you need the final velocity (which is the velocity after the cart hits the wall).

Answer: the velocity of the cart after it hits the wall.

6 0

3 years ago

Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees

Alexandra [31]

Answer:

T ambient = 10 degrees

Explanation:

Using Newton's Law of Cooling:

T(t) = Tamb + (Ti - Tamb)*e^(-kt) ..... Eq 1

Ti = 100

We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1

70 = Tamb + (100 - Tamb)*e^(-10k) ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k) ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)

Squaring both sides we get:

((70 - Tamb) / (100 - Tamb))^2 = e^(-20k) .... Eq 4

Substitute Eq 4 into Eq 3

50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2

After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

5000 - 50*Tamb = 4900 - 40*Tamb

Tamb = 100 / 10 = 10 degrees

6 0

3 years ago