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3 years ago

Photosynthesis occurs in a plant cell's

Physics

1 answer:

Fynjy0 [20]3 years ago

5 0

Answer:

Photosynthesis occurs in the organelle, chloroplast .

Cellular respiration occurs in a plants or animals mitochondria.

Explanation:

Hope this helps! :)

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Ax = 4.3 m

Dimas [21]

Answer:

the answer will be option no b plss mark me brainliest

7 0

3 years ago

PLZ HELP I DON'T UNDERSTAND!! a boy is playing catch with his friend. He throws the ball straight up. When it leaves his hand, t

algol13

Explanation:

The mass of a ball, m = 2 kg

It is traveling with a speed of 10 m/s

The ball's kinetic energy just as it leaves the boy's hand is calculated as follows :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2\times (10)^2\\\\=100\ J$

The ball's kinetic energy just as it leaves the boy's hand is 100 J. The potential energy of the ball when it reaches the highest point is same as the kinetic energy as it leaves the boy's hand.

Hence, the required kinetic and potential energy is 100 J.

6 0

3 years ago

If an object of mass 70kg falls from a height of 500 m, what is the maximum velocity of the object?

Umnica [9.8K]

Answer:

H = 1/2 * g * t^2 since initial velocity is zero

v = g * t where v is the final velocity

t = v / g

H = 1/2 g * v^2 / g^2 = 1/2 v^2 / g

v = (2 * H * g)^1.2

v = (2 * 500 * 9.8)^1/2 = 99 m/s

Check: t = v / g = 99 / 9.8 = 10.1 sec

H = 1/2 * 9.8 * 10.1^2 = 500 m

7 0

3 years ago

Three balls with the same radius 21 cm are in water. Ball 1 floats, with

Masteriza [31]

Answer:

Explanation:

A )

The ball floats with half of it exposed above the water level . So it must have density half that of water . In other words its density must have been 500 kg / m³

B )

Tension in the ball will be equal to net force acting on the ball

Net force on the ball = buoyant force - weight .

4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1000 - 893 )

= 40.65 x 10⁻⁶ N .

C )Tension in the 3 rd ball will be equal to net force acting on the ball

Net force on the ball = weight - buoyant force

= 4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1320 - 1000 )

= 121.6 x 10⁻⁶ N .

7 0

3 years ago

A Carnot engine operates between two heat reservoirs at temperatures TH and TC. An inventor proposes to increase the efficiency

Dovator [93]

Answer:

A) Efficiency = 1 - (TC/TH)

B) Comparing this with the efficiency of the original carnot engine, the efficiency is the same

Explanation:

The formula for efficiency of an original carnot engine is;

e = 1 - T(C) /T(H) ——— eq 1

In like manner, for a composite engine, the efficiency is;

e(12) = (W1 + W2)/Q(H1)

Where W1 is work done by 1st engine; W2 is work done by second engine and Q(H1) is the heat input to the first engine.

Now the total work done is;

W = Q(H) + Q(C)

Where Q(H) is the heat input and Q(C) is the heat released.

Thus,

e(12) = [Q(H1) + Q(C1) + Q(H2) + Q(C2)] / Q(H1)

Now, from the earlier e(12) equation compared to this, QH2 = -QC1

Thus;

e(12) = [Q(H1) + Q(C1) - Q(C1) + Q(C2)] / Q(H1)

So e(12) = [Q(H1) + Q(C2)] / Q(H1)

So e(12) = 1 + [Q(C2)/Q(H1)] ———eq 2

Also,

Q(C2) /Q(H2) = (-Tc/T')

Where T' is intermediate temperature.

So, simplifying that,

Q(C2) = -Q(H2) (Tc/T')

This is also equal to Q(C1) (TC/T')

But Q(C1) is also equal to;

-Q(H1) (T'/TH)

Thus; Q(C2) is now written as;

Q(C2) = -Q(H1) (T'/TH)(TC/T')

So T' will cancel out to remain;

Q(C2) = -Q(H1)(TC/TH)

Replacing this with Q(C2) in eq 2 to obtain;

e(12) = 1 + [[-Q(H1)(TC/TH)] /Q(H1)]

e(12) = 1 - TC/TH

8 0

3 years ago