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stiks02 [169]
3 years ago
10

What can turn light energy into chemical energy

Physics
1 answer:
Vikki [24]3 years ago
8 0

Chlorophyll traps the light energy given off by the Sun and converts it to chemical energy required to drive photosynthesis

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A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0
1 year ago
A car is traveling 100 km/hr. How many hours will it take to cover a distance of 850 km?
Evgesh-ka [11]

Answer:

8.5 hours

Explanation:

7 0
3 years ago
Read 2 more answers
Please help, and show steps. Thank you very much!
Vikentia [17]
V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)
6 0
3 years ago
The alternative to nuclear fission reactors in a nuclear fusion reactor. Explain why it is much more difficult to get a fusion r
Sliva [168]

Answer: The major challenges are as

1) understanding of the plasma: Plasma is a soup like mixture of subatomic particles of different atoms nuclei and electrons that are shattered apart by the temperature at which plasma is formed. further research is needed to understand the behavior of plasma so that it can be put to a proper use.

2) Confinement of plasma: Once we get the plasma we need to hold it so that we can obtain heat from it to drive a steam turbine but the sheer temperature of plasma is in millions of Celsius thus currently making it impossible to confine conventionally. Scientists use a loop of electric and magnetic fields to keep it in circulatory like manner so that it can be studied.

3) finally to obtain electricity from the plasma it should be stable to produce electricity. But currently to obtain pressure, temperature so that we have a sustained supply is highly difficult in technical and economical aspects.

Inertial confinement: In order to get the nuclei of atoms close enough for fusion this type of method used compression of the nuclei into highly small volumes.This is accomplished by use of lasers which are directed towards the fuel pellets that implode and travel towards other nuclei making fusion possible. It's main advantage is that it requires lesser time to initiate fusion but the disadvantage being that a large power is used to fire the lasers and the lasers should all hit the small target.

Magnetic Confinement: In this method we use a magnetic and electric fields in a properly designed space to keep the plasma in motion. In motion the nuclei of the atoms come close enough to initiate fusion.It's advantage being less power is required to start the process as compared to inertial confinement and the disadvantage being that plasma confinement is currently not properly understood.

5 0
3 years ago
A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c
tensa zangetsu [6.8K]

Answer:

E = \frac{3kQ^2}{5R}

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

V = \frac{kq}{r}

now we can say that

q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)

q = \frac{Qr^3}{R^3}

now electric potential is given as

V = \frac{k\frac{Qr^3}{R^3}}{r}

V = \frac{kQr^2}{R^3}

now work done to bring a small charge from infinite to the surface of this sphere is given as

dW = V dq

dW = \frac{kQr^2}{R^3} dq

here we know that

dq = \frac{3Qr^2dr}{R^3}

now the total energy of the sphere is given as

E = \int dW

E = \int_0^R  \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})

E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)

E = \frac{3kQ^2}{5R}

7 0
3 years ago
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