Why are chloroplasts essentially to the process of photosynthesis to occurs

If you quadruple the temperature of a black body, by what factor will the total energy radiated per second per square meter incr

arlik [135]

In modern physics, as it was called "Stefan-Boltzmann law", the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's temperature T

as:

$P\alpha T^4$

where: P is the power (total energy radiated per second per square meter) and T is the temperature of a black body.

then we can make a ratio between the state of before quadruple (with subscript 1) and after (with subscript 2) as:

$\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{T_{2}^4}$

As

$T_{2}=4T_{1}$

Then

$\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{(4T_{1})^4}$

then

$P_{2}=256*P_{1}$

The factor will the total energy radiated per second per square meter increase = 256

5 0

3 years ago

Calculate the average velocity of a dancer who moves 5 m toward the left of the stage over the course of 15 s. ** Velocity = dis

Alex73 [517]

Answer:

B

Explanation:

Velocity=disp/time

V=5m/15s

V=1/3 m/s

3 0

3 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s