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2 years ago

Do you think a hurricane can be considered to be a heat machine? Why?

Physics

2 answers:

Colt1911 [192]2 years ago

8 0

Answer:

YES

Explanation:

When the surface water is warm, the storm sucks up heat energy from the water, just like a straw sucks up a liquid. This creates moisture in the air. If wind conditions are right, the storm becomes a hurricane. This heat energy is the fuel for the storm. So therefore yes i do

Leona [35]2 years ago

5 0

Answer:

Explanation:

Hurricanes always do not bring heat. For example, Hurricane Sandy brought snow.

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A sample is brought to the laboratory and it is determined that one-eighth of the original

vlabodo [156]

Answer:

Oh please dont be so dramatic

Explanation:

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2 years ago

If the distance between the bounces were 1.95 m instead of 1.30 m, but the height remained at 1.30 m, which of your answers to P

Oksanka [162]

Answer:

Below is the calculation of all the required quantities. From the calculations, Vx must change, the velocity before hitting the ground must change and the angles also must change.

Explanation:

Below is an attachment that contains the calculation steps.

Thank you for reading.

7 0

3 years ago

Which condition is a neuron in when the outside of the neuron has a net positive charge and the inside has a net negative charge

White raven [17]

The condition is a neuron in when the outside of the neuron has a net positive charge and the inside has a net negative charge (due to accumulation of more sodium ions) is C. resting potential. The resting membrane potential of a neuron is approximately -70 mV (mV=millivolt)

8 0

3 years ago

Read 2 more answers

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

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

3 0

3 years ago

If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of

kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass m₁ and m₂ a distance r apart is

F = G m₁ m₂ / r²

where G = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

F = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

F ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)g, where g = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

F = G m₁ m₂ / (2r)² = 1/4 G m₁ m₂ / r²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0

3 years ago