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lakkis [162]
3 years ago
5

According to the law of conservation of energy, which statement must be

Physics
1 answer:
alina1380 [7]3 years ago
7 0

Answer: THE ANSWER IS C!

The total energy of a system can decrease only if energy leaves the system.

Explanation: Apex!

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Explain how Rutherfords' gold foil experiment explains how this experiment confirms the nuclear model of the atom and the idea t
alekssr [168]

Explanation:

When Rutherford did the Gold Foil experiment, he expected that most of his alpha particles will go straight through, without any major deflections. He noticed that, during the experiment, most alpha particles went straight through, but was surprised to see that some got deflected by major angles. So he came up with a theory which states that every atom has got a nucleus, and also stated that most parts of the atom are empty, as most of the particles went through.

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6 0
3 years ago
A 1.0 kg object moving at 4.5 m/s has a wavelength of:
Lisa [10]
By wave particle  duality.

Wavelength , λ = h / mv

where h = Planck's constant = 6.63 * 10⁻³⁴ Js,  m = mass in kg,  v = velocity in m/s.
m = 1kg,  v = 4.5 m/s

λ = h / mv

λ = (6.63 * 10⁻³⁴) /(1*4.5)

λ ≈  1.473 * 10⁻³⁴  m

Option D.
7 0
3 years ago
What is meant by input and output work​
N76 [4]

Answer:

the x and y values

Explanation:

because on the table the x is the input and y is the output

5 0
2 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:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c)  </em>\frac{4492.18}{v_{car} ^{2} }

<em></em>

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} = <em>6738.27 J</em>

<em></em>

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} = <em>61.908 J</em>

<em></em>

<em></em>

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
What is the abbreviation for mojave desert?
vagabundo [1.1K]
There is no abbreviation
3 0
3 years ago
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