A mass m on a frictionless plane, inclined at an angle θ with respect to the horizontal, is connected by a cord that runs parallel to the inclined plane and is wrapped around a flywheel of radius r and moment of inertia

[tex]i=\frac{3mr^{2}}{4}[/tex]

what is the acceleration of the mass down
the plane?

Answers

Answer 1

The acceleration of the mass down the plane is determined as (4mg sinθ)/(3mr²).

Conservation of angular momentum

The acceleration of the mass down the plane is determined by applying the principle of conservation of angular momentum.

Fr = Iα

where;

F is weight of the object parallel to the planer is the radius of the flywheelI is moment of inertiaα is angular acceleration

(mg sinθ)r = Iα

(mg sinθ)r = I(ar)

(mg sinθ) = I(a)

[tex]a = \frac{mg \times sin(\theta)}{I} \\\\a = \frac{mg \times sin(\theta)}{3mr^2/4} \\\\a = \frac{4mg \times sin(\theta)}{3mr^2}[/tex]

Thus, the acceleration of the mass down the plane is determined as (4mg sinθ)/(3mr²).

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Related Questions

A body of mass 5kg is ejected vertically from the ground when a force of 600N acts on it for 0.1s.Calculate the velocity with which the body leaves the ground.​

Answers

1200

Explanation:

F = m x (v / t)

v / t = F / m

v = (F / m) / t

v = (600 / 5) / 0.1

v = 120 / 0.1

v = 1200 m / s

Why do comets only have tails when they are near the sun?.

Answers

When far from the sun, a comet is like a stone rolling around the universe. But when it approaches the sun, the heat evaporates the comet's gases, causing it to emit dust and micro particles, electrons and ions.

_____ is a measure of how closely packed particles of matter are in certain amount of space​

Answers

I think It’s Density

Sorry If I am wrong

Answer:

Density

Explanation:

The more dense something is, the more packed the particles are. The equation for density is m/v.

What is the wavelength in nm of a light whose first order brightband forms a diffraction angle of 19.0°, and the diffraction grating has 600.0 lines per mm? round to the nearest whole number. nm

Answers

The wavelength of light will be 195.34 nm.

What is wavelength?

The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.

The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.

The wavelength from the diffraction grating formula is found as;

[tex]\rm m \lambda= d sin\theta \\\\ \lambda= \frac{dsin \theta }{m}\\\\ \lambda= \frac{600 \times sin 19^0 }{1} \\\\ \lambda= 195.34\ nm[/tex]

Hence the wavelength of light will be 195.34 nm.

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Answer:543nm

Explanation:

becuase

Electromagnets project student guide
y’all i need helppppo

Answers

Answer:

I don't know

Explanation:

you need then search here

A pair of opposite electric charges of equal magnitude is called a(n)

Answers

Answer:

Dipole

Explanation:

A pair of opposite electric charges of equal magnitude is called a(n)

Answer:

Dipole

Explanation:

A flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω. It is connected to a 21. 2-v battery at the instant t = 0. Consider the moment when the current is 2. 50 a

Answers

For a  flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω, the rate of energy being delivered is mathematically given as

P= 53 W

What rate is energy being delivered by the battery?

Generally, the equation for the Battery power  is mathematically given as

P = I (dt)V

Therefore

P= 2.50 A * 21.2V

P= 53 W

In conclusion, rate of energy being delivered

P= 53 W

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A flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω. It is connected to a 21. 2-v battery at the instant t = 0.The power developed will be 53 Watt.

What is electric power?

Electric power is the product of the voltage and current. Its unit is the watt. It is the rate of the electric work done.

The given data in the problem is;

V is the voltage = 21.2 volt (V)

I is the electric current = 2.50 ampere (A)

The formula for the power is given as;

[tex]\rm P= V I \\\\\ P= 2.50 \times 21.2 \\\\ P=53 \ watt[/tex]

Hence electric power developed will be 53 Watts.

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Consider the following processes: The temperature of two identical gases are increased from the same initial temperature to the same final temperature. Reversible processes are used. For gas A the process is carried out at constant volume while for gas B it is carried out at constant pressure. The change in entropy:_________.
i. is greater for B
ii. is the same for A and B
iii. is greater for A only if the initial temperature is high
iv. is greater for A
v. is greater for A only if the initial temperature is low

Answers

The entropy at constant pressure is always greater than that at constant volume and the change in entropy for the two processes is greater for gas B.

Second law of thermodynamics

According to the second law of thermodynamics, entropy in the isolated system always stays the same or increases.

Since gas does work in isobaric (constant pressure) change processes, the entropy at constant pressure will be greater than that at constant volume.

Thus, for gas A the process is carried out at constant volume while for gas B that is carried out at constant pressure. The change in entropy is greater for gas B.

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if a solution contains 30 g of
potassium chloride (KCI) per 100 mL of water at 60°C, the
solution would be considered saturated.

true
false

Answers

true bc it’s just true

You calibrate a set of automobile springs using a 25 kg car battery, attached to a rope that runs over a pulley, as shown in the diagram.

You find that the spring is pulled out by 0.06 meters. Compute the spring constant, k, in N/m.
Just type in the numeric part of your answer, to the nearest 0.1 N/m.

Answers

For an automobile spring using a 25 kg car battery, the spring constant is mathematically given as

K=4.08.3N/m

What is the spring constant?

Question Parameter(s):

a 25 kg car battery

The springs are pulled out by 0.06 meters

Generally, the equation for the Force  is mathematically given as

F=mg

Therefore

F=(25*9.8)

(25*9.8)=K(0.06)

K=4.08.3N/m

In conclusion, the spring constant

K=4.08.3N/m

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What is the frequency of a photon with an energy of 4. 56 x 10^-19 j

Answers

Answer:

ν = E/h = 4.56x10-19 J / 6.626x10-34 Jsec-1

ν = 6.88x1014 s-1

Explanation:

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

What is a frequency?

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

[tex]\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma = 6.88 \times 10^{14}\ s^{-1}[/tex]

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

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A compression spring with a spring constant of 200 N/m catches a ball that's falling with
kinetic energy of 0.5 J. How far will the spring compress to stop the ball?
Pls help It would be great if someone can show the steps too thank u

Answers

Answer:

distance of compression: 0.07071 m

Explanation:

[tex]\sf energy = \dfrac{1}{2} kx^2[/tex]

[tex]\rightarrow \sf 0.5= \dfrac{1}{2} (200)x^2[/tex]

[tex]\rightarrow \sf 0.5= (100)x^2[/tex]

[tex]\rightarrow \sf 5\ *\ 10^{-3}= x^2[/tex]

[tex]\rightarrow \sf x = \sqrt{5\ *\ 10^{-3}}[/tex]

[tex]\rightarrow \sf x =0.07071 \ m[/tex]

Explain diffraction at a single slit (light)

Answers

Answer:

At some point on say, the receiving screen, light emanating from the left side of the slit will be out of phase (a difference of 1/2 wavelengths) from   light coming from the center of the slit.

Thus for every point that is left of the center of the slit, there will be a point on the right side of the slit that is out of phase,

There will be no light on the screen at that particular point and thus there will be a dark fringe there.

That is the basic explanation for the appearance of dark and bright fringes on the receiving screen.

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