PHYSICS COLLEGE

Recall your experimental setup from Lab 05A: a constant force was applied to a disc by attaching a mass to a light string wrapped around a mass-less pulley and hanging the mass over the edge of the apparatus. In the lab, you used energy conservation arguments to derive an expression for the angular velocity of the disc after the mass had fallen a distance x . Your goal now is to use kinematics and dynamics to confirm your expression. Use the following symbols throughout this question: m is the mass of the hanging mass, M is the mass of the disc, r is the radius of the pulley, R is the radius of the disc, x is the distance the mass has fallen, and g is the acceleration due to gravity. What is the linear acceleration of the mass after it has fallen a distance x

Answers

Answer 1

Answer:

Answer:

w = $\sqrt{(2gy)/(r^2 + (1)/(2) R^2 ) }$

Explanation:

For this exercise let's start by applying Newton's second law to the mass with the string

W - T = m a

In this case, as the system is going down, we will assume the vertical directional down as positive.

T = W - m a

Now we apply Newton's second law for rotational motion to the pulley of radius r. We will assume the positive counterclockwise rotations

∑ τ = I α

T r = I α

the moment of inertia of the disk is

I = ½ M R²

angular and linear acceleration are related

a = α r

we substitute

T r = (½ m R²) (a / r)

T = ½ m ( $(R)/(r)$ )² a

we write our two equations

T = W - m a

T = ½ m ( $(R)/(r)$ )² a

we solve the system of equations

W - m a = ½ m (\frac{R}{r} )² a

m g = m a [ 1 + ½ (\frac{R}{r} )² ]

a = $(g)/( 1 + (1)/(2) ((R)/(r))^2 )$

this acceleration is constant throughout the trajectory, so with the angular and lineal kinematics relations

w² = w₀² + 2 α θ

v² = v₀² + 2 a y

as the system is released its initial angular velocity is zero

w² = 0 + 2 α θ

v² = 0 + 2 a y

we look for the angular acceleration

a =α r

α = a / r

α = $(g)/(r (1 + (1)/(2) ((R)/(r))^2 )$

we look for the angle, remember that they must be measured in radians

θ = s / r

in this case we approximate the arc to the distance

s = y

θ = y / r

we substitute

w = $\sqrt{2 (g)/( r( (1)/(2) ((R)/(r))^2 ) (y)/(r) }$

w = $\sqrt{(2gy)/(r^2 + (1)/(2) R^2 ) }$

for the simple case where r = R

w = $\sqrt{ (2gy)/( (3)/(2) R^2 ) }$

w = $\sqrt{ (4)/(3) (gy)/(R^2) }$

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A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V. Use ϵ0 = 8.85×10⁻¹² C²/N⋅m². The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 n at an angle of 32.0° above the horizontal. the coefficient of kinetic friction between the box and the surface is 0.350. what is the acceleration of the box?

Answers

The acceleration of the box is 0.81 m/sec².

What is acceleration?

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration

According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:

$\sum F= ma$ (1)

we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:

- the horizontal component of the force exerted by the rope, which is equal to

$F_x = F cos\theta = 250*cos 32 = 212 N$

the frictional force, acting in the opposite direction, which is equal to

$F_f = \mu *mg = 171.7 N$

By applying Newton's law (1), we can calculate the acceleration of the

box,

$F_x - F_f = ma\na = 0.81 m/sec^2$

The acceleration of the box is 0.81 m/sec².

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An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)

Answers

Given :

∅ = 60⁰

u = 4 m/s

g = 10m/s²

to find :

T = ?

Solution :

as per formula,

$t = (2u \: sin \theta)/(g)$

now put the value : $t \: = (2 * 4 * sin \: 60)/(10)$

as we know $sin60 \: = ( √(3) )/(2)$

therefore,

$t \: = (8 * ( √(3) )/(2) )/(10)$

as we solve this we get,

$t \: = ( 2√(3) )/(5)$

that's t = 0.69 sec

$\sf\fbox\red{\:I \:hope \:it's \:helpful \:to \:you}$

0.8 seconds

Explanation:

time of flight = 2u/g

u=4m/s

g=10

= 8/10

= 0.8 sec

just a trial...not sure!!!

The interatomic spring stiffness for tungsten is determined from Young's modulus measurements to be 90 N/m. The mass of one mole of tungsten is 0.185 kg. If we model a block of tungsten as a collection of atomic "oscillators" (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two "springs", and each "spring" is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above. Use these precise values for the constants: ℏ = 1.0546 10-34 J · s (Planck's constant divided by 2π) Avogadro's number = 6.0221 1023 molecules/mole kB = 1.3807 10-23 J/K (the Boltzmann constant)

Answers

Answer:

Explanation:

solution below

Final answer:

The quantum of energy for one atomic oscillator in tungsten, given the effective interatomic spring stiffness of 360 N/m, the mass of one tungsten atom as 3.074 x 10^-25 kg, and the reduced Planck's constant of 1.0546 x 10^-34 J · s, can be calculated to be approximately 1.33 x 10^-21 J.

Explanation:

To calculate the quantum of energy for one atomic oscillator in tungsten, we will consider the model of an atom being connected to two springs, both having an effective interatomic spring stiffness of four times the given value (90 N/m). This value thus becomes 360 N/m.

One mole of tungsten has a mass of 0.185 kg, thus the mass of one atom can be determined by dividing this value by Avogadro's number (6.0221 x 10^23 molecules/mole), which gives approximately 3.074 x 10^-25 kg.

The quantum of energy, or the energy of one quantum (the smallest possible energy increment), is given by the formula E = ħω, where ħ is the reduced Planck's constant (1.0546 x 10^-34 J · s) and ω is the angular frequency, given by sqrt(k/m), where k is the spring constant and m is the mass.

Substituting the known values into these equations gives ω= sqrt((360)/(3.074 x 10^-25)) and E= (1.0546 x 10^-34) x sqrt((360)/(3.074 x 10^-25)), which results in a quantum of energy of approximately 1.33 x 10^-21 J.

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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Answers

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_(1) +q_(2) =-q_(3)$ -----eqution 1

where,

$q_(1)$ is the heat absorbed by the solid at 0⁰C

$q_(2)$ is the heat absorbed by the liquid at 0⁰C

$q_(3)$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *(1 mole H_(2) O)/(18g) = 13.167 moles of H_(2)O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_(1) = 13.167 moles *6.01(KJ)/(mole) = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_(2) = -q_(3)$

Step 4: calculate the final temperature of the water

$79.13KJ+M_(sample) *C*\delta {T_(sample)} =-M_(water) *C*\delta {T_(water)$

Substitute in the values; we will have,

$79.13KJ + 237*4.18(J)/(g^(o)C)*(T_(f)-218}) = -350*4.18(J)/(g^(o)C)*(T_(f)-100})$

79.13 kJ + 990.66J* $(T_(f)-218})$ = -1463J* $(T_(f)-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_(f)-218})$ = -1.463KJ* $(T_(f)-100})$

$79.13 + 0.99066T_(f) -215.96388= -1.463T_(f)+146.3$

collect like terms,

2.45366 $T_(f)$ = 283.133

∴ $T_(f) =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

A 17.0 resistor and a 6.0 resistor are connected in series with a battery. The potential difference across the 6.0 resistor is measured as 15 V. Find the potential difference across the battery.

Answers

Answer:

V= 57.5 V

Explanation:

If the resistors are in the linear zone of operation, the potential difference across them, must obey Ohm's law:

$V = I*R$

For the 6.0 Ω resistor, if the potential difference across it is 15 V, we can find the current flowing through it as follows:

$I = (V)/(R) = (15 V)/(6.0 \Omega) = 2.5 A$

In a series circuit, the current is the same at any point of it, so the current through the battery is I = 2.5 A
The equivalent resistance of a series circuit is just the sum of the resistances, so, in this case, we can write the following equation:

$R_(eq) = R_(1) +R_(2) = 17.0 \Omega + 6.0 \Omega = 23.0 \Omega$

Applying Ohm's Law to the equivalent resistance, we can find the potential difference through it, that must be equal to the potential difference across the battery, as follows:

$V = I* R_(eq) = 2.5 A * 23.0 \Omega = 57.5 V$

Someone claps his or her hands is an example of… motion energy to sound energy sound energy to potential energy motion energy to radiant energy

Answers

motion energy to sound energy i think

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