PHYSICS COLLEGE

A 50n brick is suspended by a light string from a 30kg pulley which may be considered a solid disk with radius 2.0m. the brick is released from rest and falls to the floor below as the pulley rotates. it takes 4 seconds for the brick to hit the floor. i) what is the tension in newtons in the string well the brick is falling? ii) what is the magnitude of the angular momentum in kg*m^2/s of the pulley at the instant the brick hits the floor?

Answers

Answer 1

Answer:

Brick is held at a position which is at height 2 m from the floor

Now it is released from rest and hit the floor after t = 4 s

Now the acceleration of the brick is given by

$y = v_i* t + 0.5 at^2$

$2 = 0 + 0.5 * a * 4^2$

$a = 0.25 m/s^2$

Now in order to find the tension in the string

we can use Newton's law

$F_(net) = ma$

$mg - T = ma$

$50 - T = (50)/(9.8)*0.25$

$T = 48.72 N$

part b)

Now for the pulley

moment of inertia= $(1)/(2)mr^2$

m = 30 kg

R = 2 m

I = $(1)/(2)*30*2^2$

I = 60 kg m^2

Now the angular speed just before brick collide with the floor

$w = (v)/(r)[\tex]</p><p>here we have</p><p>[tex]v = v_i + a* t$

$v = 0 + 0.25 * 4$

v = 1 m/s

Now we will have

L = angular momentum = I w = $I*(v)/(R)$

L = 60 * $(1)/(2)$

L = 30 kg m^2/s

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The position of a particle is given by the function x=(5t3−8t2+12)m, where t is in s. at what time does the particle reach its minimum velocity?
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)

Q18: A cube of aluminum has an edge length of 20 cm. Aluminum has adensity of 2.7 g/cm and a specific heat of 0.217 cal/ g.°С. When
the internal energy of the cube increases by 47000 cal its temperature
increases by:
A
B
C
D
E
5 °C
10 °C
20 °C
100 °C
200 °C

Answers

The change in temperature of this cube of aluminum is equal to: B. 10°C

Given the following data:

Edge length, L = 20 cm.

Density of Aluminum = 2.7 g/cm

Specific heat capacity (C) of aluminum = 0.217 Cal/g°С

Internal energy = 47000 calories.

To find the change in temperature of this cube of aluminum:

First of all, we would determine the volume of this cube of aluminum.

$Volume \;of \;a \;cube = L^3\n\nVolume \;of \;a \;cube = 20^3\n\nVolume \;of \;a \;cube = 8000\; cm^3$

Next, we calculate the mass of this cube of aluminum:

$Mass = Density * Volume\n\nMass = 2.7 * 8000$

Mass = 21,600 grams.

Now, we can find the change in temperature of this cube of aluminum:

Mathematically, the quantity of heat energy is given by the formula;

$Q = mc\theta$

Where:

Q represents the quantity of heat energy.

m represents the mass of an object.

c is the specific heat capacity.

∅ is the change in temperature.

Substituting the parameters into the formula, we have;

$47000 = 21600 * 0.217 * \theta\n\n47000 = 4687.2 \theta\n\n \theta =(47000)/(4687.2) \n\n \theta = 10.03$

Change in temperature = 10°C

Answer:

10 °C

Explanation:

From the question given above, the following data were obtained:

Egde length (L) of aluminum = 20 cm

Density of Aluminum = 2.7 g/cm³

Specific heat capacity (C) of aluminum = 0.217 cal/ g°С

Heat (Q) energy = 47000 cal

Change in Temperature (ΔT) =?

Next, we shall determine the volume of the aluminum. This can be obtained as follow:

Egde length (L) of aluminum = 20 cm

Volume (V) of aluminum =?

V = L³

V = 20³

V = 8000 cm³

Thus, the volume of the aluminum is 8000 cm³

Next, we shall determine the mass of the aluminum. This can be obtained as follow:

Density of Aluminum = 2.7 g/cm³

Volume of Aluminum = 8000 cm³

Mass of aluminum =.?

Density = mass/volume

2.7 = mass /8000

Cross multiply

Mass of aluminum = 2.7 × 8000

Mass of Aluminum = 21600 g

Finally, we shall determine the change in temperature of the aluminum as follow:

Specific heat capacity (C) of aluminum = 0.217 Cal/g°С

Heat (Q) energy = 47000 Cal

Mass (M) of Aluminum = 21600 g

Change in Temperature (ΔT) =?

Q = MCΔT

47000 = 21600 × 0.217 × ΔT

47000 = 4687.2 × ΔT

Divide both side by 4687.2

ΔT = 47000 / 4687.2

ΔT = 10 °C

Therefore, the increase in the temperature of the aluminum is 10 °C.

A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which it will resonate. Express your answers using two significant figures separated by commas.

Answers

Answer:

160 Hz , 240 Hz , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

$f_1=(320)/(4)\ Hz$

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅= 5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz , 240 Hz , 400 Hz

Final answer:

The resonant frequencies of a string depend on its length, tension, and linear mass density. For a string resonating in four loops at 320 Hz, three possible smaller frequencies could be 160 Hz, 106.7 Hz, and 80 Hz.

Explanation:

When a string resonates, it vibrates at certain frequencies called its resonant frequencies. The resonant frequencies of a string depend on factors such as its length, tension, and linear mass density. In this case, the string resonates in four loops at a frequency of 320 Hz.

Three other possible resonant frequencies at which the string could vibrate with smaller loops include:

160 Hz: This is half the frequency of the given resonant frequency, which means the string vibrates with twice the number of loops.
106.7 Hz: This is one third of the given resonant frequency, which means the string vibrates with three times the number of loops.
80 Hz: This is one fourth of the given resonant frequency, which means the string vibrates with four times the number of loops.

Learn more about Resonant frequencies here:

brainly.com/question/32273580

#SPJ3

Which sling can the crane use to lift the 1000kg pipe?A.
800kg rated sling
B. 1000kg rated sling
C. 2000kg rated sling
D. Band C

Answers

Answer:

C. 2000kg rated sling

Explanation:

ensures better safety and can carry twice more mass than current mass.

A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separation s could you produce in a typical atom by applying

Answers

Answer:

s = 6.25 10⁻²² m

Explanation:

Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by

p = q s

s = p / q

let's calculate

s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹

s = 0.625 10⁻²¹ m

s = 6.25 10⁻²² m

We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.

Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.

Answers

Answer:

178.75 N

Explanation:

The force necessary to start moving the crate must be equal to or more than the frictional force (resistive force) acting on the crate but moving in an opposite direction to the frictional force.

So, we find the frictional force, Fr:

Fr = -μmg

Where μ = coefficient of friction

m = mass

g = acceleration due to gravity

The frictional force is negative because it acts against the direction of motion of the crate.

Fr = -0.57 * 32 * 9.8

Fr = - 178.75 N

Hence, the force necessary to move the crate must be at least equal to but opposite in direction to this frictional force.

Therefore, this force is 178.75 N

Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?

Answers

Answer:

475 N/C

Explanation:

As we know that, the electric field in parallel plate capacitor is same (constant) throughout. And is potential gradient.

So, Electric field is given by

Electric field = potential gradient

$Electric FIeld = (Change\: in\: Potential)/(Distance)$

Here, the potential change is 3.8V and the distance from negative plate to positive plate is 1.6 cm. The potential from negative plate to the center is (1.6/2)cm i.e., 0.8 cm.

But we have to take distance in SI units So, distance= $0.8 * 10^(-2) m$

So, Electric field is

$Electric\: field=(3.8V)/(0.8 * 10^(-2)m )$

$Electric\: field=475 V/m$

So, electric field is 475 Volts per meter.

Note : Also we can say 475 Newtons per coulomb

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