Answer:
The center of mass velocity is [tex]v = 32.25 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the cylinder is [tex]m = 2.50 \ kg[/tex]
The radius is [tex]r = 2.18 \ cm = 0.0218 \ m[/tex]
The length is [tex]l = 2.7 \ cm = 0.027 \ m[/tex]
The height of the plane is h = 79.60 m
and the distance covered is [tex]d = 4.64 \ m[/tex]
The center of mass velocity o the cylinder when it reaches the bottom is mathematically represented as
[tex]v = \sqrt{\frac{4gh}{3} }[/tex]
substituting values
[tex]v = \sqrt{ \frac{4 * 9.8 * 79.60}{3} }[/tex]
[tex]v = 32.25 \ m/s[/tex]
Which kind of energy is found in an atom's nucleus? (A) Nuclear (B) Elastic (C) Thermal (D) Electromagnetic
Answer:
The answer is option A.
NuclearHope this helps you
Answer:
Answer is A Nuclear
Explanation:
Just answered this question on my test
a jet fighter accelerates at 17.7 m/s^2 , increasing its velocity from 119 m/s to 233 m/s. how much time does that take?
Answer:
6.44 s
Explanation:
Given:
v₀ = 119 m/s
v = 233 m/s
a = 17.7 m/s²
Find: t
v = at + v₀
(233 m/s) = (17.7 m/s²) t + (119 m/s)
t = 6.44 s
Find the displacement. Will give brainliest!
Answer:
1000 m upwards
Explanation:
Displacement Formula: Average Velocity = Displacement/Total Time
Simply plug in our known variables and solve:
100 m/s = x m/10 seconds
100 m/s(10 s) = x m
m = 1000
What length (in m) should an oboe have to produce a fundamental frequency of 294 Hz on a day when the speed of sound is 350 m/s? It is open at both ends.
Answer:
L = 0.60 m
The length in metres should be 0.60 m
Explanation:
A pipe open at both ends can have a standing wave pattern with resonant frequency;
f = nv/2L ........1
Where;
v = velocity of sound
L = length of pipe
n = 1 for the fundamental frequency f1
Given;
Fundamental frequency f1 = 294 Hz
Velocity v = 350 m/s
n = 1
From equation 1;
Making L the subject of formula;
L = nv/2f1
Substituting the given values;
L = 1×350/(2×294)
L = 0.595238095238 m
L = 0.60 m
The length in metres should be 0.60 m
is tantalum least reactive or more
Answer:
it is more reactive in high temperature than in low temperature.
Which compound is composed of oppositely charged ions?
Answer:
Option A. Li2O
Explanation:
To know which of the compound contains oppositely charged ions, let us determine the nature of each compound. This is illustrated below:
Li2O is an ionic compound as it contains a metal (Lithium, Li) and non metal (oxygen, O). Ionic compounds are charactized by the presence of aggregate positive and negative charge ions. This is true because they are formed by the transfer of electron(s) from the metallic atom to the non-metallic atom.
2Li —> 2Li^+ + 2e
O2 + 2e —> O^2-
2Li + O2 + 2e —> 2Li^+ + O^2- + 2e
2Li + O2 —> 2Li^+ O^2- —> Li2O
OF2 is a covalent compound as it contains non metals only (i.e oxygen, O and fluorine, F). Covalent compounds are characterised by the presence of molecules. This is true because they are formed from the sharing of electron(s) between the atoms involved.
PH3 is a covalent compound as it contains non metals only (i.e phosphorus, P and hydrogen, H).
SCl2 is a covalent compound as it contains non metals only (i.e sulphur, S and chlorine, Cl).
From the above information, we can see that only Li2O contains oppositely charged ions.
Answer:
A
Explanation:
Just took the test
A small cylinder made of a diamagnetic material is brought near a bar magnet. One end of the cylinder is brought near the North pole of the magnet and is repelled. What happens when the other end of the cylinder is brought near the North pole of the magnet? Group of answer choices
Answer:
it attracts
Explanation:
since in a magnetic body there are two poles
(north and south poles)if the first pole was repeled when brought near the North Pole therefore the other end is going to attarct because the first end was also a North Pole while the second end will be a south pole
Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of 100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)
Answer:
C is the correct answer
Explanation:
Near the surface of Earth, an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (magnitude and sign) would have to be placed on a penny that has a mass of 3 g to cause it to rise into the air with an upward acceleration of 0.19 m/s2?
Answer:
q = 2.997*10^-4C
Explanation:
In order to find the required charge that the penny have to have, to acquire an upward acceleration, you take into account that the electric force on the penny must be higher than the weight of the penny.
You use the second Newton law to sum both electrical and gravitational forces:
[tex]F_e-W=ma\\\\qE-mg=ma[/tex] (1)
Fe: electric force
W: weight of the penny
q: required charge = ?
m: mass of the penny = 3g = 0.003kg
E: magnitude of the electric field = 100N/C
g: gravitational acceleration = 9.8m/s^2
a: acceleration of the penny = 0.19m/s^2
You solve the equation (1) for q, and replace the values of the other parameters:
[tex]q=\frac{ma+mg}{E}=\frac{m(a+g)}{E}\\\\q=\frac{(0.003kg)(0.19m/s^2+9.8m/s^2)}{100N/C}\\\\q=2.997*10^{-4}C[/tex]
It is necessary that the penny has a charge of 2.997*10^-4 C, in order to acquire an upward acceleration of 0.19m/s^2
If an object is determined to have a negative charge of 1.6 micro Coulomb, you can conclude that the object has an excess of
Answer:
The object has an excess of [tex]10^{13}[/tex] electrons.
Explanation:
When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:
[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]
The object has an excess of [tex]10^{13}[/tex] electrons.
Match each term to the best description. ::
1. Coherent
2. Diffraction
3. Grating
4. Interference
5. Specular dot
a. Composed of numerous narrowly spaced parallel slits or grooves
b. Having the same wavelength, frequency, and in-phase
c. Interaction of waves where they meet in space
d. The bending of waves near a boundary or as a wave passes through an opening
e. The zeroth order direct reflection fringe
A vector quantity has direction, a scalar quantity does not.
Explanation:
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A piano tuner hears a beat every 2.20 s when listening to a 266.0 Hz tuning fork and a single piano string. What are the two possible frequencies (in Hz) of the string? (Give your answers to at least one decimal place.)
Answer:
The lower frequency is [tex]f_1 = 265.55 \ Hz[/tex]
The higher frequency is [tex]f_2 = 266.4546 \ Hz[/tex]
Explanation:
From the question we are told that
The period is [tex]T = 2.20 \ s[/tex]
The frequency of the tuning fork is [tex]f = 266.0 \ Hz[/tex]
Generally the beat frequency is mathematically represented as
[tex]f_b = \frac{1}{T}[/tex]
substituting values
[tex]f_b = \frac{1}{2.20}[/tex]
[tex]f_b = 0.4546 \ Hz[/tex]
Since the beat frequency is gotten from the beat produced by the tuning fork and and the string then
The possible frequency of the string ranges from
[tex]f_1 = f- f _b[/tex]
to
[tex]f_2 = f + f_b[/tex]
Now substituting values
[tex]f_1 = 266.0 - 0.4546[/tex]
[tex]f_1 = 265.55 \ Hz[/tex]
For [tex]f_2[/tex]
[tex]f_2 = 266 + 0.4546[/tex]
[tex]f_2 = 266.4546 \ Hz[/tex]
A 2.0-m wire carrying a current of 0.60 A is oriented parallel to a uniform magnetic field of 0.50 T. What is the magnitude of the force it experiences
Answer:
The force experienced is 0.6 N
Explanation:
Given data
length of wire L= 2 m
current in wire I= 0.6 A
magnetic field B= 0.5
The force experienced can be represented as
[tex]F= BIL[/tex]
[tex]F= 0.5*0.6*2\\\F= 0.6 N[/tex]
The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the bal
Complete Question
The complete question is gotten from OpenStax
A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s ? The player’s mass is 70.0 kg, and air resistance is negligible.
The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball
Answer:
The time it will take is [tex]t = 1.4907 \ s[/tex]
Explanation:
From the question we are told that
The force experienced by the player is [tex]F = 126 \ N[/tex]
The distance of the ball from the player is [tex]d = 2.00 \ m[/tex]
The initial velocity is u = 0 m/s because the player stopped
From the Newton law the acceleration of the player is mathematically evaluated as
[tex]a = \frac{F}{m }[/tex] [i,e F = ma ]
substituting values
[tex]a = \frac{126}{70}[/tex]
[tex]a = 1.8 \ m/s^2[/tex]
Now from the equation of motion we have that
[tex]s = ut + \frac{1}{2} at^2[/tex]
substituting values
[tex]2.0 = 0 + \frac{1}{2} * 1.8 * t^2[/tex]
[tex]t = \sqrt{ \frac{2.0}{0.9} }[/tex]
[tex]t = 1.4907 \ s[/tex]
A 6.50-m-long iron wire is 1.50 mm in diameter and carries a uniform current density of 4.07 MA/m^2. Find the voltage between the two ends of the wire.
Answer:
V = 0.45 Volts
Explanation:
First we need to find the total current passing through the wire. That can be given by:
Total Current = I = (Current Density)(Surface Area of Wire)
I = (Current Density)(2πrL)
where,
r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m
L = Length of Wire = 6.5 m
Therefore,
I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]
I = 1.25 x 10⁻⁴ A
Now, we need to find resistance of wire:
R = ρL/A
where,
ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm
A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²
Therefore,
R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)
R = 0.36 Ω
From Ohm's Law:
Voltage = V = IR
V = (1.25 x 10⁻⁴ A)(0.36 Ω)
V = 0.45 Volts
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Answer:
is this a company name.? java is a computer software right..
A small ferryboat is 4.70 m wide and 6.10 m long. When a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into the river. What is the weight of the truck
Answer:
M = 1433.5 kg
Explanation:
This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,
B = ρ g V
with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition
Σ F = 0
B-W = 0
B = W
body weight
W = M g
the volume is
V = l to h
rho_liquid g (l to h) = M g
M = rho_liquid l a h
we calculate
M = 1000 4.7 6.10 0.05
M = 1433.5 kg
At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)
Answer:
The temperature of silver at this given resistivity is 2971.1 ⁰C
Explanation:
The resistivity of silver is calculated as follows;
[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]
where;
Rt is the resistivity of silver at the given temperature
Ro is the resistivity of silver at room temperature
α is the temperature coefficient of resistance
To is the room temperature
T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature
[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]
Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m
When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;
[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]
Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C
If a negative point charge is placed at P without moving the original charges, the net electrical force the charges ±Q will exert on it is
Answer:
The particle P moves directly upwards
Explanation:
Lets designate the negative point charge at point P as particle P
The +Q charge will exert an attractive force on the particle P.
The -Q charge will exert a repulsive force on the particle P
The +Q charge exerts an upwards and leftward force on particle P
The -Q charge exerts an upwards and rightward force on particle P
Since the charges are equidistant from the particle P, and are of equal magnitude, the rightward force and the leftward force will cancel out, leaving just the upward force on the particle P.
The effect of the upward force is that the particle P moves directly upwards
A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnitude 0.7 G directed at 70 below the horizontal north direction. Find the current in the loop.
Answer:
The current is [tex]I = 6.68 \ A[/tex]
Explanation:
From the question we are told that
The radius of the loop is [tex]r = 6 \ cm = 0.06 \ m[/tex]
The earth's magnetic field is [tex]B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T[/tex]
The number of turns is [tex]N =1[/tex]
Generally the magnetic field generated by the current in the loop is mathematically represented as
[tex]B = \frac{\mu_o * N * I}{2 r }[/tex]
Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth
[tex]B = B_e[/tex]
=> [tex]B_e = \frac{\mu_o * N * I }{ 2 * r}[/tex]
Where [tex]\mu[/tex] is the permeability of free space with value [tex]\mu _o = 4\pi * 10^{-7} N/A^2[/tex]
[tex]0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}[/tex]
=> [tex]I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}[/tex]
[tex]I = 6.68 \ A[/tex]
The current in the loop will be "6.68 A".
Magnetic fieldAccording to the question,
Radius of loop, r = 6 cm or,
= 0.06 m
Earth's magnetic field, [tex]B_e[/tex] = 0.7 G or,
= 0.7 × [tex]\frac{1\times 10^{-4}}{1 G}[/tex]
= 0.7 × 10⁻⁴ T
Number of turns, N = 1
We know the relation,
→ B = [tex]\frac{\mu_0\times N\times I}{2r}[/tex]
or,
B = [tex]B_e[/tex]
then,
→ [tex]B_e[/tex] = [tex]\frac{\mu_0\times N\times I}{2r}[/tex]
By substituting the values,
0.7 × 10⁻⁴ = [tex]\frac{4 \pi\times 10^{-7}\times 1\times I}{2\times 0.06}[/tex]
hence,
The current will be:
I = [tex]\frac{2\times 0.06\times 0.7\times 10^{-4}}{4 \pi\times 10^{-7}\times 1}[/tex]
= 6.68 A
Thus the above approach is correct.
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Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?
The complete question is;
In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.
Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?
Answer:
F_top = 385.36 N
Explanation:
We are given;
mass;m = 52 kg
Time;t = 4.3 s
Diameter;d = 16m
So,Radius;r = 16/2 = 8m
The formula for the centrifugal force is given as;
F_c = mω²R
Where;
R = radius
Angular velocity;ω = 2πf
f = frequency = 1/t = 1/4.3 Hz
F_c = 53 × (2π × 1/4.3)² × 8 = 905.29 N.
The force at top would be;
F_top = F_c - mg
F_top = 905.29 - (9.81 × 53) N
F_top = 385.36 N
The force at the top of ride will be "385.36 N".
Force and mass:According to the question,
Rider's mass, m = 52 kg
Time, t = 4.3 s
Diameter, d = 16 m
Radius, r = [tex]\frac{16}{2}[/tex] = 8 m
Frequency, f = [tex]\frac{1}{t}[/tex] = [tex]\frac{1}{4.3}[/tex] Hz
We know the formula,
Centrifugal force, [tex]F_c[/tex] = mω²R
or,
Angular velocity, ω = 2πf
By substituting the values in the above formula,
[tex]F_c = 53(2\pi \times (\frac{1}{4.3})^2\times 8 )[/tex]
[tex]= 905.29[/tex] N
hence,
The top force will be:
→ [tex]F_{top} = F_c[/tex] - mg
By substituting the values,
[tex]= 905.29-(9.81\times 53)[/tex]
[tex]= 385.36[/tex] N
Thus the above response is correct.
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The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?
Answer:
The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.
Explanation:
The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.
The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.
Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.
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If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with speed vo along the same line from the opposite direction toward the source, then the frequency of the sound heard by the observer is fo = c + vo c − vs fs where c is the speed of sound, about 332 m/s. (This is the Doppler effect.) Suppose that, at a particular moment, you are in a train traveling at 32 m/s and accelerating at 1.3 m/s2. A train is approaching you from the opposite direction on the other track at 48 m/s, accelerating at 1.9 m/s2, and sounds its whistle, which has a frequency of 439 Hz. At that instant, what is the perceived frequency that you hear? (Round your answer to one decimal place.) Hz
Answer: The frequency heard is 562.7 Hz.
Explanation: Doppler Effect happens when there is shift in frequency during a realtive motion between a source and the observer of that source.
It can be calculated as:
[tex]f_{o} = f_{s}(\frac{c+v_{o}}{c-v_{s}} )[/tex]
where:
c is the speed of light (c = 332m/s)
all the subscripted s is related to the Source (frequency, velocity);
all the subscripted o is related to the Observer (frequency, velocity);
As the source is moving towards the observer and the observer is moving towards the source, the velocities of each are opposite related to direction.
So, the frequency perceived by the observer:
[tex]f_{o} = 439(\frac{332+32}{332-48} )[/tex]
[tex]f_{o} = 439(\frac{364}{284} )[/tex]
[tex]f_{o} = 439(1.282 )[/tex]
[tex]f_{o}[/tex] = 562.7 Hz
At this condition, the observer hears the train's horn in a perceived frequency of 562.7 Hz
A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 15.2 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Find the net charge on the shell.
Answer:
The net charge on the shell is 30x10^-9C
Explanation:
Pls see attached file
Find the change in the force of gravity between two planets when the distance between them becomes 10 times smaller. g
Answer:
F' = F/16
So, the force of gravity has become 16 times less than initial value.
Explanation:
The force of gravity between two planets, is given by the following formula:
F = Gm₁m₂/r² ----------- equation 1
where,
F = Force of gravity between two planets
G = Gravitational Constant
m₁ = Mass of one planet
m₂ = Mass of other plant
r = Distance between two planets
Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:
F' = Gm₁m₂/(4r)²
F' = (1/16) (Gm₁m₂/r²)
using equation 1:
F' = F/16
So, the force of gravity has become 16 times less than initial value.
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to which of the following quantities?
a. The flux of the magnetic field through a surface which has the loop as its boundary.
b. The negative of the time rate of change of the flux of the magnetic field through a surface which has the loop as its boundary.
c. The line integral of the magnetic field around the closed loop.
d. The flux of the electric field through a surface which has the loop as its boundary.
Answer:
(C). The line integral of the magnetic field around a closed loop
Explanation:
Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.
This can be written mathematically as;
[tex]EMF = -\frac{\delta \phi _B}{\delta t}[/tex]
[tex](\frac{\delta \phi _B}{\delta t} )[/tex] is the rate of change of the magnetic flux through a surface bounded by the loop.
ΔФ = BA
where;
ΔФ is change in flux
B is the magnetic field
A is the area of the loop
Thus, according to Faraday's law of electric generators
∫BdL = [tex]\frac{\delta \phi _B}{\delta t}[/tex] = EMF
Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.
The correct option is "C"
(C). The line integral of the magnetic field around a closed loop
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to: D. The flux of the electric field through a surface which has the loop as its boundary.
In Physics, the surface integral with respect to the normal component of a magnetic field over a surface is the magnetic flux through that surface and it is typically denoted by the symbol [tex]\phi[/tex].
Faraday's Law states that the negative of the time rate of change ([tex]\Delta t)[/tex] of the flux of the magnetic field ([tex]\phi[/tex]) through a surface is directly proportional to the flux ([tex]\phi[/tex]) of the electric field through a surface which has the loop as its boundary.
Mathematically, Faraday's Law is given by the formula:
[tex]E.m.f = -N\frac{\Delta \phi}{\Delta t}[/tex]
Where:
N is the number of turns.Read more: https://brainly.com/question/15121836
A spherical balloon is inflated with helium at a rate of 140pift^3/min How fast is the balloon's radius increasing when the radius is 7ft
Answer:
dr = 0.71 ft/min
the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.
Explanation:
Given;
Rate of inflation dV = 140pift^3/min
Radius r = 7 ft
Change in radius = dr
Volume of a spherical balloon is;
V = (4/3)πr^3
The change in volume can be derived by differentiating both sides;
dV = (4πr^2)dr
Making dr the subject of formula;
dr = dV/(4πr^2)
Substituting the given values;
dr = 140π/(4π×7^2)
dr = 0.714285714285 ft/min
dr = 0.71 ft/min
the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.
Defination coulomb's law.
Answer:
a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?
Answer:
(a) 62.3 N
(b) 1.89 N
(c) 0.430 kg m²
Explanation:
(a) Find the acceleration of block B.
Δy = v₀ t + ½ at²
1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²
a = 0.90 m/s²
Draw a free body diagram of block B. There are two forces:
Weight force mg pulling down,
and tension force Tb pulling up.
Sum of forces in the -y direction:
∑F = ma
mg − Tb = ma
Tb = m (g − a)
Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)
Tb = 62.3 N
(b) Draw a free body diagram of block A. There are three forces:
Weight force mg pulling down,
Normal force N pushing up,
and tension force Ta pulling right.
Sum of forces in the +x direction:
∑F = ma
Ta = ma
Ta = (2.10 kg) (0.90 m/s²)
Ta = 1.89 N
(c) Draw a free body diagram of the pulley. There are two forces:
Tension force Tb pulling down,
and tension force Ta pulling left.
Sum of torques in the clockwise direction:
∑τ = Iα
Tb r − Ta r = Iα
(Tb − Ta) r = I (a/r)
I = (Tb − Ta) r² / a
I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)
I = 0.430 kg m²
The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is [tex]\rm 0.430 \; kg\;m^2[/tex].
Given :
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.a) First, determine the acceleration of the B block.
[tex]\rm s = ut + \dfrac{1}{2}at^2[/tex]
[tex]\rm 1.8 = \dfrac{1}{2}\times a\times (2)^2[/tex]
[tex]\rm a = 0.9\; m/sec^2[/tex]
Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.
[tex]\rm \sum F=ma[/tex]
[tex]\rm mg-T_b=ma[/tex]
[tex]\rm T_b = m(g-a)[/tex]
[tex]\rm T_b = 7\times (9.8-0.9)[/tex]
[tex]\rm T_b = 62.3\;N[/tex]
b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.
[tex]\rm \sum F=ma[/tex]
[tex]\rm T_a=ma[/tex]
[tex]\rm T_a = 2.1\times 0.9[/tex]
[tex]\rm T_a = 1.89\;N[/tex]
c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.
[tex]\rm \sum \tau = I\alpha[/tex]
[tex]\rm T_br-T_ar = I\alpha[/tex]
[tex]\rm I = \dfrac{(T_b-T_a)r^2}{a}[/tex]
Now, substitute the values of the known terms in the above expression.
[tex]\rm I = \dfrac{(62.3-1.89)(0.080)^2}{0.90}[/tex]
[tex]\rm I = 0.430 \; kg\;m^2[/tex]
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