Answer:
The magnitude of the magnetic moment of the rotating ring is [tex]2.96\times10^{-9}\ Am^2[/tex]
Explanation:
Given that,
Radius = 2.2 cm
Charge [tex]q = 6.08\times10^{-6}\ C[/tex]
Angular speed = 2.01 rad/s
We need to calculate the time period
[tex]T = \dfrac{2\pi}{\omega}[/tex]
Now, The spinning produced the current
Using formula for current
[tex] I = \dfrac{q}{T}[/tex]
We need to calculate the magnetic moment
Using formula of magnetic moment
[tex]M = I A[/tex]
Put the value of I and A into the formula
[tex]M=\dfrac{q}{T}\times A[/tex]
[tex]M=\dfrac{q\times\omega}{2\pi}\times \pi\times r^2[/tex]
Put the value into the formula
[tex]M=\dfrac{6.08\times10^{-6}\times2.01\times(2.2\times10^{-2})^2}{2}[/tex]
[tex]M=2.96\times10^{-9}\ Am^2[/tex]
Hence, The magnitude of the magnetic moment of the rotating ring is [tex]2.96\times10^{-9}\ Am^2[/tex]
The magnetic moment of the rotating ring with a radius of 2.2 cm, a total charge of 6.08 µC, and an angular velocity of 2.01 rad/s is approximately 1.176 x 10⁻⁵ A⋅m².
Explanation:The magnitude of the magnetic moment (μ) of a rotating charged ring can be found using the formula given by the product of the charge (Q) and the angular velocity (ω) and the radius (r) squared, as μ = Qωr²/2. For a uniform non-conducting ring rotating about an axis perpendicular to its plane, this relationship can be used to determine the magnetic moment generated by its rotation.
To calculate the magnetic moment for the given scenario:
Carrying out the multiplication gives the magnetic moment:
μ = (6.08 x 10-6 C)(2.01 rad/s)(0.000484 m2)/2 = 0.01176 x 10-6 C⋅m2/s
This results in a magnetic moment of approximately μ ≈ 1.176 x 10-5 A⋅m2.
Be sure to answer all parts. Indicate the number of protons, neutrons and electrons in each of these species: (a) 13 N 7 (b) 35 S 16 (c) 63 Cu 29 (d) 89 Sr 38 How many protons are in each isotope? How many neutrons are in each isotope? How many electrons are in each isotope?
Answer:
(a) No. of protons = atomic number = 7
No. of electrons = no. of protons = 7
No. of neutrons = mass no. - atomic no. = 6
(b) No. of protons = atomic number = 16
No. of electrons = no. of protons = 16
No. of neutrons = mass no. - atomic no. = 19
(c) No. of protons = atomic number = 29
No. of electrons = no. of protons = 29
No. of neutrons = mass no. - atomic no. = 34
(d) No. of protons = atomic number = 38
No. of electrons = no. of protons = 38
No. of neutrons = mass no. - atomic no. = 51
Explanation:
No. of protons = No. of electrons = atomic number
No. of neutrons = Mass no. - atomic no.
(a) 13 N 7
Here atomic number is 7, mass number is 13
No. of protons = atomic number = 7
No. of electrons = no. of protons = 7
No. of neutrons = mass no. - atomic no. = 13 - 7 = 6
(b) 35 S 16
Here atomic number is 16, mass number is 35
No. of protons = atomic number = 16
No. of electrons = no. of protons = 16
No. of neutrons = mass no. - atomic no. = 35 - 16 = 19
(c) 63 Cu 29
Here atomic number is 29, mass number is 63
No. of protons = atomic number = 29
No. of electrons = no. of protons = 29
No. of neutrons = mass no. - atomic no. = 63 - 29 = 34
(d) 89 Sr 38
Here atomic number is 38, mass number is 89
No. of protons = atomic number = 38
No. of electrons = no. of protons = 38
No. of neutrons = mass no. - atomic no. = 89 - 38 = 51
Final answer:
The isotopes ¹N⁷, ³⁵S¹⁶, ⁶³Cu₂₉, and ₈₉Sr₃₈ have 7, 16, 29, and 38 protons, respectively, with neutrons calculated by subtracting the number of protons from the mass number. Lastly, the number of electrons is equal to the number of protons in a neutral atom.
Explanation:
The number of protons, neutrons, and electrons for each isotope can be determined using the atomic number, mass number, and charge. The atomic number represents the number of protons (and also the number of electrons in a neutral atom), the mass number represents the sum of protons and neutrons, and the charge indicates how many electrons have been lost or gained.
(a) ¹N⁷: 7 protons, 6 neutrons, 7 electrons(b) ³⁵S¹⁶: 16 protons, 19 neutrons, 16 electrons(c) ⁶³Cu₂₉: 29 protons, 34 neutrons, 29 electrons(d) ₈₉Sr₃₈: 38 protons, 51 neutrons, 38 electronsEach isotope comprises an equal number of protons and electrons in their neutral states. Neutrons are the difference between the mass number and the number of protons.
A baseball approaches home plate at a speed of 43.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 56.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 1.90 ms. What is the average vector force the ball exerts on the bat during their interaction? (Let the +x-direction be in the initial direction of motion, and the +y-direction be up.)
Answer:
(4273.7 [tex]\hat{j}[/tex] - 3281.6 [tex]\hat{i}[/tex])
Explanation:
[tex]\underset{v_{i}}{\rightarrow}[/tex] = initial velocity of the baseball before collision = 43 [tex]\hat{i}[/tex] m/s
[tex]\underset{v_{f}}{\rightarrow}[/tex] = final velocity of the baseball after collision = 56 [tex]\hat{j}[/tex] m/s
m = mass of the ball = 145 g = 0.145 kg
t = time of contact of the ball with the bat = 1.90 ms = 0.0019 s
[tex]\underset{F_{avg}}{\rightarrow}[/tex] = Average force vector
Using Impulse-change in momentum equation
[tex]\underset{F_{avg}}{\rightarrow}[/tex] t = m ([tex]\underset{v_{f}}{\rightarrow}[/tex] - [tex]\underset{v_{i}}{\rightarrow}[/tex] )
[tex]\underset{F_{avg}}{\rightarrow}[/tex] (0.0019) = (0.145) (56 [tex]\hat{j}[/tex] - 43 [tex]\hat{i}[/tex])
[tex]\underset{F_{avg}}{\rightarrow}[/tex] = (4273.7 [tex]\hat{j}[/tex] - 3281.6 [tex]\hat{i}[/tex])
During the time a compact disc (CD) accelerates from rest to a constant rotational speed of 477 rev/min, it rotates through an angular displacement of 0.250 rev. What is the angular acceleration of the CD
The angular acceleration α of the CD is 1.590 rev/min².
Explanation:The angular acceleration α can be calculated using the formula:
α = [tex]\frac{\Delta \omega}{\Delta t}[/tex]
Where [tex]\(\Delta \omega\)[/tex] is the change in angular velocity and [tex]\(\Delta t\)[/tex] is the change in time. Angular velocity [tex](\(\omega\))[/tex] is related to angular displacement [tex](\(\theta\))[/tex] and time (\(t\)) by the equation:
[tex]\[ \omega = \frac{\theta}{t} \][/tex]
Given that the CD accelerates from rest [tex](\(\omega_0 = 0\))[/tex] to a constant angular velocity of 477 rev/min, and rotates through an angular displacement of 0.250 rev}, we can use the following relationships:
[tex]\[ \omega = \frac{\theta}{t} \][/tex]
[tex]\[ \omega_f = \frac{477}{1} \, \text{rev/min} \][/tex]
[tex]\[ \omega_0 = 0 \, \text{rev/min} \][/tex]
Substituting these values into the first equation, we get:
[tex]\[ \frac{477}{1} = \frac{0.250}{t} \][/tex]
Solving for t, we find t = 0.525 min.
Now, substitute these values into the angular acceleration formula:
α = [tex]\frac{\frac{477}{1} - 0}{0.525}[/tex]
α = 1.590 rev/min²
Therefore, the angular acceleration of the CD is 1.590 rev/min².
At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constant speed. At time t2 = 5.00 s, its acceleration is 4.00 i - 6.00 j m/s2 . What is the radius of the path taken by the particle if t2-t1 is less than one period?
The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.
At time [tex]t_1[/tex], the acceleration vector has direction [tex]\theta_1[/tex] such that
[tex]\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ[/tex]
which indicates the particle is situated at a point on the lower left half of the circle, while at time [tex]t_2[/tex] the acceleration has direction [tex]\theta_2[/tex] such that
[tex]\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ[/tex]
which indicates the particle lies on the upper left half of the circle.
Notice that [tex]\theta_1-\theta_2=90^\circ[/tex]. That is, the measure of the major arc between the particle's positions at [tex]t_1[/tex] and [tex]t_2[/tex] is 270 degrees, which means that [tex]t_2-t_1[/tex] is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.
Recall that
[tex]\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}[/tex]
where [tex]R[/tex] is the radius of the circle and [tex]T[/tex] is the period. We have
[tex]t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s[/tex]
and the magnitude of the particle's acceleration toward the center of the circle is
[tex]\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}[/tex]
So we find that the path has a radius [tex]R[/tex] of
[tex]7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}[/tex]
The acceleration of a particle moving in a circular path with constant speed is given by the centripetal acceleration. Since the velocity is not given, the radius cannot be determined with this information alone.
Explanation:In circular motion, the acceleration of a particle is given by the centripetal acceleration, which is equal to the square of the speed divided by the radius of the circular path (v^2/r). As the particle moves at a constant speed, we know that the magnitude of acceleration (which gives us the centripetal acceleration) remains the same at both times t1 and t2. Therefore, we can calculate the magnitude of acceleration at both points and set them equal to each other. This will allow us to solve for the radius of the path.
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What is the average de Broglie wavelength of oxygen molecules in air at a temperature of 27°C? Use the results of the kinetic theory of gases The mass of an oxygen molecule is 5.31 x 1026 kg
Answer:
[tex]\lambda = 2.57 \times 10^{-11} m[/tex]
Explanation:
Average velocity of oxygen molecule at given temperature is
[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]
now we have
M = 32 g/mol = 0.032 kg/mol
T = 27 degree C = 300 K
now we have
[tex]v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}[/tex]
[tex]v_{rms} = 483.4 m/s[/tex]
now for de Broglie wavelength we know that
[tex]\lambda = \frac{h}{mv}[/tex]
[tex]\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}[/tex]
[tex]\lambda = 2.57 \times 10^{-11} m[/tex]
A ball is thrown straight up. Taking the drag force of air into account, does it take longer for the ball to travel to the top of its motion or for it to fall back down again?
Taking the drag force of air into account, it takes longer for the ball to travel to the top of its motion compared to the time it takes to fall back down.
Explanation:When a ball is thrown straight up, the drag force of air acts against its motion. As the ball ascends, the drag force opposes its upward velocity, causing a reduction in the net force acting on the ball. This results in a slower upward acceleration and a longer time to reach the top of its motion.
On the descent, the gravitational force aids the motion, overcoming the drag force. The downward velocity increases, and the time taken to fall back down is less than the time taken to ascend.
In the absence of air resistance, the ascent and descent times would be equal. However, with air resistance considered, the ascent time is prolonged, impacting the overall motion of the ball.
Compare the electric and gravitational force between the electron and the proton in a hydrogen atom. The electric charge of the electron is -1.60x10-19C and the same but positive for the proton. The electron’s mass is 9.11x10-31 kg, while the proton’s mass is 1.67x10-27kg. Assume the distance between them is 5.3x10-11 m
Explanation:
Electric force is the force acting between two charged particles. Electric force between electron and proton is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Distance between them, [tex]r=5.3\times 10^{-11}\ m[/tex]
[tex]F=9\times 10^9\times \dfrac{-(1.6\times 10^{-19})^2}{(5.3\times 10^{-11}\ m)^2}[/tex]
[tex]F=-8.2\times 10^{-8}\ N[/tex]
Gravitational force is the force acting between two masses. The gravitational force between the electron and proton is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
Distance between them, [tex]r=5.3\times 10^{-11}\ m[/tex]
[tex]F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\ kg\times 1.67\times 10^{-27}\ kg}{(5.3\times 10^{-11}\ m)^2}[/tex]
[tex]F=3.61\times 10^{-47}\ N[/tex]
Hence, this is the required solution.
A 2 kg mass is free falling in the negative Y direction when a 10 N force is exerted in the minus X direction. What is the acceleration of the mass?
Answer:
The mass's acceleration is 5 m/s^2 in the minus X direction and 9,8 m/s^2 in the minus Y direction.
Explanation:
By applying the second Newton's law in the X and Y direction we found that in the minus X direction an external force of 10 N is exerted, while in the minus Y direction the gravity acceleration is acting:
X-direction balance force: [tex]-10 [N] = m.ax [/tex]
Y-direction balance force: [tex]-m*9,8 \frac{m}{s^2} = m.ay[/tex]
Where ax and ay are the components of the respective acceleration and m is the mass. By solving for each acceleration:
[tex]ax=(-10 [N]) / m [/tex]
[tex]ay=-m*9,8\frac{m}{s^2} / m[/tex]
Note that for the second equation above the mass is cancelled and, the Y direction acceleration is minus the gravity acceleration:
[tex]ay=-9,8\frac{m}{s^2} [/tex]
For the x component aceleration we must replace the Newton unit:
[tex]N =\frac{kg.m}{s^2} [/tex]
[tex]ax= -10 \frac{kg.m}{s^2} / (2 kg) [/tex]
[tex]ax= - 5 \frac{m}{s^2} [/tex]
The acceleration of a 2 kg mass with a 10 N force exerted in the negative X direction, while free falling, is 5 m/s² in the X direction. The gravitational acceleration in the Y direction remains 9.8 m/s².
To calculate the acceleration of the mass in the situation where a 2 kg mass free falls in the negative Y direction while a 10 N force is exerted in the negative X direction, we must use Newton's Second Law.
The law states that acceleration is equal to the net force divided by the mass: Acceleration (a) = Force (F) / Mass (m). Since the force is applied in the horizontal (X) direction and there's no opposing force mentioned, the acceleration in the X direction can be calculated. However, for the Y direction, the mass is simultaneously experiencing gravitational acceleration.
In this problem, the force exerted on the mass in the negative X direction is 10 N and the mass is 2 kg. So, we first calculate acceleration in the X direction:
ax = F / m = 10 N / 2 kg = 5 m/s²
Since the Y direction involves free fall, the acceleration there is due to gravity, which on Earth is approximately 9.8 m/s² downward or in the negative Y direction. We don't combine these accelerations directly because they are perpendicular to each other. The magnitude of the total acceleration can be found using the Pythagorean theorem, but since the Y direction acceleration is due to gravity and constant, we typically consider only the X direction when discussing the change caused by the applied force.
A rock is dropped from the top of a tower. When it is 40 meters above the ground velocity of 17 m/s. When its velocity is 24 m/s, how far is it above the ground? Select one: O a. 29.4 meters o b. 25.4 meters o c. 33.3 meters d. 10.7 meters O e. 147 meters
Answer:
Option B is the correct answer.
Explanation:
Let us consider 40 meter above ground as origin.
Initial velocity = 17 m/s
Final velocity = 24 m/s
Acceleration = 9.81 m/s
We have equation of motion v² = u² + 2as
Substituting
24² = 17² + 2 x 9.81 x s
s = 14.63 m
Distance traveled by rock = 14.63 m down.
Height of rock from ground = 40 - 14.63 = 25.37 m = 25.4 m
Option B is the correct answer.
The brakes in a car increase in temperature by ΔT when bringing the car to rest from a speed v. How much greater would ΔT be if the car initially had twice the speed? You may assume the car to stop sufficiently fast so that no heat transfers out of the brakes.\
Answer:
T would be two times the temperature That it was before the double so if the initial temperature was 20 it would now be 40
Explanation:
If the car initially had twice the speed ΔT would be 4 times the original change in temperature of the breaks.
If we assume no heat loss to the surrounding. Then the loss in kinetic energy (KE) of the car after applying the breaks is completely converted into heat energy (Q) of the breaks.
Loss in KE = Gain in Heat Energy
ΔKE = ΔQ
Originally, ΔKE = [tex]\frac{1}{2}mv^{2}[/tex] = ΔQ
Now, ΔQ is directly proportional to change in temperature
[tex]\frac{1}{2}mv^{2}[/tex] = ΔQ ∝ ΔT
If the speed is doubled, v' = 2v, then
ΔKE = [tex]\frac{1}{2}mv^{'} ^{2}[/tex] = [tex]\frac{1}{2}m(4v^{2})[/tex] = 4 × [tex]\frac{1}{2}mv^{2}[/tex] = 4ΔQ ∝ 4ΔT
Hence, the temperature increases 4 times is speed is doubled.
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. A 2150-kg truck is traveling along a straight, level road at a constant speed of 55.0 km/h when the driver removes his foot from the accelerator. After 21.0 s, the truck's speed is 33.0 km/h. What is the magnitude of the average net force acting on the truck during the 21.0 s interval?
Answer:
624.5 N
Explanation:
m = mass of the truck = 2150 kg
v₀ = initial speed of the truck = 55 km/h = 15.3 m/s
v = final speed of the truck = 33 km/h = 9.2 m/s
t = time interval = 21 s
F = magnitude of the average net force
Using Impulse-Change in momentum equation
F t = m (v - v₀ )
F (21) = (2150) (9.2 - 15.3)
F = - 624.5 N
hence the magnitude of net force is 624.5 N
The magnitude of the average net force acting on the truck during the 21.0 s interval is about 626 N
[tex]\texttt{ }[/tex]
Further explanationAcceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration ( m/s² )
v = final velocity ( m/s )
u = initial velocity ( m/s )
t = time taken ( s )
d = distance ( m )
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
mass of truck = m = 2150 kg
initial speed of truck = u = 55.0 km/h = 15⁵/₁₈ m/s
final speed of truck = v = 33.0 km/h = 9¹/₆ m/s
time taken = t = 21.0 s
Asked:
average net force = ∑F = ?
Solution:
[tex]\Sigma F = m a[/tex]
[tex]\Sigma F = m ( v - u ) \div t[/tex]
[tex]\Sigma F = 2150 ( 9\frac{1}{6} - 15 \frac{5}{18} ) \div 21.0[/tex]
[tex]\Sigma F = 2150 ( -6\frac{1}{9} ) \div 21.0[/tex]
[tex]\Sigma F = -626 \texttt{ N}[/tex]
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Kinematics
A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed 14.0 m/s, the total horizontal force on the driver has magnitude 130 N. What is the total horizontal force on the driver if the speed on the same curve is 18.0 m/s instead?
Explanation:
It is given that,
Initial speed, v₁ = 14 m/s
Initial force, F₁ = 130 N
We need to find the total horizontal force (F₂) on the driver if the speed on the same curve is 18.0 m/s instead, v₂ = 18 m/s
The centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]\dfrac{F_1}{F_2}=\dfrac{mv_1^2/r}{mv_2^2/r}[/tex]
[tex]\dfrac{F_1}{F_2}=\dfrac{v_1^2}{v_2^2}[/tex]
[tex]F_2=\dfrac{v_2^2\times F_1}{v_1^2}[/tex]
[tex]F_2=\dfrac{(18\ m/s)^2\times 130\ N}{(14\ m/s)^2}[/tex]
[tex]F_2=214.8\ N[/tex]
So, if the speed is 18 m/s, then the horizontal force acting on the car is 214.8 N. Hence, this is the required solution.
Final answer:
The total horizontal force on the driver when the car's speed increases to 18.0 m/s can be found using the proportionality between the centripetal force and the square of the velocity. By comparing the initial and final conditions, we can calculate the new total horizontal force required for the increased speed.
Explanation:
The question involves finding the total horizontal force on a driver when the speed of the car increases on the same curve. The horizontal force experiences by the driver in a curved path, which is part of a horizontal circle, can be determined by using the formula for centripetal force (Fc = mv²/r). We know from the initial condition that the horizontal force is 130 N when the car is traveling at 14.0 m/s. Since the force depends on the square of the velocity, we can use proportions to find the new force when the speed increases to 18.0 m/s.
Thus, (F2/F1) = (v₂²/v₁²). Substituting the given values, we get (F2/130 N) = (18.0 m/s)² / (14.0 m/s)². Calculating this provides us with the new force F2 which represents the total horizontal force on the driver at the higher speed of 18.0 m/s.
If an electric field is defined as E = Q/r2 what would be the equivalent gravitation field equation?
Explanation:
The electric field is given by :
[tex]E=k\dfrac{Q}{r^2}[/tex]
Where
k is electrostatic constant, [tex]k=\dfrac{1}{4\pi \epsilon_o}=9\times 10^9\ Nm^2/C^2[/tex]
Q is the magnitude of charge
r is the distance from the charge
The gravitational field equation is given by :
[tex]g=\dfrac{GM}{r^2}[/tex]
G is universal gravitational constant
M is the mass
r is the distance
Final answer:
The equivalent gravitational field equation to E = Q/r² is g = G×M/r², where g is the gravitational field strength, G is the gravitational constant, M is the mass, and r is the distance from the mass.
Explanation:
If an electric field is defined by the equation E = Q/r², where E represents the electric field, Q is the charge, and r is the distance from the point charge, the analogous equation for the gravitational field would be g = G×M/r². In this equation, g is the gravitational field strength, G is the gravitational constant, M is the mass creating the gravitational field, and r is the distance from the center of mass.
The concept of the gravitational field is similar to that of the electric field. Both fields decrease with the square of the distance from the source (charge or mass). The field strength represents the force applied per unit charge for the electric field, measured in newtons per coulomb (N/C), and force applied per unit mass for the gravitational field, measured in newtons per kilogram (N/kg).
A 3.5 kg gold bar at 94°C is dropped into 0.20 kg of water at 22°C What is the final temperature? Assume the specific heat of gold is 129 J/kg C Answer in units of °C
Answer:
47.17 degree C
Explanation:
mg = 3.5 kg, T1 = 94 degree C, sg = 129 J/kg C
mw = 0.2 kg, T2 = 22 degree C, sw = 4200 J/kg C
Let T be the temperature at equilibrium.
Heat given by the gold = Heat taken by water
mg x sg x (T1 - T) = mw x sw x (T - T2)
3.5 x 129 x (94 - T) = 0.2 x 4200 x (T - 22)
42441 - 451.5 T = 840 T - 18480
60921 = 1291.5 T
T = 47.17 degree C
A 65 kg astronaut in the international space station is approximately 3.81 × 10^8 m away from the moon. How much gravitation force does she feel? The mass of the Moon is 7.355 × 10^22 kg.
Answer:
[tex]F = 2.2 \times 10^{-3} N[/tex]
Explanation:
Force of gravitation between Astronaut and Moon is given by the Universal law of gravitation
[tex]F = \frac{Gm_1m_2}{r^2}[/tex]
now we will have
[tex]m_1 = 7.355 \times 10^{22} kg[/tex]
[tex]m_2 = 65 kg[/tex]
[tex]r = 3.81 \times 10^8 m[/tex]
now from above formula we have
[tex]F = \frac{(6.67 \times 10^{-11})(65)(7.355 \times 10^{22})}{(3.81 \times 10^8)^2}[/tex]
[tex]F = 2.2 \times 10^{-3} N[/tex]
A celling fan has two different angular speed setting: omega_1 = 440 rev/min and omega_2 = 110 rev/min. what is the ratio a_1/a_2 of the centripetal accelerations of a given point on a blade?
Answer:
Ratio of the centripetal accelerations = 16
Explanation:
We have ω₁ = 440 rev/min, ω₂ = 110 rev/min
We have centripetal acceleration
[tex]a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2[/tex]
Ratio of centripetal acceleration
[tex]\frac{a_1}{a_2}=\frac{r\omega_1 ^2}{r\omega_2 ^2}=\left ( \frac{440}{110}\right )^2=4^2=16[/tex]
Ratio of the centripetal accelerations = 16
The centripetal acceleration ratio of a ceiling fan at different angular speeds can be found using the formula a = rω². The ratio a₁/a₂ for angular velocities of 440 rev/min and 110 rev/min is 16.
The ratio of centripetal accelerations a₁/a₂; can be found using the formula a = rω², where a is the centripetal acceleration, r is the distance from the axis of rotation, and ω is the angular velocity.
For ω₁ = 440 rev/min and ω₂ = 110 rev/min, the ratio a₁/a₂ = (rω₁²)/(rω₂²) = (ω₁/ω₂)² = (440/110)² = 16.
Therefore, the ratio of centripetal accelerations a₁/a₂ for the two angular speeds is 16.
A bottle has a volume of 40.2 liters. a) What is its volume in cubic centimeters?
(b) In cubic meters?
Answer:
a) 40.2 liters = 40200 cubic centimeters
b) 40.2 liters = 0.0402 cubic meters
Explanation:
Volume of bottle = 40.2 liters.
a)
1 liter = 1000 cubic centimeter
40.2 liters = 40.2 x 1000 = 40200 cubic centimeters
b)
1 liter = 0.001 cubic meter
40.2 liters = 40.2 x 0.001 = 0.0402 cubic meters
A boxcar traveling at 8 m/s approaches a string of 4 identical boxcars sitting stationary on the track. The moving boxcar collides and links with the stationary cars and they all move off together along the track. What is the final speed of the cars immediately after the colision? (You may take the mass of each boxcar to be 18,537 kg.)
Answer:
[tex]v_f = 1.6 m/s[/tex]
Explanation:
As per momentum conservation we know that initial momentum of one box car must be equal to the momentum of all box cars together
here we know that there is no external force on the system of all box cars so there momentum conservation is applicable
[tex]m_1v_{i} = (m_1 + m_2 + m_3 + m_4 + m_5)v_{f}[/tex]
so above equation is momentum of one box car is equal to the momentum of all box cars together
So here we will have
[tex]v_f = \frac{m_1 v_i}{m_1 + m_2 + m_3 + m_4 + m_5}[/tex]
[tex]v_f = \frac{m(8)}{5m}[/tex]
[tex]v_f = \frac{8}{5} = 1.6 m/s[/tex]
By applying the conservation of momentum, the final speed of a boxcar linking with four stationary boxcars would be 1.6 m/s. The mass of each boxcar is 18,537 kg and the initial speed of the moving boxcar is 8 m/s before the collision.
The student's question involves finding the final speed of a set of linked boxcars immediately after a collision. This problem deals with the conservation of momentum, a fundamental concept in physics. According to the law of conservation of momentum, the total momentum of a closed system of objects (which does not interact with external forces) remains constant. In this case, because the boxcars are linking together and moving as one unit post-collision, they must conserve momentum.
To compute the final speed after the collision, we will use the formula:
m1 * v1 + m2 * v2 = (m1 + m2 + m3 + m4 + m5) * vf
where
m1 is the mass of the moving boxcarv1 is the initial velocity of the moving boxcarm2, m3, m4, m5 are the masses of the other stationary boxcarsv2 is the initial velocity of the stationary boxcars (zero in this case)vf is the final velocity of the combined cars after collisionGiven that the initial velocity of the stationary boxcars is 0 (since they are at rest) and knowing all the masses are equal:
m = 18,537 kg
v1 = 8 m/s
v2 = 0 m/s (stationary)
We can calculate the final velocity (vf) by the following:
(18,537 kg * 8 m/s) + (4 * 18,537 kg * 0 m/s) = (5 * 18,537 kg) * vf
Solving this for vf gives us:
vf = (18,537 kg * 8 m/s) / (5 * 18,537 kg)
vf = 1.6 m/s
Therefore, the final speed of the boxcars after the collision would be 1.6 m/s.
If the perturbation is a constant potential, how many terms are needed for an exact solution?
a)Infinite
b)One
c)It depends on the nature of the system.
d)Zero
e)Two
The total energy is less than the potential energy because of energy conservation. Because you can solve with different linear solutions, I would say it depends on the nature of the system.
How much work is required to stop a car of mass 1100 kg that is traveling at 27 m/s? 15kJ
30kJ
40 J
401 kJ
Answer:
Work done, W = 401 kJ
Explanation:
It is given that,
Mass of the car, m = 1100 kg
Initial, velocity of the car, u = 27 m/s
Finally it stops, v = 0
According to work energy theorem, the change in kinetic energy is equal to the work done i.e.
[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]
[tex]W=\dfrac{1}{2}\times 1100\ kg\times (0-(27\ m/s)^2)[/tex]
W = -400950 J
or
[tex]W=-401\ kJ[/tex]
-ve sign shows the direction of force and displacement are in opposite direction.
So, 401 kJ work is required to stop a car. Hence, this is the required solution.
To stop an 1100 kg car traveling at 27 m/s, the work required is calculated using the kinetic energy formula KE = ½ mv² and is 401 kJ.
The question asks about the amount of work required to stop a car with a mass of 1100 kg that is traveling at a velocity of 27 m/s. To solve this, we can use the equation for kinetic energy (KE), which is the work needed to accelerate a body of a given mass from rest to its stated velocity. Using the equation KE = ½ mv², we find that KE = ½ (1100 kg) * (27 m/s)² = 399,150 J, which is approximately 401 kJ. Thus, the correct amount of work required to stop the car is 401 kJ.
We decided to make an iced latte by adding ice to a 200 mL hot latte at 45 °C. The ice starts out at 0 C. How much ice do we need to add for the final drink to be 10 C? The latent heat of fusion of ice is 335 J/g. Approximate the latte as water. Express your answer in g, without specifying the units.
Answer:
m = 77.75 g
Explanation:
Here we know that at equilibrium the temperature of the system will be 10 degree C
so heat given by hot latte = heat absorbed by the ice
now we have
heat given by latte = [tex]m s\Delta T[/tex]
[tex]Q_1 = (200)(4.186)(45 - 10)[/tex]
[tex]Q_1 = 29302 J[/tex]
now heat absorbed by ice is given as
[tex]Q_2 = mL + ms\Delta T[/tex]
[tex]Q_2 = m(335 + 4.186(10 - 0))[/tex]
[tex]Q_2 = m(376.86)[/tex]
now by heat balance we have
[tex]Q_1 = Q_2[/tex]
[tex]29302 = m(376.86)[/tex]
[tex]m = 77.75 g[/tex]
To lower the temperature of the hot latte to 10°C, you need to add approximately 124.87 g of ice.
Explanation:To find out how much ice you need to add to a hot latte to lower its temperature to 10°C, you can use the formula Q = m × Lf, where Q is the heat transferred, m is the mass of the ice, and Lf is the latent heat of fusion of ice. First, calculate the heat that needs to be transferred:
Q = (200 mL) × (0.997 g/mL) × (45°C - 10°C) × (4.184 J/g°C)
Q = 41,817 J
Next, divide the heat transferred by the latent heat of fusion of ice to find the mass of ice needed:
m = 41,817 J ÷ 335 J/g
m ≈ 124.87 g
Therefore, you need to add approximately 124.87 g of ice to the hot latte to lower its temperature to 10°C.
The motion of a particle is defined by the relation x = 2t3 – 9t2 +12t +10, where x and t are expressed in feet and seconds, respectively. Determine the time, position, and acceleration of the particle when v = 2.00 ft/s. (Round the final answer to two decimal places.) The time, the position, and the acceleration of the particle when v = 2.00 are:
Explanation:
The motion of a particle is defined by the relation as:
[tex]x=2t^3-9t^2+12t+10[/tex]........(1)
Differentiating equation (1) wrt t we get:
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]v=\dfrac{d(2t^3-9t^2+12t+10)}{dt}[/tex]
[tex]v=6t^2-18t+12[/tex]............(2)
When v = 2 ft/s
[tex]2=6t^2-18t+12[/tex]
[tex]6t^2-18t+10=0[/tex]
t₁ = 2.26 s
and t₂ = 0.73 s
Put the value of t₁ in equation (1) as :
[tex]x=2(2.26)^3-9(2.26)^2+12(2.26)+10[/tex]
x₁ = 14.23 ft
Put the value of t₂ in equation (1) as :
[tex]x=2(0.73)^3-9(0.73)^2+12(0.73)+10[/tex]
x₁ = 14.74 ft
For acceleration differentiate equation (2) wrt t as :
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]a=\dfrac{d(6t^2-18t+12)}{dt}[/tex]
a = 12 t - 18.........(3)
Put t₁ and t₂ in equation 3 one by one as :
[tex]a_1=12(2.26)-18=9.12\ ft/s^2[/tex]
[tex]a_2=12(0.73)-18=-9.24\ ft/s^2[/tex]
Hence, this is the required solution.
The time taken by the particle is 2.26 s and the magnitude of acceleration and position of the particle at the obtained instant are [tex]9.12 \;\rm ft/s^{2}[/tex] and 14.23 ft respectively.
What is linear motion?When an object is observed to move in a straight line, then the motion of the object is known as linear motion. It can be referenced as motion in one dimension.
Given data:
The position of the particle is, [tex]x = 2t^{3}-9t^{2}+12t+10[/tex].
The velocity of the particle is, [tex]v = 2.00 \;\rm ft/s[/tex].
We know that velocity can be obtained by differentiating the position. Then,
[tex]v =\dfrac{dx}{dt}\\\\v = \dfrac{d(2t^{3}-9t^{2}+12t+10)}{dt}\\\\v=6t^{2}-18t+12[/tex]
Now when v = 2.00 ft/s. The time is,
[tex]2 = 6t^{2}-18t+12\\\\6t^{2}-18t+10=0\\\\t = 2.26 \;\rm s[/tex]
Now, the position of the particle is,
[tex]x = 2t^{3}-9t^{2}+12t+10\\\\x = (2 \times 2.26^{3})-(9 \times 2.26^{2})+(12 \times 2.26)+10\\\\x =14.23\;\rm ft[/tex]
And the magnitude of the acceleration is calculated by differentiating the velocity as,
[tex]a = \dfrac{dv}{dt}\\\\a = \dfrac{d(6t^{2}-18t+12)}{dt}\\\\a=12t-18[/tex]
Substituting t = 2.26 s as,
[tex]a = 12(2.26)-18\\\\a=9.12 \;\rm ft/s^{2}[/tex]
Thus, we can conclude that the time taken by the particle is 2.26 s and the magnitude of acceleration and position of the particle at the obtained instant are [tex]9.12 \;\rm ft/s^{2}[/tex] and 14.23 ft respectively.
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In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 30.0 rad/s and the ball is 1.30 m from the elbow joint, what is the velocity of the ball?
Answer:
39.0 m/s
Explanation:
Linear velocity = angular velocity × radius
v = ωr
v = (30.0 rad/s) (1.30 m)
v = 39.0 m/s
The linear velocity of the ball is 39 m/s.
The given parameters;
angular velocity of the ball about the elbow, ω = 30 rad/sthe distance of the ball from the elbow joint, r = 1.3 mThe velocity of the ball is the tangential velocity of the ball along the circular path of the elbow joint.
v = ωr
where;
v is the linear velocity (m/s)ω is the angular velocity (rad/s)r is the radius of the circular path (m)Substitute the given parameters and solve for the value of v;
v = (30 rad/s) x (1.3 m)
v = 39 m/s
Thus, the linear velocity of the ball is 39 m/s.
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Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/s, and the mass of rain per second striking the roof is 0.071 kg/s. (a) Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof. (b) If hailstones having the same mass as the raindrops fall on the roof at the same rate and with the same speed, how would the average force on the roof compare to that found in part (a)?
Answer: (a) 1.065 N (b) 2.13 N
Explanation:
(a) average force exerted by the rain on the roofAccording Newton's 2nd Law of Motion the force [tex]F[/tex] is defined as the variation of linear momentum [tex]p[/tex] in time:
[tex]F=\frac{dp}{dt}[/tex] (1)
Where the linear momentum is:
[tex]p=mV[/tex] (2) Being [tex]m[/tex] the mass and [tex]V[/tex] the velocity.
In the case of the rain drops, which initial velocity is [tex]V_{i}=15m/s[/tex] and final velocity is [tex]V_{f}=0[/tex] (we are told the drops come to rest after striking the roof). The momentum of the drops [tex]p_{drops}[/tex] is:
[tex]p_{drops}=mV_{i}+mV_{f}[/tex] (3)
If [tex]V_{f}=0[/tex], then:
[tex]p_{drops}=mV_{i}[/tex] (4)
Now the force [tex]F_{drops}[/tex] exerted by the drops is:
[tex]F_{drops}=\frac{dp_{drops}}{dt}=\frac{d}{dt}mV_{i}[/tex] (5)
[tex]F_{drops}=\frac{dm}{dt}V_{i}+m\frac{dV_{i}}{dt}[/tex] (6)
At this point we know the mass of rain per second (mass rate) [tex]\frac{dm}{dt}=0.071 kg/s[/tex] and we also know the initial velocity does not change with time, because that is the velocity at that exact moment (instantaneous velocity). Therefore is a constant, and the derivation of a constant is zero.
This means (6) must be rewritten as:
[tex]F_{drops}=\frac{dm}{dt}V_{i}[/tex] (7)
[tex]F_{drops}=(0.071 kg/s)(15m/s)[/tex] (8)
[tex]F_{drops}=1.065kg.m/s^{2}=1.065N[/tex] (9) This is the force exerted by the rain drops on the roof of the car.
(b) average force exerted by hailstones on the roof
Now let's assume that instead of rain drops, hailstones fall on the roof of the car, and let's also assume these hailstones bounce back up off after striking the roof (this means they do not come to rest as the rain drops).
In addition, we know the hailstones fall with the same velocity as the rain drops and have the same mass rate.
So, in this case the linear momentum [tex]p_{hailstones}[/tex] is:
[tex]p_{hailstones}=mV_{i}+mV_{f}[/tex] (9) Being [tex]V_{i}=V_{f}[/tex]
[tex]p_{hailstones}=mV+mV=2mV[/tex] (10)
Deriving with respect to time to find the force [tex]F_{hailstones}[/tex] exerted by the hailstones:
[tex]F_{hailstones}=\frac{d}{dt}p_{hailstones}=\frac{d}{dt}(2mV)[/tex] (10)
[tex]F_{hailstones}=2\frac{d}{dt}(mV)=2(\frac{dm}{dt}V+m\frac{dV}{dt})[/tex] (11)
Assuming [tex]\frac{dV}{dt}=0[/tex]:
[tex]F_{hailstones}=2(\frac{dm}{dt}V)[/tex] (12)
[tex]F_{hailstones}=2(0.071 kg/s)(15m/s)[/tex] (13)
Finally:
[tex]F_{hailstones}=2.13kg.m/s^{2}=2.13N[/tex] (14) This is the force exerted by the hailstones
Comparing (9) and (14) we can conclude the force exerted by the hailstones is two times greater than the force exerted by the raindrops.
The average force exerted by the rain on the roof of a parked car can be calculated using the equation Force = mass × acceleration. If hailstones with the same mass and speed fall on the roof, the average force on the roof would be the same.
Explanation:(a) To find the average force exerted by the rain on the roof, we can use the equation:
Force = mass × acceleration
The mass of rain per second striking the roof is given as 0.071 kg/s. Since the drops come to rest after striking the roof, the acceleration is equal to the initial velocity of the drops, which is 15 m/s. Therefore, the force is:
Force = 0.071 kg/s × 15 m/s = 1.065 N
(b) If hailstones with the same mass as the raindrops fall on the roof at the same rate and with the same speed, the average force on the roof would be the same as found in part (a). The mass and speed of the hailstones are the same as the raindrops, so the force exerted by the hailstones on the roof would also be 1.065 N.
A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.40 Ω, what is the magnitude of the induced current
Answer:
i = 0.5 A
Explanation:
As we know that magnetic flux is given as
[tex]\phi = NBA[/tex]
here we know that
N = number of turns
B = magnetic field
A = area of the loop
now we know that rate of change in magnetic flux will induce EMF in the coil
so we have
[tex]EMF = NA\frac{dB}{dt}[/tex]
now plug in all values to find induced EMF
[tex]EMF = (20)(50 \times 10^{-4})(\frac{6 - 2}{2})[/tex]
[tex]EMF = 0.2 volts[/tex]
now by ohm's law we have
[tex]current = \frac{EMF}{Resistance}[/tex]
[tex]i = \frac{0.2}{0.40} = 0.5 A[/tex]
Using Faraday's Law of electromagnetic induction and Ohm's Law, you can calculate that the magnitude of the induced current in the coil is 0.025 amperes.
Explanation:The magnitude of the induced current can be calculated using Faraday's Law of electromagnetic induction, which states that the induced electromotive force (EMF) in a closed circuit is equal to the time rate of change of magnetic flux through the circuit. Mathematically, it is denoted as:
ε =-N (ΔΦ/Δt)
where
N is the number of turns in the coil,Φ is the magnetic flux through one loop, and(ΔΦ/Δt) is the rate of change of the magnetic flux.In this problem, N=20, the initial magnetic field (B1) is 2.0 T, the final magnetic field (B2) is 6.0 T, and the time interval (Δt) is 2.0 s.
Magnetic flux (Φ) is given by the product of the magnetic field and the area each loop encloses:
Φ = BA
The change in magnetic flux (ΔΦ) is then:
ΔΦ = N * A * ΔB
where A=50 cm²=5 x 10⁻⁴ m² (converted to m² from cm²), and ΔB = B2 - B1 = 4.0 T.
So,
ΔΦ = 20 * 5 x 10⁻⁴ * 4 = 0.02 Wb (webers)
Now we can calculate the induced EMF using Faraday's Law:
ε = -(0.02 Wb) / 2.0 s = -0.01 V (volts)
The negative sign merely indicates that the induced current flows in a direction to oppose the change in the magnetic field.
Finally, we calculate the magnitude of the induced current (I) by dividing the induced EMF by the total resistance (R) of the coil using Ohm's law:
I = ε / R = 0.01V/0.40 Ω = 0.025 A (amperes)
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When the legal speed limit for the New York Thruway was increased from 55 mi/h to 65 mi/h,
how much time was saved by a motorist who drove the 700 km between the Buffalo entrance and
the New York City exit at the legal speed limit?
Answer:
[tex]\Delta t = 1.22 hours[/tex]
Explanation:
Initially legal speed to move through the drive way is 55 mi/h
so it is given as
[tex]v_1 = 55 mph = 24.48 m/s[/tex]
final legal speed is given as
[tex]v_2 = 65 mph = 28.93 m/s[/tex]
now we know that time taken to cover the distance of 700 km is given by the formula
[tex]t = \frac{d}{v}[/tex]
so initially time taken to cover that distance is
[tex]t_1 = \frac{700 \times 10^3 m}{24.48 m/s}[/tex]
[tex]t_1 = 28595 s = 7.94 h[/tex]
Now if legal speed is increased then the time taken to move through same distance is given as
[tex]t_2 = \frac{700 \times 10^3}{28.93}[/tex]
[tex]t_2 = 24196 s = 6.72 h[/tex]
so total time saved is given as
[tex]\Delta t = t_1 - t_2 [/tex]
[tex]\Delta t = 1.22 hours[/tex]
Which has frequency below the audible range? (a) Ultrasonic (b) Supersonic (c) Infrasonic (d) All of these
Answer:
option (c)
Explanation:
The frequency ranges for infrasonic is below 20 Hz.
The frequency ranges for audible is 20 Hz to 20000 Hz.
The frequency ranges for ultrasonic is above 20000 Hz.
So, infrasonic have frequency below the audible range. We cannot hear this frequency.
A large rectangular barge floating on a lake oscillates up and down with a period of 4.5 s. Find the damping constant for the barge, given that its mass is 2.30Ã10^5 kg and that its amplitude of oscillation decreases by a factor of 2.0 in 4.9 minutes.
Answer:
b = 1084.5 kg/s
Explanation:
As we know that the amplitude of damped oscillation is given as
[tex]A = A_o e^{-\frac{bt}{2m}}[/tex]
here we know that
[tex]A = \frac{A_o}{2}[/tex]
after time t = 4.9 minutes
also we know that
[tex]m = 2.30 \times 10^5 kg[/tex]
now we will have
[tex]\frac{A_o}{2} = A_o e^{-\frac{bt}{2m}}[/tex]
[tex]\frac{bt}{2m} = ln2[/tex]
[tex]b = \frac{ln2 (2m)}{t}[/tex]
[tex]b = \frac{2(ln2)(2.30 \times 10^5)}{4.9 \times 60}[/tex]
[tex]b = 1084.5 kg/s[/tex]
ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its n
Answer:
[tex]\beta = 41.68°[/tex]
Explanation:
according to snell's law
[tex]\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }[/tex]
refractive index of water n_w is 1.33
refractive index of glass n_g is 1.5
[tex]sin\alpha = \frac{n_w}{n_g}* sin30[/tex]
[tex]sin\alpha = 0.443[/tex]
now applying snell's law between air and glass, so we have
[tex]\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}[/tex]
[tex]sin\beta = \frac{n_g}{n_a} sin\alpha[/tex]
[tex]\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha][/tex]
we know that [tex]sin\alpha = 0.443[/tex]
[tex]\beta = 41.68°[/tex]
A 3.95 kg mass attached to a spring oscillates with a period of 0.365 s and an amplitude of 19.0 cm.
(a) Find the total mechanical energy of the system.
(b) Find the maximum speed of the mass.
Answer:
(a) 21.11 J
(b) 3.3 m/s
Explanation:
(a)
m = mass attached to the spring = 3.95 kg
A = amplitude of the oscillation = 19 cm = 0.19 m
T = Time period of oscillation = 0.365 s
U = Total mechanical energy
Total mechanical energy is given as
[tex]U = (0.5)mA^{2}\left ( \frac{2\pi }{T} \right )^{2}[/tex]
[tex]U = (0.5)(3.95)(0.19)^{2}\left ( \frac{2(3.14) }{0.365} \right )^{2}[/tex]
U = 21.11 J
(b)
v = maximum speed of the mass
maximum speed of the mass is given as
[tex]v = A\left ( \frac{2\pi }{T} \right )[/tex]
[tex]v = (0.19)\left ( \frac{2(3.14) }{0.365} \right )[/tex]
[tex]v [/tex] = 3.3 m/s