1. Spring constant k is 233.17 N/m.
2. Oscillation frequency f is 0.924 Hz.
3. Speed of the block after 0.42 s is 2.62 m/s.
4. Magnitude of the maximum acceleration is 23.8 m/s².
5. Magnitude of the net force at 0.42 s is 105.02 N.
[tex]\\\( \omega_i \)[/tex]is the initial angular velocity (given as 90.0 rpm), Let's solve the problem step-by-step.
1. What is the spring constant of the spring?
The spring constant k can be determined using Hooke's Law F = kx , where F is the force exerted by the spring and x is the displacement.
At equilibrium, the force exerted by the spring equals the weight of the block:
F = mg
Given:
m = 6.9 kg
g = 9.8 m/s²
x = 0.29 m
We calculate the force:
F = 6.9 kg * 9.8 m/s² = 67.62 N
Now, using F = kx :
k = F/x = 67.62 N/0.29 m = 233.17 N/m
Spring constant k is 233.17 N/m
2. What is the oscillation frequency?
The frequency [tex]\( f \)[/tex] can be found using the formula for the angular frequency [tex]\( \omega \):[/tex]
[tex]\[ \omega = \sqrt{\frac{k}{m}} \][/tex]
Given:
[tex]\[ k = 233.17 \text{ N/m} \]\[ m = 6.9 \text{ kg} \][/tex]
Calculate [tex]\( \omega \):[/tex]
[tex]\[ \omega = \sqrt{\frac{233.17 \text{ N/m}}{6.9 \text{ kg}}} = \sqrt{33.8 \text{ s⁻²}} = 5.81 \text{ rad/s} \][/tex]
The frequency[tex]\( f \)[/tex]is:
[tex]\[ f = \frac{\omega}{2\pi} = \frac{5.81 \text{ rad/s}}{2\pi} \approx 0.924 \text{ Hz} \][/tex]
Oscillation frequency f is 0.924 Hz
3. After t = 0.42 s, what is the speed of the block?
The speed of the block in simple harmonic motion can be described by:
[tex]\[ v(t) = \omega A \cos(\omega t + \phi) \][/tex]
The amplitude A and phase [tex]\( \phi \)[/tex]are determined from initial conditions.
The initial conditions:
- Initial position [tex]\( x(0) = 0 \)[/tex](at equilibrium)
- Initial velocity [tex]\( v(0) = -4.1 \text{ m/s} \)[/tex]
Since [tex]\( x(0) = A \cos(\phi) = 0 \), \(\phi = \pi/2\) (or 90°).[/tex]
Then, [tex]\( v(0) = -\omega A \sin(\phi) = -\omega A \):[/tex]
[tex]\[ -4.1 = -5.81 A \]\[ A = \frac{4.1}{5.81} \approx 0.705 \text{ m} \][/tex]
Now, calculate [tex]\( v(t) \) at \( t = 0.42 \text{ s} \):[/tex]
[tex]\[ v(0.42) = 5.81 \times 0.705 \times \cos(5.81 \times 0.42 + \frac{\pi}{2}) \]\[ v(0.42) = 5.81 \times 0.705 \times \cos(2.44 + 1.57) \]\[ v(0.42) = 4.096 \times \cos(4.01) \]\[ v(0.42) = 4.096 \times -0.64 = -2.62 \text{ m/s} \][/tex]
The speed of the block after 0.42 s is 2.62 m/s
4. What is the magnitude of the maximum acceleration of the block?
The maximum acceleration [tex]\( a_{max} \)[/tex] occurs at the maximum displacement [tex]\( A \):[/tex]
[tex]a_max} = \omega[/tex]² A
Given:
[tex]\[ \omega = 5.81 \text{ rad/s} \]\[ A = 0.705 \text{ m} \][/tex]
Calculate:
[tex]\[ a_{max} = (5.81)^2 \times 0.705 = 33.76 \times 0.705 = 23.8 \text{ m/s^2} \][/tex]
The magnitude of the maximum acceleration is 23.8 m/s²
5. At t = 0.42 s, what is the magnitude of the net force on the block?
The net force ( F(t) ) at any time ( t ) can be found using:
[tex]\[ F(t) = ma(t) \][/tex]
Where [tex]\( a(t) = -\omega^2 x(t) \). From \( x(t) = A \sin(\omega t + \phi) \):[/tex]
[tex]\[ x(0.42) = 0.705 \sin(5.81 \times 0.42 + \frac{\pi}{2}) \]\[ x(0.42) = 0.705 \sin(2.44 + 1.57) \]\[ x(0.42) = 0.705 \sin(4.01) \]\[ x(0.42) = 0.705 \times -0.64 = -0.451 \text{ m} \][/tex]
Now, [tex]\( a(0.42) = -\omega^2 x(0.42) \):[/tex]
[tex]\[ a(0.42) = - (5.81)^2 \times (-0.451) \]\[ a(0.42) = -33.76 \times -0.451 = 15.22 { m/s^2}[/tex]
Then, the net force:
[tex]\[ F(0.42) = 6.9 \text{ kg} \times 15.22 \text{ m/s^2} = 105.02 \text{ N} \][/tex]
The magnitude of the net force at 0.42 s is 105.02 N
A 65-kg ice hockey goalie, originally at rest, catches a 0.145-kg hockey puck slapped at him at a velocity of 35 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came.What would the final velocities of the goalie and the puck be in this case? Assume that the collision is completely elastic.v goalie = _____ m / sv puck = ______ m / s
Answer:
The velocity of the goalie = 0.156 m/s
The velocity of the puck is 34.85 m/s, but in the reverse direction (-34.85 m/s)
Explanation:
The total momentum before equals the totam momentum after the collision for objects with mass m1 and m2
m1*vi1 + m2 vi2 = m1 vf1 + m2 vf2
⇒ with vi = the initial velocity
⇒ with vf = the final velocity
initial momentum = final momentum
An elastic collision is one in which the total kinetic energy of the two colliding objects is the same before and after the collision. For an elastic collision, kinetic energy is conserved. That is:
0.5 m1 vi1² + 0.5 m2 vi2² = 0.5 m1 vf1² + 0.5 m2 vf2²
Combining the above equations gives a solution to the final velocities for an elastic collision of two objects:
vf1 = [(m1 - m2) vi1 + 2m2 vi2]/[m1 + m2]
vf2 = [2m1vi1 − (m1 - m2) vi2]/[m1 + m2]
⇒ with m1 = the mass of the goalie = 65 kg
⇒ with m2 = the mass of the puck = 0.145 kg
⇒ with vi1 = the initial velocity of the goalie 0 m/s
⇒ with vi2 = the initial velocity of the puck = 35 m/s
⇒ with vf1 = The final velocity of the goalie
=> vf1 = [(65 - 0.145)0 + 2(0.145)35]/[65.145]
= 0.1558 ≈ 0.156 m/s. Since it's positve this is in the direction of the puck
=> vf2 = [0 − (65 - 0.145)35]/[65.145] = -34.85 m/s
The velocity of the puck is 34.85 m/s, in the reverse direction. (Has a negative sign)
The heart produces a weak magnetic field that can be used to diagnose certain heart problems. It is a dipole field produced by a current loop in the outer layers of the heart.
a. It is estimated that the field at the center of the heart is 90 pT. What current must circulate around an 8.0-cm-diameter loop, about the size of a human heart, to produce this field?
To produce a magnetic field of 90 pT at the center of an 8.0-cm-diameter loop, a current of approximately 0.045 Amps must circulate around the loop.
Explanation:To calculate the current needed to produce a magnetic field of 90 pT at the center of an 8.0-cm-diameter loop, we can use the formula for the magnetic field strength at the center of a circular loop. The formula is given by B = (µ₀I)/(2R), where B is the magnetic field strength, µ₀ is the permeability of free space, I is the current, and R is the radius of the loop.
First, let's convert 90 pT to Tesla by dividing by 10^12. This gives us a value of 9 x 10^-11 T. We can now substitute this value into the formula along with the given radius of 4.0 cm (half of the diameter) to solve for the current.
B = (µ₀I)/(2R)
9 x 10^-11 T = (4π x 10^-7 T m/A)(I)/(2(0.04 m))
Simplifying the equation, we find:
I = (2π(0.04 m)(9 x 10^-11 T))/(4π x 10^-7 T m/A) = 0.045 A
Therefore, a current of approximately 0.045 Amps must circulate around the 8.0-cm-diameter loop to produce a magnetic field of 90 pT at its center.
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A person who weighs 685 N steps onto a spring scale in the bathroom, and the spring compresses by 0.88 cm. (a) What is the spring constant? N/m (b) What is the weight of another person who compresses the spring by 0.38 cm?
To solve this problem it is necessary to use the concepts of Force of a spring through Hooke's law, therefore,
[tex]F = kx[/tex]
Where,
k = Spring constant
x = Displacement
Initially our values are given,
[tex]F = 685N[/tex]
[tex]x = 0.88 cm[/tex]
PART A ) With this values we can calculate the spring constant rearranging the previous equation,
[tex]k = \frac{F}{x}[/tex]
[tex]k = \frac{685}{0.88*10^-2}[/tex]
[tex]k = 77840.9N/m[/tex]
PART B) Since the constant is unique to the spring, we can now calculate the force through the new elongation (0.38cm), that is
[tex]F = kx[/tex]
[tex]F = (77840.9)(0.0038)[/tex]
[tex]F = 295.79N[/tex]
Therefore the weight of another person is 265.79N.
Final answer:
To determine the spring constant (k) and the weight of another person, Hooke's law is applied. The spring constant is calculated using the weight of the first person and the distance the spring was compressed. Then, with the found spring constant, the weight for another compression distance is determined.
Explanation:
To solve both parts of the question, we make use of Hooke's law, which states that the force (F) needed to compress or extend a spring by some distance (x) is directly proportional to that distance. The law is usually formulated as F = kx, where k is the spring constant. To find the spring constant, we rearrange the formula to k = F/x.
Part A:We are given that a person weighing 685 N compresses the spring by 0.88 cm (which is 0.0088 m). Using the formula k = F/x, we plug in the values to get k = 685 N / 0.0088 m, which gives us the spring constant.
Part B: With the spring constant from part A, we can then determine the weight of another person by using the same Hooke's law. If the spring is compressed by 0.38 cm (which is 0.0038 m), we use the formula F = kx to calculate the new weight (force).
A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicular to the disk faces. A uniform force of 4.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk?
Final answer:
The angular acceleration of the disk is calculated using the torque exerted by the applied force and the moment of inertia of a uniform disk. The calculated angular acceleration is 5.33 rad/s².
Explanation:
The question involves calculating the angular acceleration of a rotating disk when a force is applied tangentially at its rim. To find the angular acceleration, we first need to calculate the torque exerted by the force and then use the moment of inertia of the disk to find the angular acceleration.
Given:
Mass of the disk (m) = 5.0 kgDiameter of the disk (d) = 30 cm = 0.30 m (radius r = d/2 = 0.15 m)Applied force (F) = 4.0 NThe torque (τ) exerted by the force is the product of the force and the radius at which it is applied (torque = force × radius), so:
τ = F × r = 4.0 N × 0.15 m = 0.60 N·m
The moment of inertia (I) for a uniform disk is given by (1/2)mr². Thus:
I = (1/2) × 5.0 kg × (0.15 m)² = 0.1125 kg·m²
Using the formula for angular acceleration (α = torque / moment of inertia), we get:
α = τ / I = 0.60 N·m / 0.1125 kg·m² = 5.33 rad/s²
Therefore, the angular acceleration of the disk is 5.33 rad/s².
A 0.32 kg turntable of radius 0.18 m spins about a vertical axis through its center. A constant rotational acceleration causes the turntable to accelerate from 0 to 24 revolutions per second in 8.0 s. [Hint: model the turntable has a uniform density disk.] What is the rotational acceleration of the turntable
Answer:[tex]18.85 rad/s^2[/tex]
Explanation:
Given
mass of turntable [tex]m=0.32 kg[/tex]
radius [tex]r=0.18 m[/tex]
[tex]N=24 rev/s[/tex]
[tex]\omega =2\pi N=2\pi 24[/tex]
[tex]\omega =48\pi rad/s[/tex]
time interval [tex]t=8 s[/tex]
using [tex]\omega =\omeag _0+\alpha t[/tex]
where [tex]\omega =final\ angular\ velocity[/tex]
[tex]\omega _0=initial\ angular\ velocity[/tex]
[tex]\alpha =angular\ acceleration[/tex]
[tex]t=time[/tex]
[tex]48\pi =0+\alpha \times 8[/tex]
[tex]\alpha =6\pi rad/s^2[/tex]
[tex]\alpha =18.85 rad/s^2[/tex]
A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° north of east. If he walked straight home,
(a)-how far would he have to walk?
(b)-In what direction would he have to walk?
Answer:
a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º
This is 8º north west
Explanation:
This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately
To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.
First vector A = 2 to 20º north west
Measured from the positive x axis is θ = 180 -20 = 160º
We use trigonometry to find the components
Cos 20 = Aₓ / A
sin 20 = [tex]A_{y}[/tex] / A
Aₓ = A cos 160 = 2 cos 160
[tex]A_{y}[/tex] = A sin160 = 2 sin160
Aₓ = -1,879 mi
[tex]A_{y}[/tex] = 0.684 mi
Second vector B = 4 mi 10º west of the south
Angle θ = 270 - 10 = 260º
cos 2600 = Bₓ / B
sin 260 = [tex]B_{y}[/tex] / B
Bₓ = B cos 260
[tex]B_{y}[/tex] = B sin 260
Bₓ = 4 cos 260
[tex]B_{y}[/tex] = 4 sin 260
Bₓ = -0.6946mi
[tex]B_{y}[/tex] = - 3,939 mi
Third vector C = 3 mi to 15 north east
cos 15 = Cₓ / C
sin15 = [tex]C_{y}[/tex] / C
Cₓ = C cos 15
[tex]C_{y}[/tex] = C sin15
Cₓ = 3 cos 15
[tex]C_{y}[/tex] = 3 sin 15
Cₓ = 2,898 mi
[tex]C_{y}[/tex] = 0.7765 mi
Now we can find the final position of the person
X = Aₓ + Bₓ + Cₓ
X = -1.879 -0.6949 + 2.898
X = 0.3241 mi
Y = [tex]A_{y}[/tex] + [tex]B_{y}[/tex] + [tex]C_{y}[/tex]
Y = 0.684 - 3.939 +0.7765
Y = -2.4785 mi
a) We use Pythagoras' theorem
R = √ (x2 + y2)
R = √ (0.3241 2 + (-2.4785) 2)
R = 2.4996 mi
R = 2.5 mi
b) let's use trigonometry
Tan θ = y / x
Tanθ = -2.4785 / 0.3241
θ = tan⁻¹ (-7,647)
θ = -82
Measured from the positive side of the x axis is Te = 360 - 82 = 278º
(90-82) south east
To return to your case you must walk in the opposite direction or Te = 98º
This is 8º north west
PROBLEM A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at t = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions. (b) What is the angular speed of the wheel at t = 2.00 s? (c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles? STRATEGY The angular acceleration is constant, so this problem just requires substituting given values into the proper equations. SOLUTION
Answer:
11 radians or 1.7507 revolutions
9 rad/s
7.27565 revolutions
Explanation:
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation
t = Time taken
Equation of rotational motion
[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=2\times 2+\frac{1}{2}\times 3.5\times 2^2\\\Rightarrow \theta=11\ rad=\frac{11}{2\pi}=1.7507\ rev[/tex]
Angle the wheel rotates in the given time is 11 radians or 1.7507 revolutions
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=2+3.5\times 2\\\Rightarrow \omega_f=9\ rad/s[/tex]
The angular speed of the wheel at the given time is 9 rad/s
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{(9\times 2)^2-2^2}{2\times 3.5}\\\Rightarrow \theta=45.71428\ rad=\frac{45.71428}{2\pi}=7.27565\ rev[/tex]
The number of revolutions if the final angular speed doubles is 7.27565 revolutions
This question involves the concepts of the equations of motion for angular motion.
(a) The wheel rotates through the angle "11 rad" (OR) "1.75 rev".
(b) The angular speed at t = 2 s is "9 rad/s".
(c) The angular displacement for double speed is "45.71 rad" (OR) "7.28 rev".
(a)
The angular displacement can be found using the second equation of motion for angular motion.
[tex]\theta = \omega_it+\frac{1}{2}\alpha t^2[/tex]
where,
θ = angular displacement = ?
ωi = initial angular speed = 2 rad/s
t = time interval = 2 s
α = angular acceleration = 3.5 rad/s²
Therefore,
[tex]\theta = (2\ rad/s)(2\ s)+\frac{1}{2}(3.5\ rad/s^2)(2\ s)^2\\\theta = 4\ rad\ +\ 7\ rad[/tex]
θ = 11 rad
[tex]\theta = (11\ rad)(\frac{1\ rev}{2\pi\ rad})[/tex]
θ = 1.75 rev
(b)
The angular speed can be found using the first equation of motion for angular motion.
[tex]\omega_f = \omega_i+\alpha t\\\omega_f = 2\ rad/s + (3.5\ rad/s^2)(2\ s)\\[/tex]
ωf = 9 rad/s
(c)
The angular displacement can be found using the third equation of motion for angular motion after we double the value of ωf.
[tex]2\alpha \theta = \omega_f^2-\omega_i^2\\2(3.5\ rad/s^2)\theta = (18\ rad/s)^2-(2\ rad/s)^2\\\\\theta = \frac{320\ rad^2/s^2}{7\ rad/s^2}\\\\[/tex]
θ = 45.71 rad
[tex]\theta = (45.71\ rad)(\frac{1\ rev}{2\pi\ rad})[/tex]
θ = 7.28 rev
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The attached picture shows the angular equations of motion.
A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic field of magnitude 0.10 T . When the field is oriented normal to the slab's rectangular face, a Hall emf of 20 mV is measured across the slab's width. The probe is then placed in a magnetic field of unknown magnitude B, and a Hall emf of 69 mV is measured. Determine B assuming that the angle θ between the unknown field and the plane of the slab's rectangular face is
(a) θ = 90
(b) θ = 60
Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf, [tex]V_{Hall} = 20\ mV = 0.02\ V[/tex]
Magnetic Field, B = 0.10 T
Hall emf, [tex]V'_{Hall} = 69\ mV = 0.069\ V[/tex]
Now,
Drift velocity, [tex]v_{d} = \frac{V_{Hall}}{B}[/tex]
[tex]v_{d} = \frac{0.02}{0.10} = 0.2\ m/s[/tex]
Now, the expression for the electric field is given by:
[tex]E_{Hall} = Bv_{d}sin\theta[/tex] (1)
And
[tex]E_{Hall} = V_{Hall}d[/tex]
Thus eqn (1) becomes
[tex]V_{Hall}d = dBv_{d}sin\theta[/tex]
where
d = distance
[tex]B = \frac{V_{Hall}}{v_{d}sin\theta}[/tex] (2)
(a) When [tex]\theta = 90^{\circ}[/tex]
[tex]B = \frac{0.069}{0.2\times sin90} = 0.345\ T[/tex]
(b) When [tex]\theta = 60^{\circ}[/tex]
[tex]B = \frac{0.069}{0.2\times sin60} = 0.398\ T[/tex]
Beaker A contains 100 mL of water at a temperature of 25 °C. Beaker B contains 100 mL of water at a temperature of 60 °C. Which of the following statement is true?
a. Beaker A has lower kinetic energy than beaker B
b. Beaker A has higher thermal energy than beaker B
c. Beaker A has higher potential energy than beaker B
d. Beaker A has lower potential energy than beaker B
e. Beaker A has higher kinetic energy than beaker B
Answer:
Only option A is correct. Beaker A has lower kinetic energy than beaker B.
Explanation:
Step 1: Data given
Beaker 1 has a volume of 100 mL at 25 °C
Beaker B has a volume of 100 mL at 60 °C
Thermal energy = m*c*T
Thermal energy beaker A = 100 grams*4.184 * 25°C
Thermal energy beaker B = 100 grams *4.184*60°C
⇒ Since both beakers contain the same amount of water, the thermal energy depends on the temperature.
Since beaker B has a higher temperature, it has a higher thermal energy than beaker A
When we heat a substance, its temperature rises and causes an increase in the kinetic energy of its constituent molecules. Temperature is, in fact, a measure of the kinetic energy of molecules.
This means beaker B has a higher kinetic energy than beaker A
Potential energy doesn't depend on temperature. this means the potential energy of beaker A and beaker B is the same.
a. Beaker A has lower kinetic energy than beaker B. This is correct.
b. Beaker A has higher thermal energy than beaker B. This is false.
c. Beaker A has higher potential energy than beaker B. This is false.
d. Beaker A has lower potential energy than beaker B. This is false
e. Beaker A has higher kinetic energy than beaker B. This is false.
Beaker A has lower kinetic energy than Beaker B because the temperature of a substance is directly proportional to its average kinetic energy. Thus, the water in Beaker B, which is at a higher temperature, has higher kinetic energy than the cooler water in Beaker A.
Explanation:The correct answer to your question is: a. Beaker A has lower kinetic energy than Beaker B. This is because the temperature of a substance is directly proportional to its average kinetic energy. Kinetic energy refers to the energy of movement, so in this case, it's the energy of the water molecules moving within each beaker. In Beaker B, the water is at a higher temperature, meaning the water molecules are moving more quickly, and therefore have higher kinetic energy. On the other hand, in Beaker A, the water is at a lower temperature, so the water molecules are moving more slowly, which means they have lower kinetic energy.
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A bungee jumper, whose mass is 85 kg, jumps from a tall building. After reaching his lowest point, he continues to oscillate up and down, reaching the low point two more times in 6.8 s. Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant.
Answer:
K= 290.28 N/m
Explanation:
Given: mass= 85 kg
the time taken to reach point two more times in 6.8 s.
2×t= 6.8 sec
t= 6.8/2= 3.4 sec
then, the time period for oscillation is
[tex]t= 2\pi\sqrt{\frac{m}{k} }[/tex]
Here K= spring constant
m= mass of jumper
⇒[tex]K= \frac{4\pi^2m}{t^2}[/tex]
now plugging the values we get
[tex]K= \frac{4\pi^2\times85}{3.4^2}[/tex]
K= 290.28 N/m
The Eiffel Tower is built of wrought iron approximately 300 m tall.Estimate how much its height changes between January (averagetemperature of 2o C) and July (average temperature of25o C). Ignore the angles of the iron beams and treatthe tower as a vertical beam.
The concepts required to solve this problem are the thermodynamic expressions of expansion and linear expansion.
In mathematical terms the dilation of a body can be expressed as
[tex]\Delta L = L_0 \alpha \Delta T[/tex]
Where,
[tex]L_0 =[/tex] Initial Length
[tex]\alpha =[/tex] Thermal coefficient of linear expansion
[tex]\Delta T =[/tex] Change in Length
Our values are given as
[tex]L_0 = 300m[/tex]
[tex]T_f = 25\°C[/tex]
[tex]T_i = 2\°C[/tex]
[tex]\alpha = 12*10^{-6}C^{-1} (Iron)[/tex]
Replacing at the equation we have,
[tex]\Delta L = L_0 \alpha \Delta T[/tex]
[tex]\Delta L = (300)(12*10^{{-6})(25-2)[/tex]
[tex]\Delta L = 0.0828m[/tex]
Therefore the change in the height is 8.28cm
The Eiffel Tower's height is subject to changes due to thermal expansion. In fact, it's around 8.3 cm taller in July than in January, with the temperature rise contributing to this change.
Explanation:The height change of the Eiffel Tower due to thermal expansion can be estimated using the linear expansion formula: ΔL = α*L*ΔT. Here, ΔL is the change in length, α is the thermal expansion coefficient, L is the original length, and ΔT is the change in temperature. Wrought iron has an expansion coefficient of approximately 12*10^-6 °C^-1. The height of the Eiffel Tower is approximately 300 m. The difference in average temperatures is approximately 23°C (25°C-2°C).
Let's plug these values into the formula: ΔL = 12*10^-6 °C^-1 * 300m * 23°C, which gives us a change in height of approximately 0.083m, or 8.3 cm. Therefore, the Eiffel Tower is an estimated 8.3 cm taller in July than it is in January, due to the effects of thermal expansion.
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Which of the following is characteristic of a nuclear reaction?
I. Electrons in atomic orbitals are involved in the breaking and forming of bonds
II. Elements are converted from one to another
III. Reaction rates are typically not affected by catalysts
IV. Atoms are rearranged by the breaking and forming of chemical bonds
a. I and II
b. I, II, and III
c. Ill and IV
d. II and III
e. I, II, III and IV
Answer:
d
Explanation:
Nuclear reactions unlike chemical reactions do not involve electrons or bonds.The main use for the reaction is neutrons. In nuclear there is only turning the element into another element and releasing a large amount of energy. So catalyst are also redundant in nuclear reactions.
Therefore, points III and 3 are true, and rest are all wrong. therefore option d is correct
In a nuclear reaction, elements are converted from one to another and the reaction rates are typically not affected by catalysts. Characteristics such as electrons in atomic orbitals being involved in breaking and forming bonds, and atoms being rearranged by breaking and forming of chemical bonds, pertain to chemical reactions, not nuclear ones.
Explanation:The correct characteristics of a nuclear reaction are options II and III. In a nuclear reaction, elements are indeed converted from one to another (II). This happens as the structure of an atom's nucleus changes. Furthermore, nuclear reaction rates are typically not affected by catalysts (III). This contrasts with chemical reactions, where catalysts can significantly speed up reaction rates. As for options I and IV, they are characteristic of chemical reactions, not nuclear ones. In chemical reactions, electrons in atomic orbitals are involved in the breaking and forming of bonds (I), and atoms are rearranged by the breaking and forming of chemical bonds (IV). Therefore, the answer is (d) II and III.
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A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2 . a) After 2 seconds of acceleration, how many radians of rotation has the wheel completed? b) After 3 seconds of acceleration, what is the wheel’s kinetic energy due to rotation?
Answer:
(A) Angular speed 40 rad/sec
Rotation = 50 rad
(b) 37812.5 J
Explanation:
We have given moment of inertia of the wheel [tex]I=25kgm^2[/tex]
Initial angular velocity of the wheel [tex]\omega _0=10rad/sec[/tex]
Angular acceleration [tex]\alpha =15rad/sec^2[/tex]
(a) We know that [tex]\omega =\omega _0+\alpha t[/tex]
We have given t = 2 sec
So [tex]\omega =10+15\times 2=40rad/sec[/tex]
Now [tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad[/tex]
(b) After 3 sec [tex]\omega =10+15\times 3=55rad/sec[/tex]
We know that kinetic energy is given by [tex]Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J[/tex]
A 15-cm-tall closed container holds a sample of polluted air containing many spherical particles with a diameter of 2.5 μm and a mass of 1.9 x 10^−14 kg. How long does it take for all of the particles to settle to the bottom of the container?
To determine how long it takes for particles to settle at the bottom of a container, principles of sedimentation in physics are applied. The calculation considers gravitational forces, buoyancy, and drag forces. However, exact time estimation requires additional specific information on fluid characteristics.
Explanation:The question revolves around the concept of sedimentation, which is a physical process where particulate matter settles down at the bottom of a container due to gravity. To calculate how long it takes for spherical particles with a diameter of 2.5 μm (micrometers) and a mass of 1.9 x 10⁻¹⁴ kg to settle at the bottom of a 15-cm-tall container, we need to understand the principles of particle sedimentation in fluids, a topic studied in physics. This involves the forces acting on the particles, including gravitational forces, buoyancy, and drag forces, along with the properties of the fluid and the particles themselves. However, without a specific formula or further details on the characteristics of the fluid (e.g., viscosity) and assuming we ignore any air resistance or buoyancy effects, it's difficult to provide an exact time for sedimentation.
Therefore, it takes approximately [tex]\(4657.76\)[/tex] seconds for all the particles to settle to the bottom of the container.
To find the time it takes for all the particles to settle to the bottom of the container, we can use Stokes' law, which relates the settling velocity of particles to their size and density.
Stokes' law states: [tex]\( v = \frac{{2gr^2(\rho_p - \rho)}}{{9\eta}} \),[/tex] where:
- v is the settling velocity of the particle,
- g is the acceleration due to gravity [tex](\( 9.8 \, \text{m/s}^2 \)),[/tex]
- r is the radius of the particle [tex](\( 1.25 \times 10^{-6} \, \text{m} \)),[/tex]
- [tex]\( \rho_p \)[/tex]is the density of the particle[tex](\( \frac{m}{V} = \frac{m}{\frac{4}{3}\pi r^3} = \frac{1.9 \times 10^{-14}}{\frac{4}{3}\pi(1.25 \times 10^{-6})^3} \)),[/tex]
- [tex]\( \rho \)[/tex] is the density of the fluid [tex](\( \rho = 1.2 \, \text{kg/m}^3 \)),[/tex] and
- [tex]\( \eta \)[/tex] is the viscosity of the fluid[tex](\( \eta = 1.81 \times 10^{-5} \, \text{Pa s} \)).[/tex]
After substituting the values into Stokes' law and solving for the settling velocity, we found it to be approximately[tex]\(3.22 \times 10^{-5} \, \text{m/s}\).[/tex]
Then, using the formula [tex]\(t = \frac{h}{v}\), where \(h\)[/tex]is the height of the container (0.15 m), we calculated the time it takes for the particles to settle to the bottom:
[tex]\[ t \approx 4657.76 \, \text{s} \][/tex]
wo balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L = 0.550 m. One is attached at one end of the rod and the other at the middle of the rod. If the rod is held by the open end and rotates in a circular motion with angular speed of 45.6 revolutions per second,
Answer:
[tex]T_1 =677224.40\ N[/tex]
Explanation:
given,
mass of the both ball = 5 Kg
length of rod = 2 L
where L = 0.55 m
angular speed = 45.6 rev/s
ω = 45.6 x 2 π
ω = 286.51 rad/s
v₁ = r₁ ω₁
v₁ =0.55 x 286.51 = 157.58 m/s
v₂ = r₂ ω₂
v₂ = 1.10 x 286.51 = 315.161 m/s
finding tension on the first half of the rod
r₁ = 0.55 r₂ = 2 x r₁ = 1.10
[tex]T_1 = m (\dfrac{v_1^2}{r_1}+\dfrac{v_2^2}{r_2})[/tex]
[tex]T_1 = 5 (\dfrac{157.58^2}{0.55_1}+\dfrac{315.161^2}{1.1})[/tex]
[tex]T_1 =677224.40\ N[/tex]
In large buildings, hot water in a water tank is circulated through a loop so that the user doesn’t have to wait for all the water in long piping to drain before hot water starts coming out. A certain recirculating loop involves 40-m-long, 1.2-cm-diameter cast iron pipes with six 90° threaded smooth bends and two fully open gate valves. If the average flow velocity through the loop is 2 m/s, determine the required power input for the recirculating pump. Take the average water temperature to be 60°C and the efficiency of the pump to be 76 percent. The density and dynamic viscosity of water at 60°C are rho = 983.3 kg/m3, μ = 0.467 × 10–3 kg/m·s. The roughness of cast iron pipes is 0.00026 m. The loss coefficient is KL = 0.9 for a threaded 90° smooth bend and KL = 0.2 for a fully open gate valve. (Round the final answer to three decimal paces.)
Answer:
The power input is 0.102 kW
Solution:
As per the question:
Length of the loop, L = 40 m
Diameter of the loop, d = 1.2 cm
Velocity, v = 2 m/s
Loss coefficient of the threaded bends, [tex]K_{L, bend} = 0.9[/tex]
Loss coefficient of the valve, [tex]K_{L, valve} = 0.2[/tex]
Dynamic viscosity of water, [tex]\mu = 0.467\times 10^{- 3}\ kg/m.s[/tex]
Density of water, [tex]\rho = 983.3\ kg/m^{3}[/tex]
Roughness of the pipe of cast iron, [tex]\epsilon = 0.00026\ m[/tex]
Efficiency of the pump, [tex]\eta = 0.76[/tex]
Now,
We calculate the volume flow rate as:
[tex]\dot{V} = Av[/tex]
where
[tex]\dot{V}[/tex] = Volume rate flow
A = Area
v = velocity
[tex]\dot{V} = \frac{\pi}{4}d^{2}\times 2 = 2.262\times 10^{- 4}\ m^{3}/s[/tex]
For this, Reynold's N. is given by:
[tex]R_{e} = \frac{\rho vd}{\mu}[/tex]
[tex]R_{e} = \frac{983.3\times 2\times 0.012}{0.467\times 10^{- 3}} = 50533.62[/tex]
Since, [tex]R_{e}[/tex] > 4000, the flow is turbulent in nature.
Now,
With the help of the Colebrook eqn, we calculate the friction factor as:
[tex]\frac{1}{\sqrt{f}} = - 2log[\frac{\frac{\epsilon}{d}}{3.7} + \frac{2.51}{R_{e}\sqrt{f}}][/tex]
[tex]\frac{1}{\sqrt{f}} = - 2log[\frac{\frac{0.00026}{0.012}}{3.7} + \frac{2.51}{50533.62\sqrt{f}}][/tex]
f = 0.05075
Now,
To calculate the total head loss:
[tex]H_{loss} = (\frac{fL}{d} + 6K_{L, bend} + 2K_{L, valve})\farc{v^{2}}{2g}[/tex]
[tex]H_{loss} = (\frac{0.05075\times 40}{0.012} + 6\times 0.9 + 2\times 0.2)\farc{2^{2}}{2\times 9.8} = 35.71\ m[/tex]
Now,
The drop in the pressure can be calculated as:
[tex]\Delat P = \rho g H_{loss}[/tex]
[tex]\Delat P = 983.3\times 9.8\times 35.71 = 344.113\ kN/m^{2}[/tex]
Now,
to calculate the input power:
[tex]\dot{W} = \frac{\dot{W_{p}}}{\eta}[/tex]
[tex]\dot{W} = \frac{\dot{V}\Delta P}{0.76}[/tex]
[tex]\dot{W} = \frac{2.262\times 10^{- 4}\times 344.113\times 1000}{0.76} = 0.102\ kW[/tex]
A brass statue with a mass of 0.40 kg and a density of 8.00×103kg/m3 is suspended from a string. When the statue is completely submerged in an unknown liquid, the string tension is 3.2 N. What is the density of the liquid?
Answer:
ρ = 1469 kg/m³
Explanation:
given,
mass of statue = 0.4 Kg
density of statue = 8 x 10³ kg/m³
tension in the string = 3.2 N
density of the fluid = ?
Volume of the statue
[tex]V = \dfrac{0.4}{8\times 10^3}[/tex]
V = 5 x 10⁻⁵ m³
W = ρ g V
W = ρ x 9.8 x 5 x 10⁻⁵
now, tension on the string will be equal to
T = mg - W
3.2 = 0.4 x 9.8 - ρ x 9.8 x 5 x 10⁻⁵
ρ x 9.8 x 5 x 10⁻⁵ = 0.72
ρ = 1469 kg/m³
You are riding in an enclosed train car moving at 90 km/h. If you throw a baseball straight up, where will the baseball land?
a. In front of you.
b. In your hand.
c. Behind you.
d. Can't decide from the given information.
Answer:b. In your hand
Explanation: The objects from an outer perspective are moving at 90km/h all of them, even if there was a fly flying around in the train. Then if there is no change of speed of the train the moment the ball leaves the hand its relative speed with respect to the hand is zero (0), then it will land in the hand again. So the only way for it not to happen is either you move the hand or the train changes its speed, therefore the relative speed of the
ball with respect to the hand would be differente from zero and further information would be required, but since it is not stated that neither the train accelerates nor the hand is going to be moved, we can affirm the ball will land in the hand.
A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. Even without knowing how big the pressure cooker is, or what altitude it is being used at, we can make predictions about how much force the lid will experience under different conditions. Part A If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force F120 on the lid when the air inside the cooker had been heated to 120∘C? Assume that the temperature of the air outside the pressure cooker is 20∘C (room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be pa. Treat the air, both inside and outside the pressure cooker, as an ideal gas obeying pV=NkBT. Express the force in terms of given variables. View Available Hint(s) F120F 120 F_120 = nothing Part B The pressure relief valve on the lid is now opened, allowing hot air to escape until the pressure inside the cooker becomes equal to the outside pressure pa. The pot is then sealed again and removed from the stove. Assume that when the cooker is removed from the stove, the air inside it is still at 120∘C. What is the magnitude of the net force F20 on the lid when the air inside the cooker has cooled back down to 20∘C?
The net force on the pressure cooker lid depends on the temperature inside the cooker. At 120°C, the net force is found using the ideal gas law and the given parameters. After the cooker cools to 20°C, and pressure is equalized, the net force is zero.
Explanation:Part A: The net force on the pressure cooker lid at 120°C is dependent on the difference in pressure inside and outside the cooker. Using the ideal gas law, we find pressure is proportional to temperature (p = NkBT/V). The force exerted on the cooker lid is the product of this pressure and the lid's area (F = pA). Given that the pressure outside the pot is pa (atmospheric pressure) at a temperature of 20°C and the pressure inside increased to a temperature of 120°C, the net force on the lid is F120 = A(NkB(120+273)/V - pa), where T is in Kelvins.
Part B: After the pressure equalizes and the cooker has cooled down to 20°C, the pressures inside and outside the cooker are equal, therefore the net force is zero. This is because the difference in pressures is what causes the force (F20 = A(pa - pa) = 0)
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In the vicinity of Earth's orbit around the Sun, the energy intensity of sunlight is about 1200 W/m2. What is the approximate magnitude of the electric field in the sunlight? (What you calculate is actually the "root-mean-square" or "rms" magnitude of the electric field, because in sunlight the magnitude of the electric field at a fixed location varies sinusoidally, and the intensity is proportional to E2.)
Answer:
672.29 W/m²
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
I = Intensity of light = 1200 W/m²
[tex]E_m[/tex] = Maximum value electric field
Intensity of light is given by
[tex]I=\frac{1}{2}\epsilon_0cE_m^2\\\Rightarrow E_m=\sqrt{\frac{2I}{\epsilon_0c}}\\\Rightarrow E_m=\sqrt{\frac{2\times 1200}{8.85\times 10^{-12}\times 3\times 10^8}}\\\Rightarrow E_m=950.765\ N/C[/tex]
RMS value
[tex]E_r=\frac{E_m}{\sqrt2}\\\Rightarrow E_r=\frac{950.765}{\sqrt2}\\\Rightarrow E_r=672.29\ W/m^2[/tex]
The approximate magnitude of the electric field in the sunlight is 672.29 W/m²
An object moves in simple harmonic motion with amplitude 14cm and period 0.2 seconds. At time =t0 seconds, its displacement d from rest is 0cm, and initially it moves in a positive direction. Give the equation modeling the displacement d as a function of time t.
Answer:
d= 14 sin 31.41 t
Explanation:
Lets take the general equation of the motion
d= D sinωt
d=Displacement
ω=Natural angular frequency
t=Time
D=Amplitude
Given that D= 14 cm
We know that time period T
[tex]T=\dfrac{2\pi}{\omega}[/tex]
Given that T= 0.2 s
[tex]0.2=\dfrac{2\pi}{\omega}[/tex]
ω=31.41 rad/s
D= 14 cm
Therefore
d= D sinωt
d= 14 sin 31.41 t
Final answer:
The equation modeling the displacement d of an object in simple harmonic motion with an amplitude of 14 cm and a period of 0.2 seconds, starting at 0 cm and moving in a positive direction at t = 0, is d(t) = 14cos(10πt + π/2) cm.
Explanation:
The displacement d of an object in simple harmonic motion can be modeled by the function d(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant. Since the amplitude A is 14 cm and the period T is 0.2 seconds, and given that the initial displacement at time t = 0 is 0 cm and it initially moves in a positive direction, we can determine ω using the formula ω = 2π/T and φ using the fact that the cosine function must start at 0.
Therefore, the angular frequency ω = 2π/0.2 s = 10π rad/s, and since the function starts at 0 and moves in a positive direction, φ = +π/2. Hence, the equation that models the displacement d as a function of time t is d(t) = 14cos(10πt + π/2) cm.
An air-filled capacitor stores a potential energy of 6 mJ due to its charge. It is accidentally filled with water in such a way as not to discharge its plates. How much energy does it continue to store after it is filled? (The dielectric constant for water is 78 and for air it is 1.0006.
Answer:
The potential energy in the water will be 0.0769 mJ
Explanation:
We know that energy stored in the capacitor due to charge is given by [tex]U=\frac{Q^2}{2C}[/tex]
From the relation we can see that potential energy is inversely proportional to the capacitance
And we also know that capacitance is directly proportional to the dielectric constant
So the new potential energy will be [tex]=\frac{6}{78}=0.0769mJ[/tex]
So the potential energy in the water will be 0.0769 mJ
Final answer:
When a capacitor filled with air and charged is filled with water without discharging its plates, the energy stored increases due to the higher dielectric constant of water. The formula used is U' = K_water * U, leading to the capacitor storing 468 mJ of energy after being filled with water.
Explanation:
The energy stored in a capacitor is given by the formula U = ½ CV^2, where U is the stored energy, C is the capacitance and V is the voltage across the capacitor. When a dielectric is introduced between the plates of a capacitor that is charged but disconnected from any voltage supply, the capacitance of the capacitor increases by a factor equal to the dielectric constant of the material inserted, K. Since the charge Q remains constant, and since C increases while U is directly proportional to C, the energy stored in the capacitor will increase as well.
For this particular question, the initial energy is 6 mJ and the dielectric constant for water is 78. After filling the capacitor with water, the new energy stored U' can be found using the dielectric constant. However, since dielectric constant for air is approximately 1, and capacitance is directly proportional to dielectric constant, we can simply multiply the original energy by the dielectric constant of water to find the energy stored after it is filled.
U' = K_water * U
= 78 * 6 mJ
= 468 mJ
Therefore, the energy the capacitor continues to store after it is filled with water is 468 mJ.
Which of the following is prohibited by the 2nd law of thermodynamics?
(A) A device that converts heat into work with 100% efficiency
(B) A device that converts work into heat with 100% efficiency.
(C) A device that uses work to supply X heat at high temperature while taking in Y heat at lower temperature. where X>Y.
(D) A device that converts work into heat with >100% efficiency
(A) A device that converts heat into work with 100% efficiency
It clearly violates the second law of thermodynamics because it warns that while all work can be turned into heat, not all heat can be turned into work. Therefore, despite the innumerable efforts, the efficiencies of the bodies have only been able to reach 60% at present.
The nearest known exoplanets (planets beyond the solar system) are around 20 light-years away. What would have to be the minimum diameter of an optical telescope to resolve a Jupiter-sized planet at that distance using light of wavelength 600 nm? (Express your answer to two significant figures.)
To solve the problem, it is necessary to apply the concepts related to the diffraction given in circular spaces. By definition it is expressed as
[tex]\Delta \theta = 1.22\frac{\lambda}{d}[/tex]
Where,
\lambda = Wavelength
d = Optical Diameter
[tex]\theta =[/tex] Angular resolution
In turn you can calculate the angle through the diameter and the arc length, that is,
[tex]\Delta \theta = \frac{x}{D}[/tex]
Where,
x = The length of the arc
D = Distance
From known data we know that Jupiter's diameter is,
[tex]x_J = 1.43*10^8m[/tex]
[tex]D = 20*9.4608*10^{15}[/tex]
[tex]\lambda = 600*10^{-9}m[/tex]
Replacing we have that,
[tex]\frac{x}{D} = 1.22\frac{\lambda}{d}[/tex]
[tex]\frac{1.43*10^8}{20*9.4608*10^{15} } = 1.22\frac{600*10^{-9}}{d}[/tex]
Re-arrange to find d,
[tex]d = 968.5m = 0.968Km[/tex]
Therefore the minimum diameter of an optical telescope to resolve a Jupiter-sized planet is 0.968Km.
A man fell out of an airplane and barely survived. He was moving at a speed of 100m/s just before landing in deep snow on a mountain side. Experts estimated that the average net force on him was 600 N as he plowed through the snow for 10 s. What was his mass, in kg
Answer:
Mass in kg will be 60 kg
Explanation:
We have given initial speed of the man u = 100 m/sec
And final speed v = 0 m/sec
Time is given, t = 10 sec
So acceleration [tex]a=\frac{v-u}{t}=\frac{0-100}{10}=-10m/sec^2[/tex]
Force is given , F= 600 N
From Newton's law we know that F = ma
So [tex]600=m\times 10[/tex]
mass = 60 kg
So mass in kg will be 60 kg
The efficiency of a Stirling cycle depends on the temperatures of the hot and cold isothermal parts of the cycle.If you increase the upper temperature, keeping the lower temperature the same, does the efficiency increase,decrease, or remain the same?A. increaseB. decreaseC. remain the same
Answer:
A. increase
Explanation:
Stirling cycle having four processes
1.Two processes are constant temperature processes
2.Two processes are constant volume processes
The efficiency of Stirling cycle is same as the efficiency of Carnot cycle.
The efficiency of Stirling cycle given as
[tex]\eta=1-\dfrac{T_L}{T_H}[/tex]
[tex]T_L[/tex]=Lower temperature
[tex]T_H[/tex]=Upper temperature
If we increase then upper temperature while the lower temperature is constant then the efficiency of Stirling cycle will increase because the ratio of lower and upper temperature will decreases.
Therefore answer is
A. increase
Suppose a dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 65 m below. How much time elapses before the dolphin hears the echoes of the clicks? (The speed of sound in seawater is approximately 1530 m/s.)
To solve this problem it is necessary to apply the concepts related to the kinematic equations of motion.
By definition we know that speed can be expressed as
[tex]v= \frac{d}{t}[/tex]
Where,
t = time
d = distance
For this specific case we know that the speed traveled corresponds to the round trip, therefore
[tex]v = \frac{2d}{t}[/tex]
Re-arrange to find t (The time taken to here the echo)
[tex]t = \frac{2d}{v}[/tex]
Replacing with our values we have that the distance is equal to 65m
[tex]t = \frac{2*65}{1530}[/tex]
[tex]t = 0.085s[/tex]
Therefore the time before the dolphin hears the echoes of the clicks is 0.085s
An inductor is connected in series to a fully charged capacitor. Which of the following statements are true? Check all that apply. a. The stored electric field energy can be greater than the stored magnetic field energy. b. As the capacitor is discharging, the current is increasing. c. As the capacitor is charging, the current is increasing. d. The stored electric field energy can be less than the stored magnetic field energy. e. The stored electric field energy can be equal to the stored magnetic field energy.
Answer:
a. The stored electric field energy can be greater than the stored magnetic field energy.
b. As the capacitor is discharging, the current is increasing.
d. The stored electric field energy can be less than the stored magnetic field energy.
e. The stored electric field energy can be equal to the stored magnetic field energy.
Explanation:
We know that total energy of capacitor and inductor system given as
[tex]TE=\dfrac{LI^2}{2}+\dfrac{CV^2}{2}[/tex]
L=Inductance ,I =current
C=Capacitance ,V=Voltage
The total energy of the system will be remain conserve.If energy in the inductor increases then the energy in the capacitor will decrease and vice -versa.
The values of stored energy in the capacitor and inductor can be equal ,greater or less than to each other.But the summation will be constant always.
Therefore the following option are correct
a ,b , d , e
In a series LC circuit, energy oscillates between the capacitor and inductor. Statements a, b, d, and e are correct as there are moments when the electric field energy can be greater than, less than, or equal to the magnetic field energy. Statement c is incorrect because the current decreases as the capacitor charges.
When an inductor is connected in series with a fully charged capacitor, the behavior of the system can be analyzed by considering the energy stored in both components.
Statement a: This is true. The stored electric field energy can be greater than the stored magnetic field energy at any given instant depending on the phase of oscillation.Statement b: This is also true. As the capacitor discharges, the current in the circuit initially increases because the inductor resists changes in current.Statement c: This statement is false. As the capacitor is charging, the current actually decreases because energy is being stored back in the capacitor.Statement d: This is true as well. The stored electric field energy can be less than the stored magnetic field energy at any point in time due to the oscillatory nature of the energy transfer.Statement e: This statement is correct. There are instants where the stored electric field energy can be equal to the stored magnetic field energy during the oscillation cycle.Overall, the series LC circuit exhibits an oscillatory energy transfer between the electric field of the capacitor and the magnetic field of the inductor.
Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignoring Earth's rotation, how much energy is needed to get the same object into orbit at height h?Express your answer in terms of some or all of the variables h, mass of the object m, mass of Earth mE, its radius RE, and gravitational constant G.
Answer:
Explanation:
Potential energy on the surface of the earth
= - GMm/ R
Potential at height h
= - GMm/ (R+h)
Potential difference
= GMm/ R - GMm/ (R+h)
= GMm ( 1/R - 1/ R+h )
= GMmh / R (R +h)
This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.
Extra energy is needed to get the same object into orbit at height h
= Kinetic energy of the orbiting object at height h
= 1/2 x potential energy at height h
= 1/2 x GMm / ( R + h)
Two ponies of equal mass are initially at diametrically opposite points on the rim of a large horizontal turntable that is turning freely on a vertical, frictionless axle through its center. The ponies simultaneously start walking toward each other across the turntable. As they walk, what happens to the angular speed of the turntable? (A) It increases(B) It decreases(C) It stays constantConsider the ponies-turntable system in this process is the angular momentum of the system conserved?(A) Yes(B) No
Answer:A
Explanation:
Given
Two ponies are at Extreme end of turntable with mass m
suppose turntable is moving with angular velocity [tex]\omega [/tex]
Moment of inertia of Turntable and two ponies
[tex]I=I_o+2mr_0^2[/tex]
let say at any time t they are at a distance of r from center such [tex]r<r_0[/tex]
Moment of inertia at that instant
[tex]I'=I_0+2mr'^2[/tex]
[tex]I'<I_0[/tex]
conserving angular momentum as net Torque is zero
[tex]I\omega =I'\omega '[/tex]
[tex]\omega '=\frac{I}{I'}\times \omega [/tex]
[tex]\omega '>\omega [/tex]
Thus we can say that angular velocity increases as they move towards each other.
The angular speed of the turntable increases as the ponies walk towards each other due to conservation of angular momentum. The angular momentum remains constant as there are no external forces acting upon the system.
Explanation:As the two ponies of equal mass walk toward each other across the turntable, the angular speed of the system increases. This is because the moment of inertia of the system decreases as the ponies move closer to the center, and according to the principle of conservation of angular momentum, a decrease in the moment of inertia must be accompanied by an increase in angular speed to keep the system's angular momentum constant.
In this scenario, yes, the angular momentum of the system is conserved. There are no external forces acting on the system, thus the sum of the angular momenta of the ponies and the turntable stays the same.
This principle can be exemplified by the behavior of figure skaters as they increase their rotation speed by bringing their arms in closer to the body, decreasing their moment of inertia and consequently increasing their angular velocity, to conserve their angular momentum.
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