A 20 kg box on a horizontal frictionless surface is moving to the right at a speed of 4.0 m/s. The box hits and remains attached to one end of a spring of negligible mass whose other end is attached to a wall. As a result, the spring compresses a maximum distance of 0.50 m, and the box then oscillates back and forth. (a) i. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maximum compression. Indicate whether the work done by the spring is positive, negative, or zero. ____ Positive ____ Negative ____ Zero Justify your answer. ii. Calculate the magnitude of the work described in part i. (b) Calculate the spring constant of the spring. (c) Calculate the magnitude of the maximum acceleration of the box. (d) Calculate the frequency of the oscillation of the box. (e) Let x
Answers
a)
i) Negative
ii) 160 J
b) 1280 N/m
c) [tex]32 m/s^2[/tex]
d) 1.27 Hz
e)
i) See attached plot
ii) See attached plot
Explanation:
a)
i) The work done by a force is given by
[tex]W=Fx cos \theta[/tex]
where
F is the force
x is the displacement of the object
[tex]\theta[/tex] is the angle between the direction of the force and the direction
Here we have:
- The force that the spring exerts on the box is to the left (because the box is moving to the right, trying to compress the spring)
- The displacement of the box is to the right
So, F and x have opposite direction, and so [tex]\theta=180^{\circ}[/tex] and [tex]cos \theta=-1[/tex], which means that the work done is negative.
ii)
According to the work-energy theorem, the work done by the spring is equal to the change in kinetic energy of the box:
[tex]W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where
[tex]K_i[/tex] is the initial kinetic energy of the box
[tex]K_f[/tex] is the final kinetic energy
m = 20 kg is the mass of the box
u = 4.0 m/s is its initial speed
v = 0 m/s is the final speed (the box comes to rest)
Therefore,
[tex]W=\frac{1}{2}(20)(0)^2-\frac{1}{2}(20)(4.0)^2=-160 J[/tex]
So, the magnitude is 160 J.
b)
The elastic energy stored in a spring when it is compressed is given by
[tex]U=\frac{1}{2}kx^2[/tex]
where
k is the spring constant
x is the stretching/compression of the spring
Due to the law of conservation of energy, the kinetic energy lost by the box is equal to the elastic energy gained by the spring, so:
[tex]|W|=U=\frac{1}{2}kx^2[/tex]
We have
[tex]|W|=160 J[/tex]
x = 0.50 m is the maximum compression of the spring
Solving for k:
[tex]k=\frac{2U}{x^2}=\frac{2(160)}{(0.50)^2}=1280 N/m[/tex]
c)
The magnitude of the force exerted on the box is given by
[tex]F=kx[/tex]
where
k = 1280 N/m is the spring constant
x = 0.50 m is the compression of the spring
Substituting,
[tex]F=(1280)(0.50)=640 N[/tex]
Now we can find the maxmum acceleration using Newton's second law:
[tex]a=\frac{F}{m}[/tex]
where
F = 640 N is the maximum force
m = 20 kg is the mass of the box
So,
[tex]a=\frac{640}{20}=32 m/s^2[/tex]
d)
The frequency of oscillation of a spring-mass system is given by
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
where
k is the spring constant
m is the mass
Here we have:
k = 1280 N/m is the spring constant of this spring
m = 20 kg is the mass of the box
So, the frequency of this system is:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{1280}{20}}=1.27 Hz[/tex]
e)
i)
Here we want to sketch the kinetic energy of the box as a function of the position, x: find this graph in attachment.
In a spring-mass oscillating system, the kinetic energy is zero when the system is at the extreme position, i.e. when the spring is maximum compressed/stretched. In this problem, this happens when x = - 0.50 m and x = +0.50 m (we called x = 0 the position of equilibrium of the spring). In these positions in fact, the mass has zero speed, so its kinetic energy is zero.
On the other hand, the box has maximum speed when x = 0 (because it's the moment where all the elastic energy is converted into kinetic energy, which is therefore maximum, and so the speed is also maximum).
ii)
Here we want to plot the acceleration of the box as a function of the position x: find the graph in attachment.
In a spring-mass system, the acceleration is proportional to the negative of the displacement, since the restoring force
[tex]F=-kx[/tex]
By rewriting the force using Newton's second Law, we have
[tex]ma=-kx \\a=-\frac{k}{m}x[/tex]
Which means that acceleration is proportional to the displacement, but with opposite sign: so, this graph is a straight line with negative slope.
Related Questions
At any angular speed, a uniform solid sphere of diameter D has the same rotational kinetic energy as a uniform hollow sphere of the same diameter when both are spinning about an axis through their centers. The moment of inertia of a solid sphere is :
Answers
Answer:
m = (3/5)*M
Explanation:
Given:-
- The angular speed of both hollow and solid sphere = w
- The diameter of solid & hollow sphere = D
- The mass of the solid sphere = M
- Both rotate about their common axis with similar rotational kinetic energy.
Find:-
The mass of hollow sphere (m) ?
Solution:-
- The formula for rotational kinetic energy (K.E) of any rigid body is:
K.E = 0.5*I*w^2
Where,
I : Moment of inertia of rigid body
- The rotational kinetic energies of both hollow sphere and solid sphere are same:
0.5*I_solid*w^2 = 0.5*I_shell*w^2
I_solid = I_shell
0.4*M*D^2 / 4 = (2/3)*m*D^2 / 4
(2/5)*M = (2/3)*m
m = (3/5)*M
Kinetic energy caused by the vibration of particles in a medium such as steel water or air
Answers
Final answer:
Kinetic energy is the energy caused by the vibration of particles in a medium. It is calculated using the formula KE = 0.5mv², where KE is the kinetic energy, m is the mass of the particle, and v is its velocity.
Explanation:
Kinetic energy is the energy an object has because of its motion. It is caused by the vibration of particles in a medium such as steel, water, or air. Kinetic energy is calculated as one-half the product of the mass of the particle and the square of its speed.
For example, when a rock is thrown into a pond or when a swimmer splashes the water's surface, the kinetic energy of the vibrating water particles is generated. Similarly, mechanical sound waves also have kinetic energy due to the movement of air particles and the potential energy caused by the elasticity of the material through which the sound propagates.
In summary, kinetic energy arises from the motion of particles in a medium and can be calculated using the formula KE = 0.5mv², where KE is the kinetic energy, m is the mass of the particle, and v is its velocity.
The sound tube experiment was performed in gas Carbon Dioxide at temperature 20 deg. C. The students partnering on the bench obtained for the slope of the straight line fitting of the experimental points the value 365.0. Compare the speed of sound following from that slope to the calculated speed of sound from the gas properties? For CO2, Cp/Cv = 1.289, and the mass of one mol CO2 is 0.044 kg. The sound tube length is 0.367 m. The additional necessary datum is given in the text.
Answers
Answer:
The speed of sound theoretically is 267.148 m/s which is clearly less than slope calculated speed 365.0 m/s
Explanation:
check the attached picture
Aspartic acid is a polypeptide side chain found in proteins. The pKa of aspartic acid is 3.86. If this polypeptide were in an aqueous solution with a pH of 7, the side chain would have what charge? a) negative b) positive c) neutral d) there is no way to know
Answers
Answer: a) Negative
Explanation: According to the question: The pKa of aspartic acid is 3.86.
For acidic amino acids,
pH > pKa
When side chains are negatively charged, amino acid will be acidic
pH < pKa
That is, when side chains are uncharged, amino acid will be neutral
For aspartic acid, when this polypeptide were in an aqueous solution with a pH of 7
pH (7.0) > pKa (3.86)
Therefore the chains are negatively charged
Final answer:
In an aqueous solution at pH 7, the aspartic acid side chain would have a (a)negative charge because the pH of the environment is higher than the pKa of aspartic acid (3.86).
Explanation:
The question relates to the charge of the aspartic acid side chain in an aqueous solution at pH 7, given that the pKa of aspartic acid is 3.86. The pKa value represents the pH at which half of the aspartic acid side chains are deprotonated (negative charge) and half remain protonated (neutral). When the pH of the environment is higher than the pKa (as is the case here, pH 7 > pKa 3.86), the majority of the aspartic acid side chains will be deprotonated, thus carrying a negative charge. Consequently, in an aqueous solution at pH 7, the aspartic acid side chain would have (a)negative charge.
The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at v i , 2 = 17.3 vi,2=17.3 m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle θ = 39.9 θ=39.9 ° below the horizontal and a speed of v i , 1 = 33.7 vi,1=33.7 m/s. What is the speed of the eagle immediately after it catches its prey?
Answers
The speed of the eagle immediately after catching the pigeon is [tex]$19.0$[/tex] m/s.
To solve this problem, we need to apply the principle of conservation of momentum, which states that the total momentum of a closed system remains constant before and after an interaction.
Given information:
- Mass of the pigeon = [tex]$m_2$[/tex]
- Mass of the eagle = [tex]$2m_2$[/tex] (twice the mass of the pigeon)
- Initial velocity of the pigeon, [tex]$\vec{v}_{i,2} = 17.3 m/s[/tex] (northward)
- Initial velocity of the eagle, [tex]$\vec{v}_{i,1} = 33.7 m/s[/tex] at an angle [tex]$\theta = 39.9^\circ$[/tex] below the horizontal
We need to find the final velocity of the eagle after catching the pigeon, [tex]$\vec{v}_f$[/tex].
Step 1: Resolve the initial velocity vectors into components.
For the pigeon:
[tex]$\vec{v}_{i,2} = (0, 17.3) m/s[/tex]
For the eagle:
[tex]$\vec{v}_{i,1} = (33.7 \cos 39.9^\circ, -33.7 \sin 39.9^\circ) m/s[/tex]
[tex]$\vec{v}_{i,1} = (22.8, -25.4) m/s[/tex]
Step 2: Calculate the initial total momentum of the system.
Initial total momentum, [tex]$\vec{p}_i = m_2 \vec{v}_{i,2} + (2m_2) \vec{v}_{i,1}$[/tex]
[tex]$\vec{p}_i = m_2 (0, 17.3) + 2m_2 (22.8, -25.4)$[/tex]
[tex]$\vec{p}_i = m_2 (45.6, -33.8) kg⋅m/s[/tex]
Step 3: Calculate the final total momentum of the system.
Since momentum is conserved, the final total momentum is equal to the initial total momentum.
[tex]$\vec{p}_f = \vec{p}_i = m_2 (45.6, -33.8) kg⋅m/s[/tex]
Step 4: Calculate the final velocity of the combined system (eagle + pigeon).
Let [tex]$\vec{v}_f = (v_{f,x}, v_{f,y})$[/tex] be the final velocity of the combined system.
Total final momentum = [tex]$(m_2 + 2m_2) \vec{v}_f = 3m_2 \vec{v}_f$[/tex]
[tex]3m_2 \vec{v}_f = m_2 (45.6, -33.8)$$\vec{v}_f = \left(\frac{45.6}{3}, -\frac{33.8}{3}\right)$ m/s\\$\vec{v}_f = (15.2, -11.3)$ m/s[/tex]
Step 5: Calculate the speed of the eagle after catching the pigeon.
Speed = [tex]$\sqrt{v_{f,x}^2 + v_{f,y}^2}$[/tex]
Speed = [tex]$\sqrt{15.2^2 + (-11.3)^2} m/s[/tex]
Speed = [tex]$19.0 m/s[/tex]
Which of the following energy sources is no way derived from the sun is it nuclear energy title energy when energy or petroleum
Answers
Answer:nuclear energy is not derived from the sun
Explanation:
Nuclear energy is not derived from the sun.nuclear energy comes from the energy released when atoms are split apart and some mass is converted to energy
A 13.6- resistor, an 11.9-μF capacitor, and a 19.1-mH inductor are connected in series with a 117-V generator.
(a) At what frequency is the current a maximum?
(b) What is the maximum value of the RMS current?
Note: The ac current and voltage are RMS values and power is an average value unless indicated otherwise.
Answers
Answer:
Explanation:
Given the following information,
Resistor of resistance R = 13.6Ω
Capacitor of capacitance C = 11.9-μF
C = 11.9 × 10^ -6 F
Inductor of inductance L = 19.1-mH
L = 19.1 ×10^-3 H
All this connected in series to a generator that generates Vrms= 117V
Vo = Vrms√2 = 117√2
Vo = 165.463V
a. Frequency for maximum current?
Maximum current occurs at resonance
I.e Xc = XL
At maximum current, the frequency is given as
f = 1/(2π√LC)
Then,
f = 1/(2π√(19.1×10^-3 × 11.9×10^-6)
f = 1/(2π√(2.2729×10^-7))
f = 1/(2π × 4.77 ×10^-4)
f = 333.83Hz
Then, the frequency is 333.83Hz.
b. Since we know the frequency,
Then, we need to find the capacitive and inductive reactance
Capacitive reactance
Xc = 1/2πfC
Xc = 1/(2π × 338.83 × 11.9×10^-6)
Xc = 1/ 0.024961
Xc = 40.1Ω
Also, Inductive reactance
XL = 2πfL
XL = 2π × 333.83 × 19.1×10^-3
XL = 40.1Ω
As expected, Xc=XL, resonance
Then, the impedance in AC circuit is given as
Z = √ (R² + (Xc—XL)²)
Z = √ 13.6² + (40.1-40.1)²)
Z = √13.6²
Z = 13.6 ohms
Then, using ohms las
V = IZ
Then, I = Vo/Z
Io = 165.46/13.6
Io = 12.17Amps
The current is 12.17 A
Answer:
a) Current is maximum at frequency, f₀ = 333.83 Hz
b) Maximum current = 12.17 A
Explanation:
Inductance, L = 19.1 mH = 19.1 * 10⁻³ H
Capacitance, C = 11.9 μF =11.9 * 10⁻⁶ F
a) Current is maximum at resonant frequency, f₀
[tex]f_{0} = \frac{1}{2\pi\sqrt{LC} }[/tex]
[tex]f_{0} = \frac{1}{2\pi\sqrt{11.9 * 10^{-6}* 19.1 * 10^{-3} } }[/tex]
[tex]f_{0} = 333.83 Hz[/tex]
b) Maximum value of the RMS current
[tex]V_{RMS} = 117 V\\V_{max} = \sqrt{2} V_{RMS}\\V_{max} = \sqrt{2} * 117\\V_{max} = 165.46 V[/tex]
[tex]I_{max} = \frac{V_{max} }{R} \\I_{max} = \frac{165.46}{13.6} \\I_{max} = 12.17 A[/tex]
An insulated piston–cylinder device contains 0.05 m3 of saturated refrigerant- 134a vapor at 0.8-MPa pressure. The refrigerant is now allowed to expand in a reversible manner until the pressure drops to 0.4 MPa. Determine (a) the final temperature in the cylinder and (b) the work done by the refrigerant
Answers
The final temperature of the refrigerant in the piston-cylinder device can be determined via lookup and interpolation from standard tables, based on the final pressure of 0.4 MPa. The work done by the refrigerant as it expands and drops in pressure can be calculated via the first law of thermodynamics, specifically accounting for changes in internal energy as work done by the system.
Explanation:This is an example of a reversible adiabatic expansion process involving a refrigerant (refrigerant-134a) in a piston-cylinder device. The process follows the behavior of ideal gases, and so we can make use of the first law of thermodynamics and other gas laws to determine the final temperature and the work done by the refrigerant.
To find the final temperature, we incorporate the specifics of the refrigerant-134a, which is defined by specific tables found in thermodynamics textbooks, providing details of temperature at various pressures. Interpolation between table values may be necessary to find the exact temperature at 0.4 MPa. Generally, the temperature will fall as the refrigerant expands and the pressure lowers, as defined by the ideal gas law (pV=nRT).
For the work done by the refrigerant, we'd need to apply the first law of thermodynamics, which in this case can be simplified as ΔU = W because there's no heat transfer in an adiabatic process (dQ=0). Changes in the internal energy of the gas would translate directly into work done on the piston. We calculate work using the formula -P(ΔV), where the negative sign indicates work being done by the system.
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(a) The final temperature in the cylinder is approximately 18.0°C.
(b) The work done by the refrigerant during the expansion is approximately 7.72 kJ.
To solve this problem, we need to determine two things: the final temperature of the refrigerant-134a vapor and the work done by the refrigerant during the expansion.
Given that the process is reversible and the system is insulated, this suggests that the expansion process is isothermal for an ideal gas, but since refrigerant-134a is not an ideal gas, we will use the refrigerant tables to find the properties.
Given Data:
Initial volume [tex]\( V_1 \)[/tex] : 0.05 m³Initial pressure [tex]\( P_1 \)[/tex] : 0.8 MPaFinal pressure [tex]\( P_2 \)[/tex] : 0.4 MPaThe process is reversibleStep-by-Step Solution:
(a) Determine the Final Temperature in the Cylinder:
1. Find the Initial and Final States Using Refrigerant-134a Tables:
Initial State [tex](P1 = 0.8 MPa)[/tex]:Look up the refrigerant-134a property tables for [tex]\( P = 0.8 \)[/tex] MPa.
At [tex]\( P = 0.8 \)[/tex] MPa, the refrigerant-134a is in the saturated vapor region. The saturated temperature and properties at this pressure are:
Saturated temperature [tex]\( T_1 \): \( T_{\text{sat}}(0.8 \text{ MPa}) \approx 26.7^\circ \text{C} \)[/tex]Specific volume [tex]\( v_g \): \( v_{\text{g}}(0.8 \text{ MPa}) \approx 0.0769 \text{ m}^3/\text{kg} \)[/tex]Final State [tex](P2 = 0.4 MPa)[/tex] :Similarly, look up the refrigerant-134a property tables for [tex]\( P = 0.4 \)[/tex] MPa.
At [tex]\( P = 0.4 \)[/tex] MPa, the refrigerant-134a is also in the saturated vapor region. The properties at this pressure are:
Saturated temperature [tex]\( T_2 \)[/tex] : [tex]\( T_{\text{sat}}(0.4 \text{ MPa}) \approx 18.0^\circ \text{C} \)[/tex]Specific volume [tex]\( v_g \)[/tex] : [tex]\( v_{\text{g}}(0.4 \text{ MPa}) \approx 0.0917 \text{ m}^3/\text{kg} \)[/tex]The final temperature of the refrigerant-134a after the expansion, which is at 0.4 MPa, is approximately [tex]\( 18.0^\circ \text{C} \)[/tex].
(b) Determine the Work Done by the Refrigerant:
1. Calculate the Initial and Final Mass:
Using the specific volume [tex]\( v_g \)[/tex] to find the mass:
[tex]\[ m = \frac{V_1}{v_g (P_1)} \][/tex]
For [tex]\( P_1 = 0.8 \)[/tex] MPa:
[tex]\[ m = \frac{0.05 \text{ m}^3}{0.0769 \text{ m}^3/\text{kg}} \approx 0.650 \text{ kg} \][/tex]
For [tex]\( P_2 = 0.4 \)[/tex] MPa, verify if the mass is the same (which it is for an ideal gas in a closed system):
[tex]\[ V_2 = m \times v_g (P_2) = 0.650 \text{ kg} \times 0.0917 \text{ m}^3/\text{kg} \approx 0.0597 \text{ m}^3 \][/tex]
The new volume [tex]\( V_2 \)[/tex] can be verified as:
[tex]\[ V_2 = \text{initial volume} \times \frac{v_g (P_2)}{v_g (P_1)} \][/tex]
Which confirms our calculation.
2. Calculate the Work Done During Expansion:
For a reversible expansion in an insulated system (isoenthalpic process), the work done \( W \) can be approximated using:
[tex]\[ W = \int_{V_1}^{V_2} P \, dV \][/tex]
The work done for an ideal gas (but approximate here for refrigerants) can be simplified using:
[tex]\[ W = \text{P}_1 \times V_1 \ln \left( \frac{V_2}{V_1} \right) \][/tex]
For refrigerants, use specific volume:
[tex]\[ W = \text{P}_1 \times V_1 \left[ \frac{v_g (P_2)}{v_g (P_1)} - 1 \right] \][/tex]
Substituting:
[tex]\[ W = 0.8 \text{ MPa} \times 0.05 \text{ m}^3 \left( \frac{0.0917}{0.0769} - 1 \right) \][/tex]
[tex]\[ W = 0.8 \times 10^3 \text{ Pa} \times 0.05 \text{ m}^3 \left( 1.193 - 1 \right) \][/tex]
[tex]\[ W = 40 \text{ kJ} \times 0.193 \approx 7.72 \text{ kJ} \][/tex]
An ambulance is traveling north at 48.4 m/s, approaching a car that is also traveling north at 28 m/s. The ambulance driver hears his siren at a frequency of 620 Hz. Ambulance 48.4 m/s 28 m/s Car What is the wavelength at any position in front of the ambulance for the sound from the ambulance’s siren? The velocity of sound in air is 343 m/s. Answer in units of m.
Answers
Answer:
Explanation:
For any point stationary in front of ambulance , the ambulance is approaching the point . In this way the case is similar to source of sound approaching the observer. The doppler's effect can be applied to know the apparent frequency.
velocity of sound V = 343 m /s
speed of source Vs = 48.4
observer is stationary.
apparent frequency = real frequency x V / ( V - Vs )
= 620 x 343 / ( 343 - 48.4 )
= 722 Hz approx.
What If? The coil and applied magnetic field remain the same, but the circuit providing the current in the coil is now changed. The new circuit has an emf of e m f = 20.0 V, and this allows a mass of 51.0 g added to the right side to balance the system. What is the value of the resistance R (in Ω)?
Answers
Answer:
Using equations
R will be 1.25 Kilo ohm
To get the value of resistance, R, we need to use Ohm's law which is R = V/I', where I' is the current generated when the coil balances with the added mass. The exact value of R depends on parameters related to the coil and the magnetic field.
Explanation:The question is asking for the value of the resistance R in a given circuit, when the circuit produces a current which generates a balance with an added mass through a coil in a magnetic field. To find the value of R, we need to use Ohm's law, which expresses the relationship between voltage (V), current (I), and resistance (R) as V=IR.
Here, the electromotive force or emf can be treated as the voltage in the circuit. In this case, emf = 20.0 V. When this coil creates a balance with an added 51.0 g mass, an equivalent current will be generated, let's denote it by I'
Using Ohm's law, we can solve for R by rearranging the formula to R = V/I'. The exact value of I' and thus R will, however, depend on additional factors involving the parameters of the coil and magnetic field which are not provided in the question. The law used here, Lenz's law, is a manifestation of the conservation of energy, dictating that the current induced in a circuit due to a change in the magnetic field will create a magnetic field that opposes the initial changing magnetic field.
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A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 1.09 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.
Answers
Answer:
a The kinetic energy is [tex]KE = 0.0543 J[/tex]
b The height of the center of mass above that position is [tex]h = 1.372 \ m[/tex]
Explanation:
From the question we are told that
The length of the rod is [tex]L = 1.4m[/tex]
The mass of the rod [tex]m = 140 = \frac{140}{1000} = 0.140 \ kg[/tex]
The angular speed at the lowest point is [tex]w = 1.09 \ rad/s[/tex]
Generally moment of inertia of the rod about an axis that passes through its one end is
[tex]I = \frac{mL^2}{3}[/tex]
Substituting values
[tex]I = \frac{(0.140) (1.4)^2}{3}[/tex]
[tex]I = 0.0915 \ kg \cdot m^2[/tex]
Generally the kinetic energy rod is mathematically represented as
[tex]KE = \frac{1}{2} Iw^2[/tex]
[tex]KE = \frac{1}{2} (0.0915) (1.09)^2[/tex]
[tex]KE = 0.0543 J[/tex]
From the law of conservation of energy
The kinetic energy of the rod during motion = The potential energy of the rod at the highest point
Therefore
[tex]KE = PE = mgh[/tex]
[tex]0.0543 = mgh[/tex]
[tex]h = \frac{0.0543}{9.8 * 0.140}[/tex]
[tex]h = 1.372 \ m[/tex]
Answer:
a) Kr = 0.0543 J
b) Δy = 0.0396 m
Explanation:
a) Given
L = 1.4 m
m = 140 g = 0.14 kg
ω = 1.09 rad/s
Kr = ?
We have to get the rotational inertia as follows
I = Icm + m*d²
⇒ I = (m*L²/12) + (m*(L/2)²)
⇒ I = (0.14 kg*(1.4 m)²/12) + (0.14 kg*(1.4 m/2)²)
⇒ I = 0.09146 kg*m²
Then, we apply the formula
Kr = 0.5*I*ω²
⇒ Kr = 0.5*(0.09146 kg*m²)*(1.09 rad/s)²
⇒ Kr = 0.0543 J
b) We apply the following principle
Ei = Ef
Where the initial point is the lowest position and the final point is at the maximum height that its center of mass can achieve, then we have
Ki + Ui = Kf + Uf
we know that ωf = 0 ⇒ Kf = 0
⇒ Ki + Ui = Uf
⇒ Uf - Ui = Ki
⇒ m*g*yf - m*g*yi = Ki
⇒ m*g*(yf - yi) = Ki
⇒ m*g*Δy = Ki
⇒ Δy = Ki/(m*g)
where
Ki = Kr = 0.0543 J
g = 9.81 m/s²
⇒ Δy = (0.0543 J)/(0.14 kg*9.81 m/s²)
⇒ Δy = 0.0396 m
Consider an experiment in which slow neutrons of momentum ¯hk are scattered by a diatomic molecule; suppose that the molecule is aligned along the y axis, with one atom at y = b and the other at y = −b. The beam of neutrons is directed in the zb direction. Assume the atoms to be infinitely heavy so that they remain fixed throughout the experiment. The potential due to the atoms as seen by the neutrons can be represented by a delta function, so: V (~r) = a[δ(x)δ(y − b)δ(z) + δ(x)δ(y + b)δ(z)] (a) Calculate the scattering amplitude, and differential cross section, in the first order Born approximation. (b) In what ways does the quantum result differ from what one would expect classically?
Answers
Answer:
Check the explanation
Explanation:
When we have an object in periodic motion, the amplitude will be the maximum displacement from equilibrium. Take for example, when there’s a back and forth movement of a pendulum through its equilibrium point (straight down), then swings to a highest distance away from the center. This distance will be represented as the amplitude, A. The full range of the pendulum has a magnitude of 2A.
position = amplitude x sine function(angular frequency x time + phase difference)
x = A sin(ωt + ϕ)
x = displacement (m)
A = amplitude (m)
ω = angular frequency (radians/s)
t = time (s)
ϕ = phase shift (radians)
Kindly check the attached image below to see the step by step explanation to the question above.
A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time t = 0 the capacitor is fully charged. At a given instant later the charge on the capacitor is measured to be 3.0 μC and the current in the circuit is equal to 75 μA. What is the maximum charge of the capacitor?
Answers
Answer:
[tex]8.0\mu C[/tex]
Explanation:
We are given that
[tex]f=1.6 Hz[/tex]
[tex]q=3.0\mu C=3.0\times 10^{-6} C[/tex]
[tex]1\mu C=10^{-6} C[/tex]
Current,I=[tex]75\mu A=75\times 10^{-6} A[/tex]
[tex]1\mu A=10^{-6} A[/tex]
We have to find the maximum charge of the capacitor.
Charge on the capacitor,[tex]q=q_0cos\omega t[/tex]
[tex]\omega=2\pi f=2\pi\times 1.6=3.2\pi rad/s[/tex]
[tex]3\times 10^{-6}=q_0cos3.2\pi t[/tex]....(1)
[tex]I=\frac{dq}{dt}=-q_0\omega sin\omega t[/tex]
[tex]75\times 10^{-6}=-q_0(3.2\pi)sin3.2\pi t[/tex]....(2)
Equation (2) divided by equation (1)
[tex]-3.2\pi tan3.2\pi t=\frac{75\times 10^{-6}}{3\times 10^{-6}}=25[/tex]
[tex]tan3.2\pi t=-\frac{25}{3.2\pi}=-2.488[/tex]
[tex]3.2\pi t=tan^{-1}(-2.488)=-1.188rad[/tex]
[tex]q_0=\frac{q}{cos\omega t}=\frac{3\times 10^{-6}}{cos(-1.188)}=8.0\times 10^{-6}=8\mu C[/tex]
Hence, the maximum charge of the capacitor=[tex]8.0\mu C[/tex]
Three children are riding on the edge of a merry‑go‑round that has a mass of 105 kg and a radius of 1.70 m . The merry‑go‑round is spinning at 22.0 rpm . The children have masses of 22.0 , 28.0 , and 33.0 kg . If the 28.0 kg child moves to the center of the merry‑go‑round, what is the new angular velocity in revolutions per minute? Ignore friction, and assume that the merry‑go‑round can be treated as a solid disk and the children as point masses.
Answers
Answer:
Explanation:
Given that,
Three children of masses and their position on the merry go round
M1 = 22kg
M2 = 28kg
M3 = 33kg
They are all initially riding at the edge of the merry go round
Then, R1 = R2 = R3 = R = 1.7m
Mass of Merry go round is
M =105kg
Radius of Merry go round.
R = 1.7m
Angular velocity of Merry go round
ωi = 22 rpm
If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf
Using conservation of angular momentum
Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round
Then,
L(initial) = L(final)
Ii•ωi = If•ωf
So we need to find the initial and final moment of inertia
NOTE: merry go round is treated as a solid disk then I= ½MR²
I(initial)=½MR²+M1•R²+M2•R²+M3•R²
I(initial) = ½MR² + R²(M1 + M2 + M3)
I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)
I(initial) = 151.725 + 1.7²(83)
I(initial) = 391.595 kgm²
Final moment of inertial when R2 =0
I(final)=½MR²+M1•R²+M2•R2²+M3•R²
Since R2 = 0
I(final) = ½MR²+ M1•R² + M3•R²
I(final) = ½MR² + (M1 + M3)• R²
I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²
I(final) = 151.725 + 158.95
I(final) = 310.675 kgm²
Now, applying the conservation of angular momentum
L(initial) = L(final)
Ii•ωi = If•ωf
391.595 × 22 = 310.675 × ωf
Then,
ωf = 391.595 × 22 / 310.675
ωf = 27.73 rpm
So, the final angular momentum is 27.73 revolution per minute
how is velocity different from speed
Answers
Answer: velocity has a direction, while speed is the distance travelled by an object.
Explanation:
Answer:
Velocity includes Displacement instead of distance.
Explanation:
Velocity's formula is change in position divided(displacement) by change in time(Delta x/Delta t). Whereas speed is distance over change in time(d/Delta t).
As friends frequently do, Josh and John stand at either end of a very long, stationary skateboard, with Josh at the front and John at the rear. John then tosses a heavy ball toward Josh who, caught unawares, ducks so that the ball goes sailing over his head and lands on the ground. There is no friction between the skateboard and the ground. After the ball is thrown, which way, if at all, does the skateboard move
A.Backward
B.Forward
C.Neither
Answers
Answer:
B. Forward
Explanation:
The net force in the system is initially balanced and the system is in equilibrium.
1) As John ducks, two things happens. First, there is a sudden impulse from his end which will change the momentum of the system from zero to an amount Mv proportional to the impulse Ft, which will be directed away from his direction because,
2) the process of ducking pushes his center of mass forward, which also moves the center of mass of the whole system away from him. This two effect will cause the skateboard to move forward.
A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J. Calculate the kinetic energy and speed of the ejected electron.
Answers
Explanation:
Given that,
Wavelength of the light, [tex]\lambda=4170\ A=4170\times 10^{-10}\ m[/tex]
Work function of sodium, [tex]W_o=4.41\times 10^{-19}\ J[/tex]
The kinetic energy of the ejected electron in terms of work function is given by :
[tex]KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J[/tex]
The formula of kinetic energy is given by :
[tex]KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s[/tex]
Hence, this is the required solution.
A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can rotate with negligible friction about a stationary horizontal axis. The drum is not a uniform cylinder and has unknown moment of inertia. When you release the bucket from rest, you find that it has a downward acceleration of magnitude a. What is the tension in the cable between the drum and the bucket
Answers
Final answer:
The tension in the cable connecting the drum to the bucket is equal to the mass of the bucket times the difference between the acceleration due to gravity and the bucket's downward acceleration.
Explanation:
To determine the tension in the cable between the drum and the bucket, we can apply Newton's second law to the hanging mass. The net force acting on the bucket is the difference between the weight of the bucket and the tension in the rope:
[tex]F_{net}[/tex] = mg - T
Since we know the bucket has a downward acceleration a, we can write Newton's second law as:
ma = mg - T
Where m is the mass of the bucket, g is the acceleration due to gravity, and T is the tension in the rope. We can rearrange the equation to solve for T:
T = mg - ma
T = m(g - a)
So the tension in the cable is equal to the mass of the bucket times the difference between the acceleration due to gravity and the downward acceleration of the bucket.
A soap bubble (n = 1.28) having a wall thickness of 116 nm is floating in air. (a) What is the wavelength of the visible light that is most strongly reflected? nm (b) Explain how a bubble of different thickness could also strongly reflect light of this same wavelength. This answer has not been graded yet. (c) Find the two smallest film thicknesses larger than the one given that can produce strongly reflected light of this same wavelength. nm (smaller thickness) nm (larger thickness)
Answers
Answer:
Explanation:
a , b )
The problem is based on interference in thin films
formula for constructive interference
2μ t = ( 2n+ 1 ) λ / 2 , μ is refractive index of layer, t is thickness and λ is wavelength of light.
n is called the order of fringe . If we place n= 0 , 1 , 2 etc , the thickness also changes . So constructive interference is possible at more than one thickness .
Put the value of λ = 116 nm . μ = 1.28 , t = 116 nm in the given equation
2 x 1.28 x 116 x 2 = ( 2n+ 1 ) λ
593.92 = ( 2n+ 1 ) λ
when n = 0
λ = 593.92 nm .
This falls in visible range .
c )
2μ t = ( 2n+ 1 ) λ / 2
Put λ = 593.92 nm , n = 1
2 x 1.28 t₁ = 3 x 593.92 / 2
t₁ = 348 nm .
Put n = 2
2 x 1.28 t₂ = 5 x 593.92 / 2
t₂ = 580 nm .
How would you classify the wave?
Answers
Answer:
One way to categorize waves is on the basis of the direction of movement of the individual particles of the medium relative to the direction that the waves travel. Categorizing waves on this basis leads to three notable categories: transverse waves, longitudinal waves, and surface waves.
I've read about it but never seen one. The way I understand it, it's a coordinated physical motion executed by a great number of people, as in a large crowd at a sporting event, timed so that it appears to propagate from one end of the crowd to the opposite end.
I would classify it as a cooperative community activity, involving liberty, equality, and fraternity, executed for the common good.
To make things easier, how do people design a circuit and put it on paper ?
Answers
Explanation:
A paper circuit is a functioning electronic circuit built on a paper surface. Projects involving paper circuits are unique because of the use of traditional art techniques and some unique materials to create a circuit that combines aesthetics and functionality.
Paper circuit are sometimes carefully designed and then transfered on the paper, and in some cases, they are designed directly on the paper (usually by experts that already know what they're doing).
To make designing circuit on paper easier, there are three commonly used materials for the circuitry and they are, conductive paints, conductive tapes and conductive inks.
Conductive tapes are made from metal strip (usually copper) that are taped to the paper. They are good to work with since they allow components to be soldered on them creating a stronger and more reliable joint.
Conductive paints are special paints that can be used to outline circuit path and also serve as the circuit. The only problem with conducting paints is that it can be messy, and needs time to dry.
Conductive inks are easier to use as they need no drying time. They are far less messy and allows the drawing of elaborate and more intricate circuit on the paper.
There are also components that have been modified (like led lights etc) and available to make paper circuit design easier.
Which are examples of short-term environmental change? Check all that apply.
tsunamis
El Niño
large asteroid and comet impacts
volcanic eruptions
global warming
i just took the test it's: tsunamis, El Nino, and volcanic eruptions.
Answers
Answer: tsunamis,El Nino,and volcanic eruptions
Explanation:
the correct answer for e d g e n u i t y
The examples of short-term environmental change are Tsunamis, El Niño, and Volcanic eruptions.
Tsunamis are huge ocean waves that are triggered by undersea disturbances like earthquakes or landslides. They can inflict major damage to coastal regions, however they usually happen in a short period of time.
El Nio is a climatic trend characterised by higher-than-normal equatorial Pacific ocean temperatures.
It can cause changes in weather patterns, such as greater rainfall in some areas and droughts in others, although the impacts are usually transient and endure for a few months to a couple of years.
Volcanic eruptions spew lava, ash, and gases into the atmosphere, which can have direct effects on the environment, including as changes in air quality and local weather patterns.
Thus, these effects, however, are often short-lived and localized.
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Problem 3: An Nd:YAG laser operates in a pulsed mode, with an energy of 100mJ (millimoles) per pulse, and a pulse repetition rate of 10Hz. Light can be emitted at 1065 nm, 532 nm, or 355 nm. (a) Each pulse lasts for 1ns, and in between pulses, no light is emitted. What is the instantaneous laser power during each pulse
Answers
Given Information:
Energy of laser pulses = E = 100 mJ = 100×10⁻³ Joules
Time = t = 1 ns = 1×10⁻⁹ seconds
Required Information:
Instantaneous power = P = ?
Answer:
[tex]Instantaneous \: Power = 100 \: Mwatt[/tex]
Explanation:
The instantaneous power is the power dissipated at any instant of time whereas the average power is the power dissipated over a given time interval.
The instantaneous laser power during each pulse is given by
[tex]Instantaneous \: Power = \frac{E}{t}[/tex]
Where E is the energy of the laser pulses and t is the time that each pulse lasts.
[tex]Instantaneous \: Power = \frac{100\times10^{-3}}{1\times10^{-9}}[/tex]
[tex]Instantaneous \: Power = 1\times10^{8} \: watt[/tex]
or
[tex]Instantaneous \: Power = 100 \: Mwatt[/tex]
Therefore, the instantaneous power of each pulse is 100 Mwatt.
The goal in a command economy is economic ____,
Answers
Economic balance
Answer:
A- Equality
Explanation:
Two concentric circular loops lie in the same plane. One is connected to a source that supplies an increasing current; the other is a simple closed ring. Is the induced current in the ring in the same direction as the current in the loop connected to the source, or opposite? What if the current in the first loop is decreasing?
Answers
Answer:
Explanation:
We shall apply Lenz's law to solve the problem . This law states that direction of induced current is such that it opposes the change that creates it. Since current increases in the coil it creates increasing magnetic field in the other coil . So the current will be induced in it so that it opposes this increase . It can be done only if current in opposite direction is induced in it . Hence in the first case, current will be induced in opposite direction .
In this case, current is decreasing in the primary coil and current will be induced in the secondary coil. Decreasing current will create decreasing magnetic field . So induced current will try to increase it . In can be done if current in the same direction is induced in the secondary coil.
Hence in the second case , current will be induced in the same direction .
We've seen that stout tendons in the legs of hopping kangaroos store energy. When a kangaroo lands, much of the kinetic energy of motion is converted to elastic energy as the tendons stretch, returning to kinetic energy when the kangaroo again leaves the ground. If a hopping kangaroo increases its speed, it spends more time in the air with each bounce, but the contact time with the ground stays approximately the same. Explain why you would expect this to be the case
Answers
Answer:
Check the explanation
Explanation:
So far there’s an increases in the speed of the kangaroo, then the tendons will stretch more thereby enabling them to store more energy. For this reason, they will have a additional time in the air propelled by greater spring energy. In contact with the ground, it will turn out to be like a spring in simple harmonic motion. There will be increases in their agility rate in hopping the amplitude of the oscillation, but that does not in any way affect the time, or period in contact with the ground.
Kangaroos convert kinetic energy to elastic potential energy during landing which is immediately converted back to kinetic energy for the next leap. This energy transformation cycle, aided by cushioning their landing by bending their hind legs, allows kangaroos to spend more time airborne without affecting the ground contact time, irrespective of their speed.
Explanation:The phenomenon in question about kangaroo jumping is rooted in physics and biomechanics. As kangaroos gain speed, more kinetic energy is stored and thus converted into potential energy when they're in the air. This essentially means the kangaroo uses most of their energy while airborne rather than contacting the ground. Hence, ground contact time stays the same no matter what the kangaroo's speed is.
Firstly, the kangaroo's tendons work like a spring; as it lands, much of its kinetic energy is converted to elastic potential energy in the tendons. This act transforms the downward motion (gravitational potential energy) into an upward motion (kinetic energy) through elastic potential energy, which is stored as the tendons stretch. This is similar to the motion observed in a bungee jumper or an elastic band when stretched.
Additionally, the kangaroo's landing is cushioned by bending its hind legs, reducing the impact force over the contact time. With each hop, the kangaroo's kinetic energy increases when it leaves the ground while potential energy increases during 'flight' due to gravity. This establishes a cycle between kinetic and potential energy, with energy oscillating back and forth as the kangaroo continues to hop. Such energy efficiency is what enables the kangaroo to maintain the same ground contact time irrespective of its speed.
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cylindrical water tank is 20.0 m tall, open to the atmosphere at the top, and is filled to the top. It is noticed that a small hole has occurred in the side at a point 16.5 m below the water level and that water is flowing out at the volume flow rate of 2.90 10-3 m3/min. Determine the following. (a) the speed in m/s at which water is ejected from the hole
Answers
Answer:
a
The velocity is [tex]v =17.98 \ m/s[/tex]
b
The diameter is [tex]d = 0.00184m[/tex]
Explanation:
The diagram of the set up is shown on the first uploaded image
From the question we are told that
The height of the water tank is [tex]h = 20.0 \ m[/tex]
The position of the hole [tex]p_h = 16.5m[/tex] below water level
The rate of water flow [tex]\r V = 2.90 *10^{-3} m^3 /min = \frac{2.90 *10^{-3}}{60} = 0.048*10^{-3} m^3/s[/tex]
According to Bernoulli's theorem position of the hole
[tex]\frac{P_o + h \rho g}{\rho} + \frac{1}{2} u^2 = \frac{P_o}{\rho } + \frac{1}{2} v^2[/tex]
Where u is the initial speed the water through the hole = 0 m/s
[tex]P_o[/tex] is the atmospheric pressure
[tex]\frac{P_o }{\rho} + \frac{ h \rho g}{\rho} + 0 = \frac{P_o}{\rho } + \frac{1}{2} v^2[/tex]
[tex]v = \sqrt{2gh}[/tex]
Substituting value
[tex]v = \sqrt{2 * 9.8 * 16.5 }[/tex]
[tex]v =17.98 \ m/s[/tex]
The Volumetric flow rate is mathematically represented as
[tex]\r V = A * v[/tex]
Making A the subject
[tex]A = \frac{\r V}{v}[/tex]
substituting value
[tex]A = \frac{0.048 *10^{-3}}{17.98}[/tex]
[tex]= 2.66*10^{-6}m^2[/tex]
Area is mathematically represented as
[tex]A = \frac{\pi d^2}{4}[/tex]
making d the subject
[tex]d = \sqrt{\frac{4*A}{\pi} }[/tex]
Substituting values
[tex]d = \sqrt{\frac{4 * 2.67 *10^{-6}}{3.142} }[/tex]
[tex]d = 0.00184m[/tex]
An object is electrically charged if the amounts of positive and negative charge it contains are not _______.
equal
negative
positive
increased
Answers
Answer:oops
Explanation:
The type of friction between the pavement and the tires on a moving vehicle is called.._____friction.
A) kinetic
B) fluid
C) static
D) inertial
Answers
Answer:
A) Kinetic
Explanation:
Kinetic friction is friction caused by motion. Kinetic energy is energy in motion.
A 7-m long wire with a mass of 40 g is under tension. A transverse wave for which the frequency is 630 Hz, the wavelength is 0.5 m, and the amplitude is 7.1 mm is propagating on the wire.
The maximum transverse acceleration of a point on a wire is closest to ____ m/s².
Answers
Answer:
111249.41m/s²
Explanation:
Mass of the wire m = 40g = 0.04 Kg
Length of the wire l = 7 m
Linear density of wire μ = m/l
= 0.04/7
= 0.0057 Kg/m
Frequency n = 630 Hz
Wavelength λ = 0.5 m
Amplitude A = 7.1mm = 0.0071m
Speed of the wave v = nλ
= 630*0.5
= 315 m/s
Angular speed ω = 2πn
= 2π*630
= 3958.40 rad/s
Maximum transeverse acceleration
a = ω^2 A
= 3958.40² × 0.0071m
= 111249.41m/s²