Answer:
0.639 V
Explanation:
The volatge of the cell containing both Ag/AgCl reference electrode and
[tex]Fe^{2+}/Fe^{3+}[/tex] electrode = 0.693 V
Thus,
[tex]E_{cathod}-E_{anode} =0.693 V[/tex]
E_{anode}=0.197 V
Note: potential of the silver-silver chloride reference electrode (0.197 V)
⇒E_{Cthode}= 0.693+0.197 = 0.890V
To calculate the voltage of the cell containing both the calomel reference
electrode and [tex]Fe^{2+}/Fe^{3+}[/tex] electrode as follows
Voltage of the cell = [tex]E_{cathod}-E_{anode}[/tex]
E_{anode}= calomel electrode= 0.241 V
Voltage of the cell = 0.890-0.241 = 0.639 V
Therefore, the new volatge is = 0.639 V
A solid 0.6350 kg ball rolls without slipping down a track toward a vertical loop of radius R=0.8950 m.
What minimum translational speed vmin must the ball have when it is a height H=1.329 m above the bottom of the loop in order to complete the loop without falling off the track?
Assume that the radius of the ball itself is much smaller than the loop radius R. Use g=9.810 m/s2 for the acceleration due to gravity.
The minimum translational speed the ball must have at a height of 1.329 m to complete the loop without falling off the track is approximately 6.61 m/s.
How can you solve the minimum translational speed the ball must have?
E p(top) = K e(bottom)
E p(top) = m * g * (R + H)
where:
m is the mass of the ball (0.6350 kg)
g is the acceleration due to gravity (9.810 m/s²)
R is the radius of the loop (0.8950 m)
H is the height above the bottom of the loop (1.329 m)
Calculate the minimum kinetic energy at the bottom:
Since the ball needs enough speed at the bottom to reach the top again, the minimum kinetic energy at the bottom is equal to the potential energy at the top:
K e(bottom) = E p(top) = m * g * (R + H)
Find the minimum translational speed:
K e = 1/2 * m * vmin²
where vmin is the minimum translational speed we're looking for. Solving for vmin:
v min = √(2 * K e / m) = √(2 * m * g * (R + H) / m)
v min = √(2 * g * (R + H))
Plug in the values and calculate:
v min = √(2 * 9.810 * (0.8950 + 1.329))
v min ≈ 6.61 m/s
Therefore, the minimum translational speed the ball must have at a height of 1.329 m to complete the loop without falling off the track is approximately 6.61 m/s.
A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that reach a point on a far-away screen such thatrays from the slit make an angle of 1.0° with the normal. Thedifference in phase for waves from the top and bottom of the slitis:
A) 0
B) 0.55 rad
C) 1.1 rad
D) 1.6 rad
E) 2.2 rad
To solve this exercise it is necessary to use the concepts related to Difference in Phase.
The Difference in phase is given by
[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]
Where
[tex]\delta =[/tex] Horizontal distance between two points
[tex]\lambda =[/tex] Wavelength
From our values we have,
[tex]\lambda = 500nm = 5*10^{-6}m[/tex]
[tex]\theta = 1\°[/tex]
The horizontal distance between this two points would be given for
[tex]\delta = dsin\theta[/tex]
Therefore using the equation we have
[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]
[tex]\Phi = \frac{2\pi(dsin\theta)}{\lambda}[/tex]
[tex]\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}[/tex]
[tex]\Phi= 1.096 rad \approx = 1.1 rad[/tex]
Therefore the correct answer is C.
The phase difference for waves from the top and bottom of the slit can be calculated by using the formula for calculating phase difference. With the provided values for wavelength, angle and slit width, the calculated phase difference is 0.55 rad.
Explanation:The phase difference of the waves is directly related to the path difference between them and can be calculated by using the formula:
Φ = 2 π × (d / λ) × sin(θ).
Where φ is the phase difference, π is Pi, d is the slit width, λ is the wavelength and θ is the incident angle.
Let's plug the provided numbers into our formula:
Φ = 2 π × (5.0x10-6 m / 500x10-9 m) × sin(1.0°).
Φ = 2 π × 10 × sin(1.0°)= 0.55 rad.
So, the correct answer is (B) 0.55 rad.
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You want to launch a stone using the elastic band of a slingshot. The force that the elastic band applies to an object is given by F = −α∆s4 , where α = 45 N m4 and ∆s is the displacement of the elastic band from its equilibrium position. You load the slingshot with the stone and pull back on the stone horizontally, stretching the the elastic band 20 cm. How much work do you do on the stone-elastic band system?
Answer:
0.00288 J
Explanation:
We know that
W= Fds
F = −α∆s^4
α = 45 N/m^4 and ∆s = displacement
W= −α∆s^4ds
integrating both the sides from s= 0 to 0.2
W= 45/5×0.2^5= 0.00288 J
The workdone on the stone-elastic band system is mathematically given as
W= 0.00288 J
What work do you do on the stone-elastic band system?Question Parameter(s):
F = −α∆s4 , where α = 45 N m4
∆s is the displacement of the elastic band
Generally, the equation for the Workdone is mathematically given as
W= Fds
Thereofre
F = −α *ds^4
Where
α = 45
ds = displacement
In conclusion
W= 45/5*0.2^5
W= 0.00288 J
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8. A baseball batter angularly accelerates a bat from rest to 20 rad/s in 40 ms (milliseconds). If the bat’s moment of inertia is 0.6 kg m2, then findA) the torque applied to the bat andB) the angle through which the bat moved.
Answer:
A) τ = 300 N*m
B) θ = 0.4 rad = 22.9°
Explanation:
Newton's second law:
F = ma has the equivalent for rotation:
τ = I * α Formula (1)
where:
τ : It is the moment applied to the body. (Nxm)
I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)
α : It is angular acceleration. (rad/s²)
Data
I = 0.6 kg*m² : moment of inertia of the bat
Angular acceleration of the bat
We apply the equations of circular motion uniformly accelerated :
ωf= ω₀ + α*t Formula (2)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (s)
Data
ω₀ = 0
ωf = 20 rad/s
t = 40 ms = 0.04 s
We replace data in the formula (2) :
ωf= ω₀ + α*t
20 = 0 + α* (0.04)
α = 20/ (0.04)
α = 500 rad/s²
Newton's second law to the bat
τ = (0.6 kg*m²) *(500 rad/s²) = 300 (kg*m/s²)* m
τ = 300 N*m
B) Angle through which the bat moved.
We apply the equations of circular motion uniformly accelerated :
ωf²= ω₀ ²+ 2α*θ Formula (3)
Where:
θ : Angle that the body has rotated in a given time interval (rad)
We replace data in the formula (3):
ωf²= ω₀²+ 2α*θ
(20)²= (0)²+ 2(500 )*θ
400 = 1000*θ
θ = 400/1000
θ = 0.4 rad
π rad = 180°
θ = 0.4 rad *(180°/π rad)
θ = 22.9°
Which of the following is a TRUE statement?
a. It is possible for heat to flow spontaneously from a hot body to a cold one or from a cold one to a hot one, depending on whether or not the process is reversible or irreversible.
b. It is not possible to convert work entirely into heat.
c. The second law of thermodynamics is a consequence of the first law of thermodynamics.
d. It is impossible to transfer heat from a cooler to a hotter body.
e. All of these statements are false.
Answer:
e. All of these statements are false.
Explanation:
As we know that heat transfer take place from high temperature to low temperature.
It is possible to convert all work into heat but it is not possible to convert all heat in to work some heat will be reject to the surrounding.
The first law of thermodynamics is the energy conservation law.
Second law of thermodynamics states that it is impossible to construct a device which convert all energy into work without rejecting the heat to the surrounding.
By using heat pump ,heat can transfer from cooler body to the hotter body.
Therefore all the answer is False.
The true statement among the given options is that the entropy of a system can be reduced by cooling it.
The correct answer to the student's question regarding true statements about thermodynamics is option (c) It is always possible to reduce the entropy of a system, for instance, by cooling it. This statement aligns with the principles of thermodynamics, which affirm that entropy, a measure of disorder or randomness, can decrease in a system if energy is removed from the system, such as by lowering its temperature. However, this does not violate the second law of thermodynamics because entropy may still increase in the overall process when considering the surroundings.
In contrast, options (a) and (b) are false because they wrongly imply irreversibility in scenarios where reversibility is possible. Specifically, it is possible to reverse an entropy increase by cooling a system and it is also possible to convert some amount of thermal energy back into mechanical energy, although not with 100% efficiency due to inherent thermodynamic losses.
The second law also articulates that heat transfer occurs spontaneously from a higher to a lower temperature body and not in the reverse direction without external work, implying that a spontaneous flow of heat from a colder to a warmer body is impossible, as is complete conversion of heat to work in a cyclical process.
The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the object moves from small R to large R (like a rocket being launched) or from large R to small R (like a comet approaching the earth).
O True O False
Answer:
True
Explanation:
The angular momentum around the center of the planet and the total mechanical energy will be preserved irrespective of whether the object moves from large R to small R. But on the other hand the kinetic energy of the planet will not be conserved because it can change from kinetic energy to potential energy.
Therefore the given statement is True.
The correct answer is option A. The statement is true because in celestial mechanics, angular momentum and total energy are conserved under gravity, regardless of whether an object moves from a smaller to a larger radius or vice versa.
Let's explain these concepts:
1. Conservation of Angular Momentum (L):
The angular momentum of an object with respect to the center of the planet is given by the product of its linear momentum (p) and the radius (r) of its orbit:
[tex]\[ L = p \times r = mvr \][/tex]
where m is the mass of the object, v is its velocity, and r is the radius of the orbit. For a central force like gravity, the torque on the object is zero, which means that the angular momentum is conserved.
2. Conservation of Total Mechanical Energy (E):
The total mechanical energy of an object in orbit is the sum of its kinetic energy (K) and potential energy (U):
[tex]\[ E = K + U = \frac{1}{2}mv^2 - \frac{GMm}{r} \][/tex]
where G is the gravitational constant, M is the mass of the planet, and m is the mass of the object.
The complete question is:
The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the object moves from small R to large R (like a rocket being launched) or from large R to small R (like a comet approaching the earth).
A. True
B. False
A thin, uniform, metal bar, 3.00 m long and weighing 90.0 N , is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.60 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s . The ball rebounds in the opposite direction with a speed of 5.00 m/s. Find the angular speed of the bar just after the collision? Why linear momentum not conserved?
Answer:
[tex]\omega_f=-2.6rad/s[/tex]
Since the bar cannot translate, linear momentum is not conserved
Explanation:
By conservation of the angular momentum:
Lo = Lf
[tex]I_B*\omega_o+I_b*(V_o/d)=I_B*\omega_f+I_b*(V_f/d)[/tex]
where
[tex]I_B =1/3*M_B*L^2[/tex]
[tex]I_b=m_b*d^2[/tex]
[tex]M_B=90/g=9kg[/tex]; [tex]m_b=3kg[/tex]; d=1.6m; L=3m; [tex]V_o=-10m/s[/tex]; [tex]V_f=5m/s[/tex]; [tex]\omega_o=0 rad/s[/tex]
Solving for [tex]\omega_f[/tex]:
[tex]\omega_f=m_b*d/I_B*(V_o-V_f)[/tex]
Replacing the values we get:
[tex]\omega_f=-2.6rad/s[/tex]
Since the bar can only rotate (it canno translate), only angular momentum is conserved.
A circuit consists of a 9.0 mH inductor coil, a 230 Ω resistor, a 12.0 V ideal battery, and an open switch-all connected in series. The switch closes at time t = 0. What is the initial rate of change of the current - I just after the switch closes? dt l=0 A/s What is the steady-state value of the current If a long time after the switch is closed? If= At what time 185% does the current in the circuit reach 85% of its steady-state value? 185% = What is the rate of change of the current when the current in the circuit is equa dt li=1;/2 its steady-state value? dI - dt li=1;/2
Answer:
Explanation:
This is the case of L-R charging circuit , for which the formula is as follows
i = i₀ ( 1 - [tex]e^{\frac{-t}{\tau}[/tex] )
Differentiating the equation on both sides
di / dt = i₀ / τ x [tex]e^\frac{-t}{\tau}[/tex]
i is current at time t , i₀ is maximum current , τ is time constant which is equal to L / R where L is inductance and R is resistance of the circuit .
τ = L / R = 9 x 10⁻³ / 230
= 39 x 10⁻⁶ s
i₀ = 12 / 230 = 52.17 x 10⁻³
di / dt (at t is zero) = 52.17 x 10⁻³ / 39 x 10⁻⁶
= 1.33 x 10³ A / s
Time to reach 85 % of steady state or i₀
.85 = [tex]1 - e^\frac{t}{\tau}[/tex]
[tex]e^\frac{-t}{\tau}[/tex] = .15
t / τ = ln .15
t = 74 μs
An RL circuit with given parameters will have its steady-state current at 0.052A, with an initial rate of change of 1333.33 A/s. The current reaches 85% of its steady-state value after approximately 91µs. The rate of change when the current is half its steady-state value can be found using the I(t) function and its derivative.
Explanation:When you close a switch in an RL circuit, the current doesn't instantly reach its maximum value. Instead, it gradually increases until it reaches a steady-state value. This steady-state value, also known as the final current (If), can be calculated using Ohm's law, seeing that If = V(source voltage)/R(resistance). In your case, If = 12.0V/230Ω = 0.052A or 52mA.
The initial rate of change of current - dI/dt at t=0, can be calculated using the formula dI/dt = V(source voltage)/L(inductance). From your problem, dI/dt = 12.0V/9.0mH = 1333.33 A/s.
To calculate the time at which the current reaches 85% of its steady-state value, you use the formula for the time constant of an RL circuit, which is T = L/R. From the problem, T = 9.0mH/230Ω = 0.000039s, and the expression for time (t) when the current is at a certain percentage (p, in this case, 85%) of its steady-state value is t = -T*ln(1-p), which simplifies to t = -0.000039s*ln(1-0.85) = 0.000090999s or approximately 91µs.
The rate of change of the current when the current is at half its steady-state value will be smaller than the initial rate of change because the rate decreases as the current approaches its steady value. For the current at half its steady-state value, we calculate the time similar to the 85% case and find the I(t) at this time using I(t) = (V/R)(1-e^-Rt/L). Then, we differentiate it to find the rate of change dI/dt at this time.
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About once every 30 minutes, a geyser known as Old Faceful projects water 11.0 m straight up into the air.
Use g = 9.80 m/s2, and take atmospheric pressure to be 101.3 kPa. The density of water is 1000 kg/m3.
(a) What is the speed of the water when it emerges from the ground? m/s
(b) Assuming the water travels to the surface through a narrow crack that extends 9.00 m below the surface, and that the water comes from a chamber with a large cross-sectional area, what is the pressure in the chamber?
Answers:
a) [tex]8820 m/s[/tex]
b) [tex]189500 Pa[/tex]
Explanation:
We have the following data:
[tex]t=30 min \frac{60 s}{1 min}=1800 s[/tex] is the time
[tex]h=11 m[/tex] is the height the water reaches vertically
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]P_{air}=101.3 kPa=101.3(10)^{3} Pa[/tex] is the pressure of air
[tex]\rho_{water}=1000 kg/m^{3}[/tex] is the density of water
Knowing this, let's begin:
a) Initial speed of waterHere we will use the following equation:
[tex]h=h_{o}+V_{o}t-\frac{g}{2}t^{2}[/tex] (1)
Where:
[tex]h_{o}=0 m[/tex] is the initial height of water
[tex]V_{o}[/tex] is the initial speed of water
Isolating [tex]V_{o}[/tex]:
[tex]V_{o}=\frac{1}{t}(h+\frac{g}{2}t^{2})[/tex] (2)
[tex]V_{o}=\frac{1}{1800 s}(11 m+\frac{9.8 m/s^{2}}{2}(1800 s)^{2})[/tex]
[tex]V_{o}=8820.006 m/s \approx 8820 m/s[/tex] (3)
b) Pressure in the chamber
In this part we will use the following equation:
[tex]P=\rho_{water} g d + P_{air}[/tex] (4)
Where:
[tex]P[/tex] is the absolute pressure in the chamber
[tex]d=9 m[/tex] is the depth
[tex]P=(1000 kg/m^{3})(9.8 m/s^{2})(9 m) + 101.3(10)^{3} Pa[/tex]
[tex]P=189500 Pa[/tex] (5)
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 39 . It has been determined that fracture results at a stress of 208 MPa when the maximum (or critical) internal crack length is 2.82 mm. a) Determine the value of for this same component and alloy at a stress level of 270 MPa when the maximum internal crack length is 1.41 mm.
Answer:
[tex]27.57713\ MPa\sqrt{m}[/tex]
Explanation:
Y = Fracture parameter
a = Crack length
[tex]\sigma[/tex] = Stress in part
Plane strain fracture toughness is given by
[tex]K_I=Y\sigma\sqrt{\pi a}\\\Rightarrow Y=\frac{K_I}{\sigma\sqrt{\pi a}}\\\Rightarrow Y=\frac{39}{270\times \sqrt{\pi 0.00282}}\\\Rightarrow Y=1.53462[/tex]
When a = 1.41 mm
[tex]K_I=Y\sigma\sqrt{\pi a}\\\Rightarrow K_i=1.53462\times 270\sqrt{\pi 0.00141}\\\Rightarrow K_I=27.57713\ MPa\sqrt{m}[/tex]
The value of plane strain fracture toughness is [tex]27.57713\ MPa\sqrt{m}[/tex]
A large, simple pendulum is on display in the lobby of the United Nations building. If the pendulum is 18.5 m in length, what is the least amount of time it takes for the bob to swing from a position of maximum displacement to the equilibrium position of the pendulum? (Assume that the acceleration of gravity is g = 9.81 m/s2 at the UN building.)
Final answer:
The minimum time for a simple pendulum of 18.5 meters to swing from maximum displacement to equilibrium position is approximately 2.160 seconds, which is one-fourth the total period of the pendulum.
Explanation:
The minimum time for a pendulum to swing from maximum displacement to the equilibrium position is one-fourth of its period. The period of a simple pendulum, which is the time for one complete to-and-fro swing, can be calculated using the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. Given a length, L, of 18.5 meters and gravity, g, of 9.81 m/s², we find:
T = 2π√(18.5/9.81) ≈ 2π√(1.886) ≈ 2π×1.373≈ 8.639 seconds
Therefore, the least amount of time it takes for the bob to swing from maximum displacement to the equilibrium position would be T/4:
T/4 = 8.639 seconds / 4 ≈ 2.160 seconds.
Final answer:
The least amount of time for a pendulum of length 18.5 m to swing from maximum displacement to equilibrium is approximately 2.17 seconds, which is a quarter of its period.
Explanation:
To find the time taken for a pendulum to swing from maximum displacement to the equilibrium position, we use the formula for the period of a simple pendulum, which is:
T = 2π√(L/g)
Where T is the period of oscillation, L is the length of the pendulum, and g is the acceleration due to gravity. However, the question asks for the time taken to swing from maximum displacement to equilibrium, which is only a quarter of the period. So, we divide the period by 4:
Time = T/4 = (1/2)π√(L/g)
Substitute the given values (L = 18.5 m and g = 9.81 m/s2):
Time = (1/2)π√(18.5/9.81)
After calculating, we find that the time taken is approximately 2.17 seconds.
A roundabout is a type of playground equipment involving a large flat metal disk that is able to spin about its center axis. A roundabout of mass 120kg has a radius of 1.0m is initially at rest. A child of mass 43kg is running toward the edge of the roundabout (meaning, running on a path tangent to the edge) at 2.7 m/s and jumps on. Once she jumps on the roundabout, they move together as a single object. Assume the roundabout is a uniform disk. 1. What is the magnitude of her angular momentum (with respect to the center of the roundabout) just before she jumps? 2. What is the angular speed of the roundabout after the jump? 3. Does the overall kinetic energy of the system increase, decrease, or remain constant? If you say it changed, explain what caused a change in energy.
Answer:
116.1 kgm²/s
1.12718 rad/s
Decreases
Explanation:
m = Mass of girl = 43 kg
M = Mass of roundabout = 120 kg
v = Velocity of roundabout = 2.7 m/s
r = Radius of roundabout = 1 m = R
I = Moment of inertia
Her angular momentum
[tex]L_i=mvr\\\Rightarrow L_i=43\times 2.7\times 1\\\Rightarrow L_i=116.1\ kgm^2/s[/tex]
Magnitude of angular momentum is 116.1 kgm²/s
Here the angular momentum is conserved
[tex]L_f=L_i\\\Rightarrow I\omega=L_i\\\Rightarrow (\frac{1}{2}MR^2+mr^2)\omega=116.1\\\Rightarrow \omega=\frac{116.1}{\frac{1}{2}\times 120\times 1^2+43\times 1^2}\\\Rightarrow \omega=1.12718\ rad/s[/tex]
Angular speed of the roundabout is 1.12718 rad/s
Initial kinetic energy
[tex]K_i=\frac{1}{2}mv^2\\\Rightarrow K_i=\frac{1}{2}43\times 2.7^2\\\Rightarrow K_i=156.735\ J[/tex]
Final kinetic energy
[tex]K_f=\frac{1}{2}I\omega^2\\\Rightarrow K_f=\frac{1}{2}\times (\frac{1}{2}\times 120\times 1^2+43\times 1^2)\times 1.12718^2\\\Rightarrow K_f=65.43253\ J[/tex]
The overall kinetic energy decreases as can be seen. This loss is converted to heat.
To find the angular momentum of the child just before she jumps onto the roundabout, consider her linear momentum and the moment of inertia of the roundabout. The angular speed of the roundabout after the jump can be found using the principle of conservation of angular momentum. The overall kinetic energy of the system remains constant.
Explanation:To find the angular momentum of the child just before she jumps onto the roundabout, we need to consider her linear momentum and the moment of inertia of the roundabout. The linear momentum of the child is given by the product of her mass and velocity. The angular momentum is then equal to the linear momentum multiplied by the distance from the center of the roundabout.
The angular speed of the roundabout after the jump can be found using the principle of conservation of angular momentum. The initial angular momentum of the system (just the roundabout) is zero, and since angular momentum is conserved, the final angular momentum after the child jumps on is equal to the angular momentum of the child just before she jumps.
The overall kinetic energy of the system remains constant. As the child jumps onto the roundabout, an external force (the ground pushing on the child) does work to change the linear momentum of the child, but no external torque acts on the system. So, the total mechanical energy (kinetic plus potential) is conserved.
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A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has a radius of 5.17 m, at what angular velocity will the riders be subjected to a centripetal acceleration whose magnitude is equal to 1.50 times the acceleration due to gravity?
Answer:
Angular velocity will be 2.843 rad/sec
Explanation:
We have given the radius r = 5.17 m
Centripetal acceleration [tex]a_c=1.5g=1.5\times 9.8=14.7m/sec^2[/tex]
We know that centripetal acceleration is given by
[tex]a_c=\frac{v^2}{r}[/tex]
And linear velocity is given by [tex]v=\omega r[/tex]
[tex]a_c=\frac{(\omega r)^2}{r}=\omega ^2r[/tex]
[tex]14.7=\omega ^2\times 5.17[/tex]
[tex]\omega =2.843rad/sec[/tex]
A student pushes a 21-kg box initially at rest, horizontally along a frictionless surface for 10.0 m and then releases the box to continue sliding. If the student pushes with a constant 10 N force, what is the box's speed when it is released?
Answer:v=3.08 m/s
Explanation:
Given
mass of student [tex]m=21 kg[/tex]
distance moved [tex]d=10 m[/tex]
Force applied [tex]F=10 N[/tex]
acceleration of system during application of force is a
[tex]a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2[/tex]
using [tex]v^2-u^2=2 as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex]v^2-0=2\times 0.476\times 10[/tex]
[tex]v=\sqrt{9.52}[/tex]
[tex]v=3.08 m/s[/tex]
As a civil engineer for your city, you have been assigned to evaluate the purchase of spring-loaded guard rails to prevent cars from leaving the road. In response to a request for proposals*, one company states their guard rails are perfect for the job. Each section of their guard rails consists of two springs, each having a force constant 3.13400 105 N/m with a maximimum distance of compression of 0.614 m. (According to the manufacturer, beyond this compression the spring loses most of its ability to absorb an impact elastically.) The largest vehicle the guardrails are expected to stop are trucks of mass 4550.000 kg. What is the maximum speed at which these guard rails alone can be expected to bring such vehicles to a halt within the stated maximum compression distance? (Assume the vehicles can strike the guard rail head on and that the springs are perfectly elastic.)
____ m/s
Given your result which section of road often features such a speed
School Zone
Large Road
Highway
Guard rails are pointless if the acceleration they create seriously injures passengers. One important safety factor is the acceleration experienced by passengers during a collision. Calculate the maximum acceleration of the vehicle during the time in which it is in contact with the guard rail.
_____ m/s^2
If the highway department considers 20 g\'s the maximum safe acceleration, is this guard rail safe in regards to acceleration?
Answer:
a) v = 7,207 m / s
, b) a = 42.3 m / s²
Explanation:
We will solve this exercise using the concept of mechanical energy, We will write it in two points before the car touches the springs and in point of maximum compression
Initial
Em₀ = K = ½ m v²
Final
[tex]Em_{f}[/tex] = 2 Ke = ½ k x²
The two is placed because each barred has two springs and each does not exert the same force
Emo = [tex]Em_{f}[/tex]
½ m v² = 2 ½ k x²
v = √(2k/m) x
v = √ (2 3,134 10⁵/4550) 0.614
v = 7,207 m / s
Let's take this speed to km / h
v = 5,096 m / s (1km / 1000m) (3600s / 1h)
v = 25.9 km / h
This speed is common in school zones
Let's use kinematics to calculate the average acceleration
vf² = v₀² - 2 a x
0 = v₀² - 2 a x
a = v₀² / 2 x
a = 7,207²/2 0.614
a = 42.3 m / s²
We buy this acceleration with the acceleration of gravity
a / g = 42.3 / 9.8
a / g = 4.3
This acceleration is well below the maximum allowed
The total cross-sectional area of the load-bearing calcified portion of the two forearm bones (radius and ulna) is approximately 2.3 cm2. During a car crash, the forearm is slammed against the dashboard. The arm comes to rest from an initial speed of 80 km/h in 5.8 ms. If the arm has an effective mass of 3.0 kg, what is the compressional stress that the arm withstands during the crash?
To solve this problem, it is necessary to use the concepts related to the Force given in Newton's second law as well as the use of the kinematic equations of movement description. For this case I specifically use the acceleration as a function of speed and time.
Finally, we will describe the calculation of stress, as the Force produced on unit area.
By definition we know that the Force can be expressed as
F= ma
Where,
m= mass
a = Acceleration
The acceleration described as a function of speed is given by
[tex]a = \frac{\Delta v}{\Delta t}[/tex]
Where,
[tex]\Delta v =[/tex] Change in velocity
[tex]\Delta t =[/tex]Change in time
The expression to find the stress can be defined as
[tex]\sigma=\frac{F}{A}[/tex]
Where,
F = Force
A = Cross-sectional Area
Our values are given as
[tex]v= 80km/h\\t=5.8*10^3s\\m = 3kg \\A = 2.3*10^{-4}m^2[/tex]
Replacing at the values we have that the acceleration is
[tex]a = \frac{\Delta v}{\Delta t}[/tex]
[tex]a = \frac{80km/h(\frac{1h}{3600s})(\frac{1000m}{1km})}{5.8*10^3}[/tex]
[tex]a = 3831.41m/s^2[/tex]
Therefore the force expected is
[tex]F = ma\\F = 3*3831.41m/s^2 \\F = 11494.25N[/tex]
Finally the stress would be
[tex]\sigma = \frac{F}{A}[/tex]
[tex]\sigma = \frac{11494.25N}{2.3*10^{-4}}[/tex]
[tex]\sigma = 49.97*10^6 Pa = 49.97Mpa[/tex]
Therefore the compressional stress that the arm withstands during the crash is 49.97Mpa
Final answer:
The compressional stress is approximately 4.99787 × 10⁷ Pa
Explanation:
The question asks to find the compressional stress that the arm withstands during a crash, where the arm comes to a stop from an initial speed of 80 km/h in 5.8 ms, with an effective mass of 3.0 kg, and the total cross-sectional area of the load-bearing calcified portion of the forearm bones being approximately 2.3 cm².
First, convert the initial speed from km/h to m/s: 80 km/h = 22.22 m/s. To find the acceleration, use the formula a = ∆v / ∆t, where ∆v = -22.22 m/s (as it comes to rest) and ∆t = 5.8 ms or 0.0058 s. This gives an acceleration of approximately -3831.03 m/s².
The force experienced due to this acceleration can be calculated using Newton's second law, F = ma, with m = 3.0 kg and a = -3831.03 m/s², resulting in a force of approximately -11493.09 N. Finally, the compressional stress (σ) is found using the formula σ = F / A, where F = 11493.09 N and A = 2.3 cm² = 0.00023 m². This yields a compressional stress of approximately 4.99787 × 10⁷ Pa.
Given: G = 6.67259 × 10−11 N m2 /kg2 . A 438 kg geosynchronous satellite orbits a planet similar to Earth at a radius 1.94 × 105 km from the planet’s center. Its angular speed at this radius is the same as the rotational speed of the Earth, and so they appear stationary in the sky. That is, the period of the satellite is 24 h . What is the force acting on this satellite? Answer in units of N.
Answer:
449.37412 N
Explanation:
G = Gravitational constant = 6.67259 × 10⁻¹¹ m³/kgs²
m = Mass of satellite = 438 kg
M = Mass of planet
T = Time period of the satellite = 24 h
r = Radius of planet = [tex]1.94\times 10^8\ m[/tex]
The time period of the satellite is given by
[tex]T=2\pi\sqrt{\frac{r^3}{GM}}\\\Rightarrow M=4\pi^2\frac{r^3}{T^2G}\\\Rightarrow M=4\pi^2\times \frac{(1.94\times 10^8)^3}{(24\times 3600)^2\times 6.67259\times 10^{-11}}\\\Rightarrow M=5.78686\times 10^{26}\ kg[/tex]
The gravitational force is given by
[tex]F=G\frac{Mm}{r^2}\\\Rightarrow F=6.67259\times 10^{-11}\times \frac{5.78686\times 10^{26}\times 438}{(1.94\times 10^8)^2}\\\Rightarrow F=449.37412\ N[/tex]
The force acting on this satellite is 449.37412 N
Gravity, is often known as gravitation. The gravitational force between the planet and the satellite can be written as 0.4332 N.
What is gravitational force?Gravity, often known as gravitation, is the universal force of attraction that acts between all matter in mechanics. It is the weakest known force in nature, and so has no bearing on the interior properties of ordinary matter.
[tex]F =G\dfrac{m_1m_2}{r^2}[/tex]
As it is given that the value of the gravitational constant is G = 6.67259 × 10−11 N m² /kg², while the radius of the planet is 1.94×10⁵. And the time taken by the satellite to revolve around the planet is 24 hours. therefore, the Mass of the planet can be written as,
The Time period of the satellite is given as:
[tex]T = 2\pi \sqrt{\dfrac{r^3}{GM}}[/tex]
Substitute the values in the formula,
[tex]24 = 2\pi \sqrt{\dfrac{(1.94 \times 10^5)^3}{6.67259 \times 10^{-11}M}}\\\\M = 5.78686 \times 10^{17}\rm\ kg[/tex]
Thus, the mass of the planet can be written as 5.5786×10¹⁷ kg.
Now, the gravitational force can be written as,
[tex]F = G\dfrac{Mm}{r^2}\\\\F = 6.67259\times 10^{-11} \times \dfrac{5.5786 \times 10^{17} \times 438}{1.94 \times 10^5}\\\\ F = 0.4332\rm\ N[/tex]
Hence, the gravitational force between the planet and the satellite can be written as 0.4332 N.
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Pete slides a crate up a ramp with constant speed at an angle of 24.3 ◦ by exerting a 289 N force parallel to the ramp. How much work has been done against gravity when the crate is raised a vertical distance of 2.39 m? The coefficient of friction is 0.33. Answer in units of J.
Answer:
W=972.83 J
Explanation:
Given,
angle of inclination of ramp = θ = 24.3°
Force exerted on the block = F= 289 N
Crate is raised to height of = 2.39 m
coefficient of friction = 0.33
where g is the acceleration due to gravity = g = 9.8 m/s²
Work done = ?
let m be the mass of the crate
Gravity force along ramp = m g sinθ
Friction force = μm g cosθ
now writing all the force
F = m g sinθ + μm g cosθ
By putting the values
F = m g (sinθ + μcosθ)
289 = 9.8 x m ( sin 24.3°+ 0.33 cos 24.3°) ( take g =10 m/s²)
289 = 9.8 m(0.71)
m = 41.53 kg
Work done will be equal to
W= m g h
W= 41.53 x 9.8 x 2.39 J
W=972.83 J
In 1994 the performer Rod Stewart drew over 3 million people to a concert in Rio de Janeiro, Brazil.
(a) If the people in the group had an average mass of 80.0 kg, what collective gravitational force would the group have on a 4.50-kg eagle soaring 3.00 Ã 10 2 m above the throng? If you treat the group as a point object, you will get an upper limit for the gravitational force.
(b) What is the ratio of that force of attraction to the force between Earth and the eagle
Answers:
a) [tex]8.009(10)^{-7} N[/tex]
b) [tex]1.8(10)^{-8} [/tex]
Explanation:
a) Accoding to the Universal Law of Gravitation we have:
[tex]F_{g}=G\frac{Mm}{d^2}[/tex] (1)
Where:
[tex]F_{g}[/tex] is the gravitational force between the eagle and the throng
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Universal Gravitational constant
[tex]M=4.5 kg[/tex] is the mass of the eagle
[tex]m=(80 kg)(3(10)^{6} people/kg)=240(10)^{6} kg[/tex] is the mass of the throng
[tex]d=300 m[/tex] is the distance between the throng and the eagle
[tex]F_{g}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} \frac{(4.5 kg)(240(10)^{6} kg)}{(300 m)^{2}}[/tex] (2)
[tex]F_{g}=8.009 (10)^{-7} N[/tex] (3) As we can see the gravitational force between the eagle and the throng is quite small.
b) The attraction force between the eagle and Earth is the weight [tex]W[/tex] of the eagle, which is given by:
[tex]W=Mg[/tex] (4)
Where [tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity on Earth
[tex]W=(4.5 kg)(9.8 m/s^{2})[/tex] (5)
[tex]W=44.1 N[/tex] (6)
Now we can find the ratio between [tex]F_{g}[/tex] and [tex]W[/tex]:
[tex]\frac{F_{g}}{W}=\frac{8.009 (10)^{-7} N}{44.1N}[/tex]
[tex]\frac{F_{g}}{W}=1.8(^{-8})[/tex] As we can see this ratio is also quite small
A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static friction µstat beltween the block and the slab is 0.70, whereas their kinetic friction coefficient µkin is 0.40. The 10 kg block is pulled by a horizontal force with a magnitude of 100 N.
Answer:
[tex]\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]
Explanation:
Normal reaction from 40 kg slab on 10 kg block
M × g = 10 × 9.8 = 98 N
Static frictional force = 98 × 0.7 N
Static frictional force = 68.6 N is less than 100 N applied
10 kg block will slide on 40 kg slab and net force on it
= 100 N - kinetic friction
[tex]=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)[/tex]
= 100 - 39.2
= 60.8 N
[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}[/tex]
[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}[/tex]
[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}[/tex]
[tex]\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }[/tex]
Frictional force on 40 kg slab by 10 kg block = 98 × 0.4
Frictional force on 40 kg slab by 10 kg block = 39.2 N
[tex]40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}[/tex]
[tex]40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}[/tex]
40 kg slab will move with = [tex]0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]
[tex]\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]
The return-air ventilation duct in a home has a cross-sectional area of 900 cm^2. The air in a room that has dimensions 5.0 m x 11.0 m ×x 2.4 m is to be completely circulated in a 50-min cycle.
1) What is the speed of the air in the duct? (Express your answer to two significant figures.)
To solve the problem it is necessary to apply the concepts related to the flow rate of a fluid.
The flow rate is defined as
[tex]Q = Av[/tex]
Where,
[tex]Q = Discharge (m^3/s)[/tex]
[tex]A = Area (m^2)[/tex]
v = Average speed (m / s)
And also as
[tex]Q = \frac{V}{t}[/tex]
Where,
V = Volume
t = time
Let's start by finding the total volume according to the given dimensions, that is to say
[tex]V = 5*11*2.4[/tex]
[tex]V = 132m^3[/tex]
The entire cycle must be completed in 50 min = 3000s
In this way we know that the [tex]132m ^ 3[/tex] must be filled in 3000s, that is to say that there should be a flow of
[tex]Q = \frac{V}{t}[/tex]
[tex]Q = \frac{132}{3000}[/tex]
[tex]Q = 0.044m^3/s[/tex]
Using the relationship to find the speed we have to
[tex]Q = Av[/tex]
[tex]v = \frac{Q}{A}[/tex]
Replacing with our values,
[tex]v = \frac{0.044}{900*10^{-4}m^2}[/tex]
[tex]v = 0.488m/s[/tex]
Therefore the air speed in the duct must be 4.88m/s
A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magnitude of the field is allowed to vary in time according to B = Bmax e-t/τ , where Bmax and τ are constants. The field has the constant value Bmax for t < 0. Find the emf induced in the loop as a function of time. (Use the following as necessary: A, Bmax, t, and τ.) g
Answer:
Induced emf, [tex]\epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}[/tex]
Explanation:
The varying magnetic field with time t is given by according to equation as :
[tex]B=B_{max}e^{-t/\tau}[/tex]
Where
[tex]B_{max}\ and\ t[/tex] are constant
Let [tex]\epsilon[/tex] is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:
[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]
[tex]\epsilon=-\dfrac{d(BA)}{dt}[/tex]
[tex]\epsilon=-A\dfrac{d(B)}{dt}[/tex]
[tex]\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}[/tex]
[tex]\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}[/tex]
So, the induced emf in the loop as a function of time is [tex]A\dfrac{B_{max}e^{-t/\tau}}{\tau}[/tex]. Hence, this is the required solution.
The emf induced in the loop as a function of time is determined as [tex]emf = \frac{AB_{max}}{\tau} e^{-t/\tau}[/tex].
Induced emf
The emf induced in the rectangular loop is determined by applying Faraday's law of electromagnetic induction.
emf = dФ/dt
where;
Ф is magnetic fluxФ = BA
A is the area of the rectangular loop
emf = d(BA)/dt
[tex]emf = \frac{d(BA)}{dt} \\\\emf = A \frac{dB}{dt} \\\\emf = A \times B_{max}(e^{-t/\tau})(1/\tau)\\\\emf = \frac{A B_{max}(e^{-t/\tau})}{\tau}\\\\emf = \frac{AB_{max}}{\tau} e^{-t/\tau}[/tex]
Thus, the emf induced in the loop as a function of time is determined as [tex]emf = \frac{AB_{max}}{\tau} e^{-t/\tau}[/tex].
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What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius R, with the axis along a diameter, to the center of the balsa-wood sphere?
Answer:
[tex]D_{s}[/tex] ≈ 2.1 R
Explanation:
The moment of inertia of the bodies can be calculated by the equation
I = ∫ r² dm
For bodies with symmetry this tabulated, the moment of inertia of the center of mass
Sphere [tex]Is_{cm}[/tex] = 2/5 M R²
Spherical shell [tex]Ic_{cm}[/tex] = 2/3 M R²
The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass
I = [tex]I_{cm}[/tex] + M D²
Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation
Let's start with the spherical shell, axis is along a diameter
D = 2R
Ic = [tex]Ic_{cm}[/tex] + M D²
Ic = 2/3 MR² + M (2R)²
Ic = M R² (2/3 + 4)
Ic = 14/3 M R²
The sphere
Is =[tex]Is_{cm}[/tex] + M [[tex]D_{s}[/tex]²
Is = Ic
2/5 MR² + M [tex]D_{s}[/tex]² = 14/3 MR²
[tex]D_{s}[/tex]² = R² (14/3 - 2/5)
[tex]D_{s}[/tex] = √ (R² (64/15)
[tex]D_{s}[/tex] = 2,066 R
A 44.0 kg uniform rod 4.90 m long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a wire attached to its other end. The wire makes an angle of 30.0° with the horizontal, and is bolted to the wall directly above the hinge. If the wire can withstand a maximum tension of 1450N before breaking, how far from the wall can a 69.0kg person sit without breaking the wire?
Answer:
x ≤ 3.6913 m
Explanation:
Given
Mrod = 44.0 kg
L = 4.90 m
Tmax = 1450 N
Mman = 69 kg
A: left end of the rod
B: right end of the rod
x = distance from the left end to the man
If we take torques around the left end as follows
∑τ = 0 ⇒ - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0
⇒ - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0
⇒ - (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0
⇒ x ≤ 3.6913 m
A 0.270 m radius, 510-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of 10,000 V.
Answer:
0.35701 T
Explanation:
[tex]B_i[/tex] = Initial magnetic field
[tex]B_f[/tex] = Final magnetic field
[tex]\phi[/tex] = Magnetic flux
t = Time taken = 4.17 ms
N = Number of turns = 510
[tex]\epsilon[/tex] = Induced emf = 10000 V
r = Radius = 0.27 m
A = Area = [tex]\pi r^2[/tex]
Induced emf is given by
[tex]\epsilon=-N\frac{d\phi}{dt}\\\Rightarrow \epsilon=-N\frac{B_fAcos90-B_iAcos0}{dt}\\\Rightarrow \epsilon=N\frac{B_iA}{dt}\\\Rightarrow B_i=\frac{\epsilon dt}{NA}\\\Rightarrow B_i=\frac{10000 \times 4.17\times 10^{-3}}{510\times \pi 0.27^2}\\\Rightarrow B_i=0.35701\ T[/tex]
The magnetic field strength needed is 0.35701 T
As part of a safety investigation, two 1400 kg cars traveling at 20 m/s are crashed into different barriers. Find the average forces exerted on (a) the car that hits a line of water barrels and takes 1.5 s to stop, and (b) the car that hits a concrete barrier and takes 0.10 s to stop.
Answer:
(a) 18667N
(b)280000N
Explanation:
The momentum of both car prior to collision:
M = mv = 1400*20 = 28000 kgm/s
After colliding, the average force exerting on either car to kill this momentum is
(a)[tex] F_1 =\frac{M}{t_1} = \frac{28000}{1.5} = 18667N[/tex]
(b)[tex]F_2 = \frac{M}{t_2} = \frac{28000}{0.1} = 280000N[/tex]
To find the average forces exerted on the cars, use the formula: Force = change in momentum / time interval. For the car hitting the line of water barrels, the average force is -1400 kg * m/s / 1.5 s. For the car hitting the concrete barrier, the average force is -1400 kg * m/s / 0.10 s.
Explanation:To find the average forces exerted on the cars, we can use the formula:
Force = change in momentum / time interval
(a) For the car that hits a line of water barrels and takes 1.5 s to stop:
The change in momentum is given by: Δp = m * Δv = m * (0 - 20) = -1400 kg * m/s
Then, the average force exerted on the car is: F = Δp / t = -1400 kg * m/s / 1.5 s
(b) For the car that hits a concrete barrier and takes 0.10 s to stop:
The change in momentum is given by: Δp = m * Δv = m * (0 - 20) = -1400 kg * m/s
Then, the average force exerted on the car is: F = Δp / t = -1400 kg * m/s / 0.10 s
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A noninverting op-amp circuit with a gain of 96 V/V is found to have a 3-dB frequency of 8 kHz. For a particular system application, a bandwidth of 32 kHz is required. What is the highest gain available under these conditions?
Final answer:
To find the highest available gain for an op-amp circuit with a required bandwidth, we use the Gain-Bandwidth Product (GBP), which is a constant. For a GBP of 768 kHz and a bandwidth of 32 kHz, the maximum available gain is 24 V/V.
Explanation:
The question concerns determining the highest available gain for an op-amp circuit given a required bandwidth.
The Gain-Bandwidth Product (GBP) is a constant for an op-amp and is found by multiplying the current gain with its corresponding 3-dB frequency.
With an initial gain of 96 V/V and a 3-dB frequency of 8 kHz, the GBP can be calculated as 96 V/V * 8 kHz = 768 kHz.
To meet the requirement of a 32 kHz bandwidth with the same GBP (because GBP is constant), we can calculate the maximum gain as follows:
GBP = Gain * Bandwidth, which gives us
Gain = GBP / Bandwidth.
Plugging in the numbers, we get
Gain = 768 kHz / 32 kHz, resulting in a maximum gain of 24 V/V under the conditions of a 32 kHz bandwidth.
Zirconium tungstate is an unusual material because its volume shrinks with an increase in temperature for the temperature range 0.3 K to 1050 K (where it decomposes). In fact, the volumetric coefficient of thermal expansion is –26.4 × 10–6/K. Determine the ratio ΔV/V0 for the above mentioned temperature range. Express your answer in percent.
Answer:
2.771208%
Explanation:
[tex]\Delta V[/tex] = Change of volume
[tex]V_0[/tex] = Initial volume
[tex]\Delta T[/tex] = Change in temperature = (0.3-1050)
[tex]\beta[/tex] = Volumetric coefficient of thermal expansion = [tex]-26.4\times 10^{-6}\ /K[/tex]
Volumetric expansion of heat is given by
[tex]\frac{\Delta V}{V_0}=\beta \Delta T\\\Rightarrow \frac{\Delta V}{V_0}=-26.4\times 10^{-6}\times (0.3-1050)\\\Rightarrow \frac{\Delta V}{V_0}=0.02771208[/tex]
Finding percentage
[tex]\frac{\Delta V}{V_0}=0.02771208\times 100=2.771208\%[/tex]
The ratio of change of volume to initial volume is 2.771208%
To calculate the ratio ΔV/V0 for zirconium tungstate, the volume change can be determined using the volumetric coefficient of thermal expansion. The formula for the ratio is ΔV/V0 = (Volume change/Initial volume) × 100.
Explanation:The ratio ΔV/V0 can be calculated using the formula:
ΔV/V0 = (Volume change/Initial volume) × 100
Given that the volumetric coefficient of thermal expansion for zirconium tungstate is -26.4 × 10^(-6)/K, we can use this value to calculate the volume change. The volume change can be found by multiplying the coefficient of thermal expansion by the change in temperature:
Volume change = (-26.4 × 10^(-6)/K) × (1050 K - 0.3 K)
Using this value, we can calculate the ratio ΔV/V0:
ΔV/V0 = (Volume change/Initial volume) × 100
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When you apply the torque equation ∑τ = 0 to an object in equilibrium, the axis about which torques are calculated:
a. must be located at a pivot.
b. should be located at the edge of the object.
c. must be located at the object's center of gravity.
d. can be located anywhere.
Answer:
option D.
Explanation:
The correct answer is option D.
When an object is in equilibrium torque calculated at any point will be equal to zero.
An object is said to be in equilibrium net moment acting on the body should be equal to zero.
If the net moment on the object is not equal to zero then the object will rotate it will not be stable.
In a system in equilibrium, the choice of axis about which torques are calculated can be anywhere because for such a system, the sum of torques about any point is zero.
Explanation:When applying the torque equation, ∑τ = 0, to an object in equilibrium, the chosen axis about which torques are calculated can be located anywhere on or off the object. This principle is a result of the fact that in a system in equilibrium, the sum of torques about any point is zero, not just specific points like the object's pivot, edge, or center of gravity. For example, if you have a seesaw in balance, you could calculate the torques from the center, one of the seats, or even a point suspended above it in the air. As long as the object is in equilibrium and not moving, the torques calculated from any point will sum to zero because they counteract each other in direction and magnitude to maintain the state of balance.
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A thin coil has 17 rectangular turns of wire. When a current of 4 A runs through the coil, there is a total flux of 5 ✕ 10−3 T · m2 enclosed by one turn of the coil (note that
Φ = kI,
and you can calculate the proportionality constant k). Determine the inductance in henries.
Answer:
Inductance, L = 0.0212 Henries
Explanation:
It is given that,
Number of turns, N = 17
Current through the coil, I = 4 A
The total flux enclosed by the one turn of the coil, [tex]\phi=5\times 10^{-3}\ Tm^2[/tex]
The relation between the self inductance and the magnetic flux is given by :
[tex]L=\dfrac{N\phi}{I}[/tex]
[tex]L=\dfrac{17\times 5\times 10^{-3}}{4}[/tex]
L = 0.0212 Henries
So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.
The inductance of a coil with 17 turns, which has a flux of 5 * 10^−3 T · m2 when a current of 4 A runs through it, is 0.02125 Henry.
Explanation:In this problem, we are given the total flux Φ which is equal to the product of a proportionality constant k and the current I (Φ = kI ). The proportionality constant k can be calculated by dividing the flux by the current. k = Φ/I = (5 * 10^−3 T · m2) / 4 A = 1.25 * 10^-3 H/A. However, this is the inductance for just one single turn of the coil.
Since the coil has 17 turns, the total inductance L for the entire coil is equal to the product of the proportionality constant k and the number of turns N (L = kN). Therefore, L = 1.25 * 10^-3 H/A * 17 = 0.02125 Henry (H).
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