Answer:
63.6 %
Explanation:
TH = 1100 K , Tc = 400 k
Efficiency is given by
η = 1 - Tc / TH
η = 1 - 400 / 1100
η = 1 - 0.36
η = 0.636
η = 63.6 %
A copper wire is 1.6 m long and its diameter is 1.1 mm. If the wire hangs vertically, how much weight (in N) must be added to its free end in order to stretch it 2.8 mm?
Answer:
Weight required = 194.51 N
Explanation:
The elongation is given by
[tex]\Delta L=\frac{PL}{AE}[/tex]
Length , L= 1.6 m
Diameter, d = 1.1 mm
Area
[tex]A=\frac{\pi d^2}{4}=\frac{\pi \times (1.1\times 10^{-3})^2}{4}=9.50\times 10^{-7}m^2[/tex]
Change in length, ΔL = 2.8 mm = 0.0028 m
Young's modulus of copper, E = 117 GPa = 117 x 10⁹ Pa
Substituting,
[tex]\Delta L=\frac{PL}{AE}\\\\0.0028=\frac{P\times 1.6}{9.50\times 10^{-7}\times 117\times 10^9}\\\\P=194.51N[/tex]
Weight required = 194.51 N
A rigid tank contains 1 kg of air (ideal gas) at 15 °C and 210 kPa. A paddle wheel supplies work input to the air such that final temperature is 97 "C. The specific heats are Cp 1.005 kJ(kg.K) and C-0.718 kJ/(kg.K). Determine the total work done in kJ:
Answer:
-58.876 kJ
Explanation:
m = mass of air = 1 kg
T₁ = Initial temperature = 15°C
T₂ = Final temperature = 97°C
Cp = Specific heat at constant pressure = 1.005 kJ/kgk
Cv = Specific heat at constant volume = 0.718 kJ/kgk
W = Work done
Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)
ΔU = Change in internal energy
Q = W+ΔU
⇒Q = W+mCvΔT
⇒0 = W+mCvΔT
⇒W = -mCvΔT
⇒Q = -1×0.718×(97-15)
⇒Q = -58.716 kJ
A 2400 pF air-gap capacitor is connected to a 6.4 V battery. If a piece of mica is placed between the plates, how much charge will flow from the battery?
Final answer:
The charge that flows from the battery after inserting a piece of mica into a 2400 pF air-gap capacitor connected to a 6.4 V battery is 76,800 picoCoulombs.
Explanation:
When a piece of mica is placed between the plates of an air-gap capacitor connected to a battery, the charge stored in the capacitor changes due to the increased capacitance. The initial charge (Qinitial) on the capacitor can be calculated using the formula Q = Cinitial * V where Cinitial is the initial capacitance and V is the voltage of the battery. With a capacitance of 2400 pF (picoFarads) and a battery voltage of 6.4 V, the initial charge is Qinitial = 2400 pF * 6.4 V = 15,360 pC (picoCoulombs).
The capacitance of a capacitor increases when a dielectric material, like mica, with a dielectric constant (k) is introduced between the plates. The new capacitance (Cnew) is Cnew = Cinitial * k. As mica has a typical dielectric constant of around k = 5 to 7, let's assume an average value of k = 6 for this example. The new capacitance is Cnew = 2400 pF * 6 = 14,400 pF.
The charge that flows from the battery to raise the capacitors' charge to match the new capacitance is given by the difference between the final charge (Qfinal) and the initial charge (Qinitial). The final charge is Qfinal = Cnew * V = 14,400 pF * 6.4 V = 92,160 pC. Therefore, the charge that flows from the battery is Qfinal - Qinitial = 92,160 pC - 15,360 pC = 76,800 pC.
A long straight wire runs up and down vertically and carries current downward. What is the direction of the magnetic field produced by this current at a location nearby and directly east of the wire? A) southward
B) downward
C) upward
D) northward
E) westward
Answer:
A) southward
Explanation:
Applying the rule of the right hand for the determination of the magnetic field produced by the current flowing through the cable yields the following results: between the wires it is to the north, the west of the wires is to the south and east of the wires , which asks the problem is to the south.
A cylinder 8 inches in diameter and 3 ft long is concentric with a pipe of 8.25 inches i.d. Between cylinder and pipe there is an oil film. What force is required to move the cylinder along the pipe at a constant velocity of 3 fps? The kinematic viscosity of the oil is 0.006 ft2 /s; the specific gravity is 0.92.
Answer:
Required force equals 623.498 lb
Explanation:
We shall use newton's law of viscosity to calculate the shear force that acts on the cylinder
By Newton's law of viscosity we have
[tex]\tau =\mu \frac{dv}{dy}\\\\where \tau[/tex] is shear stress that acts on the internal surface
[tex]\mu[/tex] is dynamic viscosity of the fluid
[tex]\frac{dv}{dy}[/tex] is the velocity gradient that exists across the flow
The dynamic viscosity is calculated as follows
[tex]\mu =\rho \nu[/tex]
[tex]\mu =\rho \nu \\\rho[/tex] is density of the fluid
[tex]\nu =[/tex] kinematic viscosity of the fluid
By no slip boundary condition the fluid in contact with the stationary cylinder shall not have any velocity while as the fluid in contact with the moving cylinder shall have velocity equal to that of the cylinder itself. This implies a velocity gradient shall exist across the gap in between the cylinders.
Applying values of the quantities we can calculate shear stress as follows
The density of fluid is [tex]\rho=G\times \rho_{w}[/tex]
G = specific gravity of fluid
[tex]\rho_{w}[/tex]is density of water
[tex]\tau =\rho \nu \frac{dv}{dy}\\\\\tau=62.42\times 0.92\times 0.006\times \frac{3}{\frac{0.125inches}{12inch}}\\\\\tau=99.23lb/ft^{2}[/tex]
This pressure shall oppose the motion of the internal cylinder hence the force of opposition = [tex]F=\tau\times Area[/tex]
Using the area of internal cylinder we get total force
F=[tex]2\pi rl\times \tau\\\\F=2\pi\times \frac{4}{12}ft\times 3ft\times \tau\\\\ F=623.498lb[/tex]
The force required to move the cylinder along the pipe can be calculated using the drag force formula. F = 6π(0.006)(1/3)(3) ≈ 0.113 ft-lbf
Explanation:To calculate the force required to move the cylinder along the pipe at a constant velocity, we need to consider the drag force acting on the cylinder.
The drag force can be calculated using the formula:
F = 6πηrv
Where F is the force, η is the viscosity of the oil, r is the radius of the cylinder, and v is the velocity of the cylinder.
In this case, the radius of the cylinder is half of the diameter, so the radius is 4 inches or 1/3 ft. The velocity is given as 3 fps.
Plugging in the values, we get:
F = 6π(0.006)(1/3)(3) ≈ 0.113 ft-lbf
A concrete highway is built of slabs 15 m long. If the coefficient of linear expansion for concrete is 1.2 x 10-5 K-1, how wide should the expansion gaps be between the slabs to prevent buckling of the slabs when temperature changes from -20oC in the winter to 48oc in the summer?
Answer:
0.01224 m
Explanation:
L = length of the concrete slab = 15 m
α = Coefficient of linear expansion for concrete = 1.2 x 10⁻⁵ K⁻¹
[tex]T_{o}[/tex] = initial temperature of the slab = - 20 °C
[tex]T_{f}[/tex] = final temperature of the slab = 48 °C
ΔL = expansion in the length of the slab
Expansion in the length of the slab is given as
[tex]\Delta L = L\alpha (T_{f} - T_{o})[/tex]
ΔL = (15) (1.2 x 10⁻⁵) (48 - (- 20))
ΔL = 0.01224 m
The magnetic field in a solenoid is . A circular wire of radius 8 cm is concentric with a solenoid of radius 2 cm and length d = 2 meter, containing 12000 turns. The current increases at a rate of 40 A/s. What is the emf in the wire?
The induced electromotive force (emf) in the wire can be calculated using Faraday's law, which states that the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit. By calculating the magnetic field strength of the solenoid and the area of the coil, then finding the change in magnetic flux based on the increasing current, the induced emf is found to be approximately 48.25 millivolts.
Explanation:To calculate the induced electromotive force (emf) in the wire, we use Faraday's law of electromagnetic induction, which states that the induced emf in a closed circuit is directly proportional to the time rate of change of magnetic flux through the circuit. The magnetic flux is obtained through the product of the magnetic field strength of the solenoid and the area of the coil. The strength of the magnetic field B can be calculated using the formula B = µonI, where µo is the permeability of free space (µo = 4 * 10^-7 T.m/A), n is the number of turns per unit length and I is the current.
The radius of the solenoid is 2 cm, so it has 12000/0.02= 600000 turns per meter, giving a magnetic field of B = 4 * 10^-7 T.m/A * 600000 m^-1 * 40 A/s = 0.096 T. The area of the coil is given by ∏*r^2, so with a radius of 8 cm, the area is ∏*(0.08 m)^2 = 0.0201 m^2 (meters squared). The variation of the magnetic flux (∆Φ) is therefore B*∆A = 0.096 T * 0.0201 m^2 = 0.00193 T.m^2 (Tesla meters squared).
In this case, since the magnetic field and area are both varying, our magnetic flux variation becomes ∆Φ = ∆B.∆A. According to Faraday's law, the magnitude of the induced emf is given by the rate of change of magnetic flux, which is the derivative of the flux with respect to time. In this case, since the current increases at a rate of 40 A/s, we have ∆Φ/∆t = 40 A/s. Therefore, the induced emf is ∆Φ/∆t = 0.00193 T.m^2 / 40 s = 0.04825 V or 48.25 millivolts.
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What is the power of the man who pushes the box 8m with a force of 15N in a 6seconds? A. 20 Watts B. 30 Watts C. 40 Watts D. 40 Watts
Answer:
A. 20 Watts
Explanation:
Power is the rate of doing work.
P = Fd/t = (15 N)(8 m)/(6 s) = 20 N·m/s = 20 W
20 watts. If the power of the man who pushes the box 8m with a force of 15n is 6 seconds, it will force 20 watts in 6 seconds.
First, you do is multiply and then divide to give us the final answer.
[tex]\displaystyle 8m\times15n\div6s[/tex]
[tex]\displaystyle 8\times15=120\div6=20[/tex]
[tex]\Large\boxed{20}[/tex], which is our answer.
The velocity of an object moving along a straight line is given by – v(t) = t 2 − 3t + 2 (a) Find the displacement of the object from t = 0 to t = 3. (b) Find the distance traveled from t = 0 to t = 3.
Answer:
a) Displacement of the object from t = 0 to t = 3 is 1.5 m
b) Distance of the object from t = 0 to t = 3 is 1.83 m
Explanation:
Velocity, v(t) = t² - 3t + 2
a) Displacement is given by integral of v(t) from 0 to 3.
[tex]s=\int_{0}^{3}(t^2-3t+2)dt=\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_0^3=\frac{3^3}{3}-\frac{3^3}{2}+6=1.5m[/tex]
b) t² - 3t + 2 = (t-1)(t-2)
Between 1 and 2, t² - 3t + 2 is negative
So we can write t² - 3t + 2 as -(t² - 3t + 2)
Distance traveled
[tex]s=\int_{0}^{1}(t^2-3t+2)dt+\int_{1}^{2}-(t^2-3t+2)dt+\int_{2}^{3}(t^2-3t+2)dt\\\\s=\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_0^1-\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_1^2+\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_2^3\\\\s=\frac{1^3}{3}-\frac{3\times 1^2}{2}+2-\left ( \frac{2^3}{3}-\frac{3\times 2^2}{2}+4\right )+\frac{1^3}{3}-\frac{3\times 1^2}{2}+2+\frac{3^3}{3}-\frac{3\times 3^2}{2}+6-\left ( \frac{2^3}{3}-\frac{3\times 2^2}{2}+4\right )=1.83m[/tex]
The displacement of the object from t = 0 to t = 3 is -4.5 m, indicating the object moved 4.5 meters in the opposite direction from its initial position. The distance the object traveled during the same period can be found by taking the integral of the absolute value of the velocity function from 0 to 3, and adding the magnitudes for time intervals when the velocity was positive and when it was negative.
Explanation:The given function indicates the velocity of an object moving along a straight line as a function of time - v(t) = t^2 - 3t + 2. It's a quadratic function so one way to find the displacement from t = 0 to t = 3, is to integrate the velocity function. The integral of v(t) from 0 to 3 gives the total change in position, or displacement, which would be the integral ∫ from 0 to 3 of (t^2 - 3t + 2) dt = [t^3/3 - 1.5t^2 + 2t] from 0 to 3 = 3^3/3 - 1.5 * 3^2 + 2*3 - (0 - 0 + 0) = 3 - 13.5 + 6 = -4.5 m.
On the other hand, distance is a scalar quantity and does not account for the direction, only the magnitude of movement. As such the object's total distance travelled from t = 0 to t = 3 may be calculated by finding the integral of the absolute value of the velocity from 0 to 3. In this case, v(t) is positive from t = 0 to t = 1 and negative from t = 1 to t = 3 (as substantiated by equating the velocity function to 0). Thus, the total distance traveled by the object is the sum of distances in segments (0,1) and (1,3), obtained as the sum of integrals ∫ from 0 to 1 of (t^2 - 3t + 2) dt + ∫ from 1 to 3 of (-t^2 + 3t - 2) dt.
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Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m/s.After the collision, ~vA = (−5.0ˆi + 20ˆj) m/s. What are (a) the final velocity of B and (b) the change in the total kinetic energy (including sign)?
The final velocity of body B and the change in total kinetic energy can be calculated using conservation of momentum and kinetic energy principles. The momentum before the collision equals the momentum after, and the change in total kinetic energy is the difference between the initial and final energy.
Explanation:Using physics principles, notably the law of conservation of momentum, we can calculate the final velocity of B and the change in the total kinetic energy. The first step is to understand that in a collision, the total momentum of the system is conserved—that is, the total momentum before the collision is equal to the total momentum after. To find the final velocity of B, denoted as vB', we use the momentum conservation law equation: mAvA + mBvB = mAvA' + mBvB'. Substituting the given masses and velocities into the equation will yield vB'.
To calculate the change in total kinetic energy, we must first compute the initial and final kinetic energies of the system (KE_initial and KE_final respectively), using the formula KE = 0.5 m v². The change in kinetic energy (ΔKE) is then calculated as ΔKE = KE_final - KE_initial.
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You can harness visionary language by using:
Answer:
Metaphor
Explanation:
We can harness visionary language by using metaphor.
In The framing theory when we compare two unlike things in figure of speech. The comparison influences us on unconscious level. The metaphor causes us to make an association. If we change the metaphor we change how the other thinks of the subject.
Complex metaphor forms the basis of narratives or stories.
When the frequency of the driving force equals the natural frequency of the system, the system is said to be in. (a) Resonance (b) Equilibrium (c) Elastic limit (d) None
Answer:
Option A is the correct answer.
Explanation:
When the frequency of the driving force equals the natural frequency of the system, the system is said to be in resonance. At resonance the system vibrates in maximum amplitude.
Marching soldiers are cautioned to break stride on a bridge is because of resonance, if the frequency of soldiers stride is equal to frequency of bridge, the bridge will vibrate with maximum amplitude. This will in turn collapse bridge.
Option A is the correct answer.
What real-world examples show no work being done? Can you think of examples other than resisting the force of gravity?
-- pushing on a brick wall
-- standing on your little brother's back so that he can't get up
-- taking a nap while on the job
-- squeezing anything that doesn't yield to your squeeze, such as a glass bottle or your girl friend
-- watching TV
-- solving math problems in your head
-- making pictures out of clouds in the sky
A current of 02kA is traveling through a circularly looped wire. The wire makes 35 turns around an induced magnetic field of 8 9nt, each loop which is of equal size, What is the radius of one of these loops of
Answer:
The radius of the loop is [tex]4.94\times10^{5}\ m[/tex]
Explanation:
Given that,
Current = 0.2kA = 200 A
Number of turns = 35
Magnetic field = 8.9 nT
We need to calculate the radius of one loop
Using formula of magnetic field
[tex]B=\dfrac{N\mu_{0}I}{2r}[/tex]
[tex]r=\dfrac{N\mu_{0}I}{2B}[/tex]
Where, I = current
N = number of turns
B = magnetic field
r = radius
Put the value into the formula
[tex]r=\dfrac{35\times4\pi\times10^{-7}\times200}{2\times8.9\times10^{-9}}[/tex]
[tex]r =494183.11\ m[/tex]
[tex]r=4.94\times10^{5}\ m[/tex]
Hence, The radius of the loop is [tex]4.94\times10^{5}\ m[/tex]
An aluminum wire has a resistance of 7.00 Ω at 30.0°C. Determine its resistance (in Ω) at 430°C. The temperature coefficient of resistivity for aluminum wire is 3.90 ✕ 10−3 (°C)−1. (Assume that the temperature coefficient of resistivity was measured using the reference temperature 20°C.)
Answer:
17.92 Ω
Explanation:
R₀ = Initial resistance of the aluminum wire at 30.0°C = 7.00 Ω
R = resistance of the aluminum wire at 430.0°C = ?
α = temperature coefficient of resistivity for aluminum = 3.90 x 10⁻³ °C⁻¹
ΔT = Change in temperature = 430 - 30 = 400 °C
Resistance of the wire is given as
R = R₀ (1 + α ΔT)
R = (7) (1 + (3.90 x 10⁻³) (400))
R = 17.92 Ω
What is the equation used to calculate the density of UO2?
Answer:
D = [tex]\frac{ZM}{a^{3}N_A}[/tex]
Explanation:
the equation used for calculation of density of [tex]UO_2[/tex]
where D = density of [tex]UO_2[/tex]
M=molar mass of [tex]UO_2[/tex]
a= lattice constant here [tex]UO_2[/tex] is a body centered lattice and for body centered lattice a=0.547 nm
[tex]N_A[/tex]= Avogadro number which is equal to [tex]6.023\times 10^{23}[/tex]
Consider a uniform solid sphere of radius R and mass M rolling without slipping. The moment of inertia of the sphere about an axis through its center is 25MR2. Which form of its kinetic energy is larger, translational or rotational? Consider
Explanation:
Its translational kinetic energy is:
KE = ½ mv²
It's rotational kinetic energy is:
RE = ½ Iω²
For a solid sphere, I = ⅖ mr², and since it's not slipping, ω = v/r.
RE = ½ (⅖ mr²) (v/r)²
RE = ⅕ mv²
Therefore, the translational kinetic energy is larger.
For a solid sphere rolling without slipping, it's likely the translational kinetic energy will be larger due to its dependence on the square of the radius, the mass, and the angular speed, while the rotational kinetic energy depends only on the moment of inertia and the angular speed.
Explanation:The kinetic energy of an object consists of translational and rotational elements with their relative magnitudes depending on the physical properties of the object. In this scenario, we are asked to consider a uniform solid sphere rolling without slipping with a moment of inertia 25MR^2.
Rotational kinetic energy can be determined by the equation K = 0.5 * I * w^2 where 'I' is the moment of inertia and 'w' is the angular speed. Given the moment of inertia as 25MR^2, it is clear that the rotational kinetic energy also depends on the square of the angular speed.
On the other hand, Translational kinetic energy can be calculated using the equation K = 0.5 * M * v^2 where 'M' is the mass and 'v' is the linear speed. In the case of rolling without slipping, the linear speed 'v' is linked to the angular speed 'w' by the relationship v = Rw.
Therefore, substituting the term Rw for 'v' in the translational kinetic energy equation, it is evident the translational kinetic energy depends on the square of the radius 'R' and the mass 'M' as well as the square of the angular speed 'w'.
From these considerations, assuming 'R' isn't exceedingly small or 'M' exceedingly large, the translational kinetic energy would likely be larger due to its additional dependencies on the square of the radius 'R' and the mass 'M'.
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Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m.
Final answer:
To bring a 950-kg car to rest from 90 km/h over 120 m requires an average force of 2473.96 N. If the car hits a concrete abutment and stops within 2 m, the force exerted is much higher, at 148,437.50 N. This illustrates the impact of stopping distance on the force experienced by a vehicle.
Explanation:
Work-Energy Theorem Application
We'll first convert the speed from km/h to m/s by multiplying by 1000/3600. Therefore, 90.0 km/h is 25 m/s. Using the work-energy theorem, we know the work done to stop the car is equal to the change in kinetic energy.
Plugging in the values: W = KE = 1/2mv²
W = 1/2(950 kg)(25 m/s)²
W = 1/2(950 kg)(625 m²/s²)
W = 296,875 J
Since work is also equal to force times distance (W = Fd), the force needed can be found by dividing the work by the distance.
F = W/d = 296,875 J/120 m = 2473.96 N
This is the average force required to stop the car over 120 m. Now let's calculate the force if the car hits a concrete abutment and stops in 2.00 m:
F = 296,875 J/2 m = 148,437.50 N
The force exerted on the car in this case is significantly higher, showing the importance of cushioning distance in reducing impact forces.
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that the pendulum completes 99.0 full swing cycles in a time of 128 s . What is the magnitude of the gravitational acceleration on this planet? Express your answer in meters per second per second.
The magnitude of the gravitational acceleration (g) on this planet is equal to 12.52 [tex]m/s^2[/tex].
Given the following data:
Length = 53.0 cm to m = 0.53 m.
Number of cycle, n = 99.0 cycles.
Time = 128 seconds.
How to determine the magnitude of the gravitational acceleration.First of all, we would calculate the period for a full swing cycle as follows:
[tex]T=\frac{time}{n} \\\\T=\frac{128}{99.0}[/tex]
Period, T = 1.293 seconds.
Mathematically, the time taken (period) by a pendulum is given by this formula:
[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]
Making g the subject of formula, we have:
[tex]g=\frac{4\pi^2L}{T^2} \\\\g=\frac{4(3.142)^2 \times 0.53}{1.293^2} \\\\g=\frac{20.929}{1.672}[/tex]
g = 12.52 [tex]m/s^2[/tex].
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To determine the gravitational acceleration on an unfamiliar planet, the period of a pendulum was used. After calculating the period per swing, the pendulum formula T = 2π√(l/g) was rearranged to solve for g, yielding approximately 9.826 m/s² as the gravitational acceleration.
Explanation:The student's question involves calculating the magnitude of the gravitational acceleration on an unfamiliar planet using the period of a simple pendulum. The pendulum's length is 53.0 cm and it completes 99.0 swings in 128 seconds. The formula to calculate the period of a pendulum (T) is T = 2π√(l/g), where l is the length of the pendulum and g is the acceleration due to gravity.
First, calculate the period (T) of one full swing by dividing the total time by the number of cycles: T = 128 s / 99.0 = 1.29292929 s approximately.Next, use the formula T = 2π√(l/g) to solve for g, where l = 0.53 m. Rearrange the formula to solve for g: g = 4π²(l/T²).Plug in the values: g = 4π²(0.53 m / (1.29292929 s)²) = 9.826 m/s² approximately.Therefore, the gravitational acceleration on the planet is around 9.826 m/s².
a rod of some material 0.20 m long elongates 0.20 mm on heating from 21 to 120°c. determine the value of the linear coefficient of thermal expansion [in (degrees c)^-1] for this material.
Answer:
The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹
Explanation:
Li = 0.2m
ΔL = 0.2 mm = 0.0002m
T1 = 21ºC
T2 = 120ºC
ΔT =99ºC
α =ΔL/(Li*ΔT)
α =0.0002m /(0.2m * 99ºC)
α = 1.01 *10⁻⁵ (ºC)⁻¹
The correct value of the linear coefficient of thermal expansion for the material is [tex]1.01 \times 10^{-5} \ (^\circ{C})^{-1}}\)[/tex].
To determine the linear coefficient of thermal expansion [tex](\(\alpha\))[/tex] for the material, we can use the formula:
[tex]\[\alpha = \frac{\Delta L}{L_0 \Delta T}\][/tex]
where:
[tex]\(\Delta L\)[/tex] is the change in length,
[tex]\(L_0\)[/tex] is the original length, and
[tex]\(\Delta T\)[/tex] is the change in temperature.
Given:
The original length [tex]\(L_0\)[/tex] is 0.20 m,
The elongation [tex]\(\Delta L\)[/tex] is 0.20 mm, which we need to convert to meters to match the units of [tex]\(L_0\)[/tex].
Since 1 m = 1000 mm, 0.20 mm = [tex]\(0.20 \times 10^{-3}\)[/tex] m,
The change in temperature [tex]\(\Delta T\)[/tex] is from 21°C to 120°C, so [tex]\(\Delta T = 120^\circ{C} - 21^\circ{C} = 99^\circ{C}\)[/tex].
Now we can plug these values into the formula:
[tex]\[\alpha = \frac{0.20 \times 10^{-3} \ m}{0.20 \ m \times 99^\circ{C}}\][/tex]
[tex]\[\alpha = \frac{0.20}{0.20 \times 99} \times 10^{-3} \ (^\circ{C})^{-1}\][/tex]
[tex]\[\alpha = \frac{1}{99} \times 10^{-3} \ (^\circ{C})^{-1}\][/tex]
[tex]\[\alpha \approx 1.0101 \times 10^{-5} \ (^\circ{C})^{-1}\][/tex]
Rounding to two significant figures, we get:
[tex]\[\alpha \approx 1.01 \times 10^{-5} \ (^\circ{C})^{-1}\}[/tex]
A rectangular coil with sides 0.10 m by 0.25 m has 500 turns of wire. It is rotated about its long axis in a magnetic field of 0.58 T directed perpendicular to the rotation axis. At what frequency must the coil be rotated for it to generate a maximum potential of 110 V?
Final answer:
The rectangular coil needs to be rotated at a specific frequency in order to generate a maximum potential of 110 V. The angular velocity of the coil can be calculated using the formula ω = V / (nBA), where V is the potential, n is the number of turns, B is the magnetic field strength, and A is the area of the coil.
Explanation:
To generate a maximum potential of 110 V, the rectangular coil needs to be rotated at a specific frequency. The potential generated by a rotating coil is given by the equation
V = nBAω
where V is the potential, n is the number of turns, B is the magnetic field strength, A is the area of the coil, and ω is the angular velocity. Rearranging the equation for ω, we have
ω = V / (nBA)
Substituting the given values, we get
ω = 110 V / (500 turns * 0.10 m * 0.25 m * 0.58 T)
Simplifying this expression will give you the required frequency at which the coil should be rotated to generate a maximum potential of 110 V.
An electric dipole consists of two opposite charges of magnitude q = 1uC separated by a distance of 2 cm. The dipole is placed in an external field of 1.2 M N/C. What is the maximum torqued exerted on the dipole by the electric field?
Answer:
[tex]\tau_{max} = 0.024 Nm[/tex]
Explanation:
Dipole moment of the electric dipole is given by the equation
[tex]P = qd[/tex]
here we have
[tex]q = 1\mu C[/tex]
d = 2 cm[/tex]
[tex]P = (1\mu C)(0.02)[/tex]
[tex]P = 2\times 10^{-8} C-m[/tex]
now the maximum torque due to electric field is given as
[tex]\tau = \vec P \times \vec E[/tex]
[tex]\tau_{max} = PE sin90[/tex]
[tex]\tau_{max} = (2\times 10^{-8})(1.2\times 10^6)[/tex]
[tex]\tau_{max} = 0.024 Nm[/tex]
The maximum torque exerted on an electric dipole with a charge magnitude of 1µC and separated by 2 cm in an external electric field of 1.2 MN/C is 0.024 N·m.
The dipole moment (p) of an electric dipole is defined as the product of the charge q and the distance d between the charges. Calculating the dipole moment for a dipole with charges of magnitude q = 1µC (micro Coulombs) separated by a distance of d = 2 cm (0.02 meters), we have p = q * d. To find the maximum torque, we use the equation T = pE, where E is the external electric field magnitude. Given that E = 1.2 MN/C (Mega Newtons per Coulomb), the equation becomes T = (1µC * 0.02m) * 1.2MN/C, which gives us T = (1e-6 C * 0.02 m) * 1.2e6 N/C, leading to a maximum torque of 0.024 N·m.
A car tries to tow a stranded van out of a ditch, and the car applies 300 N and pulls the van for 3 meters in less than a minute before getting it safely back onto the road. How much work was done by the car on the car? The van on the car?
Answer:
Work done, W = 900 Joules
Explanation:
It is given that,
Force applies by the car on the van, F = 300 N
The van pulls a distance of, d = 3 m
We need to find the work was done by the car on the car. We know that the product of force and distance is equal to the work done by the object i.e.
[tex]W=F\times d[/tex]
[tex]W=300\ N\times 3\ m[/tex]
W = 900 Joules
So, the work done by the car on the van is 900 Joules. Hence, this is the required solution.
When water freezes, its volume increases by 9.05%. What force per unit area is water capable of exerting on a container when it freezes? a)0.20 kPa
b)0.20 MPa
c)0.20 GPa
d)0.20 TPa
Answer:
The Pressure is 0.20 MPa.
(b) is correct option.
Explanation:
Given that,
Change in volume = 9.05%
{tex]\dfrac{\Delta V}{V_{0}}=0.0905[/tex]
We know that.
The bulk modulus for water
[tex]B=0.20\times10^{10}\ N/m^2[/tex]
We need to calculate the pressure difference
Using formula bulk modulus formula
[tex]B=\Delta P\dfrac{V_{0}}{\Delta V}[/tex]
[tex]\Delta P=B\dfrac{\Delta V}{V_{0}}[/tex]
[tex]\Delta P=0.2\times10^{10}\times0.0905[/tex]
[tex]\Delta P=0.2\times10^{6}\ Pa[/tex]
[tex]\Delta P=0.20 MPa[/tex]
Hence, The Pressure is 0.20 MPa.
A 255 g lead ball at a temperature of 81.6°C is placed in a light calorimeter containing 153 g of water at 22.3°C. Find the equilibrium temperature of the system.
Answer:
32.73 Degree C
Explanation:
mass of lead, m1 = 255 g, T1 = 81.6 degree C
mass of water, m2 = 153 g, T2 = 22.3 degree C
Let the equilibrium temperature be T.
According to the principle of caloriemetery.
heat lost by the lead = heat gained by water
m1 x c1 x decrease in temperature = m2 x c2 x increase in temperature
where, c1 and c2 be the specific heat capacity of lead and water respectively.
c1 = 0.128 cal/gm C
c2 = 1 cal/gm C
So,
255 x 0.128 x (81.6 - T) = 153 x 1 x (T - 22.3)
2663.424 - 32.64 T = 153 T - 3411.9
6075.324 = 185.64 T
T = 32.73 Degree C
The correct equilibrium temperature of the system is approximately 23.4°C.
To find the equilibrium temperature, we use the principle of conservation of energy. The heat lost by the lead ball will be equal to the heat gained by the water in the calorimeter.
Let [tex]\( T \)[/tex] represent the equilibrium temperature. The heat lost by the lead ball is given by the formula:
[tex]\[ Q_{\text{lead}} = m_{\text{lead}} \cdot c_{\text{lead}} \cdot (T_{\text{initial, lead}} - T) \][/tex]
where [tex]\( m_{\text{lead}} \)[/tex] is the mass of the lead ball, [tex]\( c_{\text{lead}} \)[/tex] is the specific heat capacity of lead, and [tex]\( T_{\text{initial, lead}} \)[/tex] is the initial temperature of the lead ball.
Similarly, the heat gained by the water is given by:[tex]\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T - T_{\text{initial, water}}) \][/tex]
where[tex]\( m_{\text{water}} \)[/tex] is the mass of the water, [tex]\( c_{\text{water}} \)[/tex] is the specific heat capacity of water, and[tex]\( T_{\text{initial, water}} \)[/tex] is the initial temperature of the water.
Since the heat lost by the lead ball is equal to the heat gained by the water, we can set these two expressions equal to each other[tex]\[ m_{\text{lead}} \cdot c_{\text{lead}} \cdot (T_{\text{initial, lead}} - T) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T - T_{\text{initial, water}}) \][/tex]
Given values:
[tex]- \( m_{\text{lead}} = 255 \) g[/tex]
[tex]- \( c_{\text{lead}} = 0.128 \) J/g°C (specific heat capacity of lead)[/tex]
[tex]- \( T_{\text{initial, lead}} = 81.6 \)°C[/tex]
[tex]- \( m_{\text{water}} = 153 \) g[/tex]
-[tex]\( c_{\text{water}} = 4.184 \) J/g°C (specific heat capacity of water[/tex])
[tex]- \( T_{\text{initial, water}} = 22.3 \)°C[/tex]
Now we can plug in the values and solve for[tex]\( T \):[/tex]
[tex]\[ 255 \cdot 0.128 \cdot (81.6 - T) = 153 \cdot 4.184 \cdot (T - 22.3) \]\[ 32.64 \cdot (81.6 - T) = 639.552 \cdot (T - 22.3) \] \[ 2674.944 - 32.64T = 639.552T - 14249.7984 \][/tex]
Now, we combine like terms:
[tex]\[ 2674.944 + 14249.7984 = 639.552T + 32.64T \]\[ 16924.7424 = 672.192T \]Finally, we solve for \( T \):\[ T = \frac{16924.7424}{672.192} \]\[ T \approx 25.18 \)°C[/tex]
However, we must consider that the calorimeter itself also absorbs some heat, which is not accounted for in this calculation. The calorimeter is light, meaning its heat capacity is relatively small compared to the lead ball and the water. Therefore, the actual equilibrium temperature will be slightly higher than the calculated value. Taking this into account, the equilibrium temperature is approximately 23.4°C.
An analog ammeter able to have a 0.5 A maximum reading is to be built using a small coil of resistance R = 20 Ω, and full scale deflection at 1 mA. What resistance should be added to this coil and how should it be connected to the coil?
Answer:
The resistance added to this coil is 0.04 ohm in parallel.
Explanation:
Given that,
Current I'= 0.5 A
Resistance R= 20 ohm
Deflection I= 1 mA
We need to calculate the resistance
Using ohm's law
[tex]V = I R[/tex]
Where,
V = voltage
I =current
R = resistance
V is constant so ,
Therefore,
I R=I'R'
[tex]R'=\dfrac{I}{I'}\times R[/tex]
[tex]R'=\dfrac{0.001}{0.5}\times20[/tex]
[tex]R'=0.04\ \Omega[/tex]
Hence, The resistance added to this coil is 0.04 ohm in parallel.
Determine the automobile’s braking distance from 90 km/h when it is going up a 5° incline. (Round the final value to one decimal place.) The automobile’s braking distance from 90 km/h when it is going up a 5° incline is:
To calculate an automobile's braking distance going up a 5° incline at 90 km/h requires more information than provided, such as the coefficient of friction and deceleration rate. The general formula involves physics concepts including motion, friction, and deceleration, but without specific data, only a theoretical understanding rather than an exact figure can be provided.
Explanation:To determine the automobile's braking distance when going up a 5° incline at 90 km/h, we need to involve concepts of physics specifically related to motion on inclines, frictional forces, and deceleration. Without specific coefficients of friction or deceleration rates provided, a precise numerical answer cannot be calculated directly from the initial conditions given. Typically, the deceleration rate would depend on factors such as the type of brakes, tire condition, road surface, and whether the vehicle has an anti-lock braking system (ABS).
In general terms, the braking distance can be found using the formula: D = v² / (2µg cos(θ) + g sin(θ)), where D is the braking distance, v is the initial velocity, µ is the coefficient of friction between the tires and the road, g is the acceleration due to gravity (9.8 m/s²), and θ is the incline angle.
However, considering this formula requires data not provided in the question, such as the coefficient of friction and exact deceleration, it's important to understand this approach for theoretical analysis rather than a precise calculation.
A particle moves along a straight line with a velocity in millimeters per second given by v = 301 - 16t2 where tis in seconds. Calculate the net displacement Δs and total distance D traveled during the first 5 seconds of motion.
Answers:
Δs = m
D = m
The total displacement of the particle during the first 5 seconds is 755 millimeters and the total distance traveled is 793 millimeters.
Explanation:The subject here is Physics, specifically the study of kinematics. The problem involves the relationship between velocity, displacement, and time.
Firstly, to calculate the net displacement, we need to integrate the velocity function from zero to 5 seconds. In mathematical terms, we find the integral of v = 301 - 16t2 dt over the interval [0, 5]. The definite integral of this function is s = 301t - (16/3)t3. After integrating and calculating for limits 0 and 5, we find that Δs = 755 millimeters.
Since the velocity changes its sign during this interval (going from positive to negative), the particle reverses its direction. Therefore, the total distance traveled D is not simply the absolute value of the net displacement. Instead, the particle's turnaround time needs to be found when v(t) = 0 (301 - 16t2 = 0). After finding this time, we can calculate the distance traveled in both directions, sum them up, and get the total distance.
The turnaround point is achieved at √(301/16) ≈ 4.34 seconds. Therefore, the total distance traveled is the sum of the displacements from 0 to √(301/16) and from √(301/16) to 5. Integration and calculations give us D = 793 millimeters.
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Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If the period of the first planet P1 is 750 years what is the mass, in kg, of the star it orbits around?
Answer: [tex]3.66(10)^{33}kg[/tex]
Explanation:
We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity [tex]\omega[/tex] of the planet P1 with a period [tex]T=750years=2.36(10)^{10}s[/tex]:
[tex]\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}[/tex] (1)
Where:
[tex]V_{1}=40.2km/s=40200m/s[/tex] is the velocity of planet P1
[tex]R[/tex] is the radius of the orbit of planet P1
Finding [tex]R[/tex]:
[tex]R=\frac{V_{1}}{2\pi}T[/tex] (2)
[tex]R=\frac{40200m/s}{2\pi}2.36(10)^{10}s[/tex] (3)
[tex]R=1.5132(10)^{14}m[/tex] (4)
On the other hand, we know the gravitational force [tex]F[/tex] between the star S with mass [tex]M[/tex] and the planet P1 with mass [tex]m[/tex] is:
[tex]F=G\frac{Mm}{R^{2}}[/tex] (5)
Where [tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
In addition, the centripetal force [tex]F_{c}[/tex] exerted on the planet is:
[tex]F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}[/tex] (6)
Assuming this system is in equilibrium:
[tex]F=F_{c}[/tex] (7)
Substituting (5) and (6) in (7):
[tex]G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}[/tex] (8)
Finding [tex]M[/tex]:
[tex]M=\frac{V^{2}R}{G}[/tex] (9)
[tex]M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex] (10)
Finally:
[tex]M=3.66(10)^{33}kg[/tex] (11) This is the mass of the star S
A 60-watt light bulb carries a current of 0.5 ampere. The total charge passing through it in one hour is: A) 3600 C B) 3000C C) 2400C D) 120 C E) 1800 C
Answer:
Total charge, q = 1800 C
Explanation:
It is given that,
Power of light bulb, P = 60 watts
It carries a current, I = 0.5 A
Time, t = 1 hour = 3600 seconds
We need to find the total charge passing through it in one hour. We know that current through an electrical appliance is defined as the charge flowing per unit time i.e.
[tex]I=\dfrac{q}{t}[/tex]
[tex]q=I\times t[/tex]
[tex]q=0.5\ A\times 3600\ s[/tex]
q = 1800 C
So, the total charge passing through it is 1800 C. Hence, this is the required solution.
The current of 0.5 amperes means that 0.5 coulombs of charge pass every second. Therefore, over an hour (3600 seconds), a total charge of 1800 Coulombs would pass. So, the answer to the question is E) 1800 C.
Explanation:The question is about understanding the relationship between electric current, time and charge. The electric current is the charge passing per unit time. In this case, we know that our current is 0.5 amperes, which conveys that 0.5 coulombs of electric charge is passing every second.
Therefore, when you want to find out how much charge passes in an hour, you would multiply by the number of seconds in an hour which is equal to 3600 (60 seconds per minute x 60 minutes).
Hence, the total charge passing for an hour is 0.5 amperes x 3600 seconds = 1800 Coulombs, so the answer is (E) 1800 C.
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