Answer:
The work and heat transfer for this process is = 270.588 kJ
Explanation:
Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k
The Pressure-Volume relation is PV = C
T = C for isothermal process
Calculating for the work done in isothermal process
W = P₁V₁ [tex]ln[\frac{P_{1} }{P_{2} }][/tex]
= mRT₁[tex]ln[\frac{P_{1} }{P_{2} }][/tex] [∵pV = mRT]
= (5) (0.287) (272.039) [tex]ln[\frac{2.0}{1.0}][/tex]
= 270.588 kJ
Since the process is isothermal, Internal energy change is zero
ΔU = [tex]mc_{v}(T_{2} - T_{1} ) = 0[/tex]
From 1st law of thermodynamics
Q = ΔU + W
= 0 + 270.588
= 270.588 kJ
To find the work and heat transfer in this process, we can apply the First Law of Thermodynamics. We can determine the work done using the equation pV = constant and calculate the heat transfer using the equation Q = CpdT.
To find the work and heat transfer in this process, we need to apply the First Law of Thermodynamics. The First Law states that the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system. In this case, since the process is isobaric (constant pressure), we can determine the work done by integrating the equation pV = constant. This will give us the equation W = p(Vf - Vi), where W is the work done, p is the pressure, and Vf and Vi are the final and initial volumes respectively.
Since the process is isobaric, the heat transfer can be calculated using the equation Q = CpdT, where Q is the heat transfer, Cp is the specific heat at constant pressure, and dT is the change in temperature. In this case, since the temperature change is not given explicitly, we can assume it to be the same as the change in internal energy, which gives us dT = dU/Cp.
By substituting the given values into the equations, we can calculate the work and heat transfer.
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A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation
Answer:
Explanation:
acceleration of test tube
= ω² R
= (2πn)² R
= 4π²n²R
n = no of rotation per second
= 3700 / 60
= 61.67
R = .10 m
acceleration
= 4π²n²R
= 4 x 3.14² x 61.67² x .10
= 14999 N Approx
Final answer:
The acceleration at the end of a test tube 10 cm from the axis of rotation in a centrifuge spinning at 3700 rpm is calculated to be 15051.2 m/s², demonstrating the significant centrifugal forces generated by such devices.
Explanation:
To calculate the acceleration at the end of a test tube that is 10 cm from the axis of rotation in a centrifuge spinning at 3700 rpm, we first need to convert the rotational speed to radians per second. The formula to convert revolutions per minute (rpm) to radians per second (ω in rad/s) is ω = (2π×rpm)/60. Thus, ω = (2π× 3700)/60 = 387.98 rad/s. Next, we use the formula for centripetal acceleration, a = ω²×r, where r is the radius of the circle (distance from the center of rotation to the point of interest) in meters. Given that r = 10 cm = 0.1 m, the acceleration a = (387.98)^2× 0.1 = 15051.2 m/s².
This centripetal acceleration is much larger than Earth's gravitational acceleration, indicating the extreme forces at play in a centrifuge's operation, which is crucial for its role in laboratory settings for sedimentation of materials.
Current passes through a solution of sodium chloride. In 1.00 s, 2.68×1016Na+ ions arrive at the negative electrode and 3.92×1016Cl− ions arrive at the positive electrode. (a) What is the current passing between the electrodes? (b) What is the direction of the current?
Explanation:
Given that,
Number of sodium ions at the negative electrode, [tex]Na^+=2.68\times 10^{16}[/tex]
Number of chloride ions at the positive electrode, [tex]Cl^-=3.92\times 10^{16}[/tex]
(a) The current flowing in the circuit is due to the positive as well as negative charges such that total charge becomes:
[tex]Q=(Na^++Cl^-)e[/tex]
[tex]Q=(2.68\times 10^{16}+3.92\times 10^{16})(1.6\times 10^{-19})[/tex]
Q = 0.01056 C
The current is given by :
[tex]I=\dfrac{Q}{t}[/tex]
[tex]I=\dfrac{0.01056}{1}=10.56\ mA[/tex]
So, the current passing between the electrodes is 10.56 mA.
(b) The direction of electric current is towards negative electrodes.
Explanation:
(a) First, we will calculate the charge of sodium ions as follows.
q = ne
= [tex]2.68 \times 10^{16} \times 1.6 \times 10^{-19} C[/tex]
= [tex]4.288 \times 10^{-3} C[/tex]
Now, charge of chlorine ions is calculated as follows.
q' = ne
= [tex]3.92 \times 10^{16} \times 1.6 \times 10^{-19} C[/tex]
= [tex]6.272 \times 10^{-3} C[/tex]
Hence, the current will be calculated as follows.
i = [tex]\frac{q}{t} + \frac{q'}{t}[/tex]
= [tex]\frac{4.288 \times 10^{-3} C}{1.00} + \frac{6.272 \times 10^{-3} C}{1.00}[/tex]
= [tex]10.56 \times 10^{-3} A[/tex]
= 10.56 mA
Therefore, current passing between the electrodes is 10.56 mA.
(b) Since, positive ions are moving towards the negative electrode. And, current is the flow of ions or electrons therefore, the direction of current is towards the negative electrode.
The dean of a university located near the ocean (who was not a science major in college) proposes building an infrared telescope right on campus and operating it in a nice heated dome so that astronomers will be comfortable on cold winter nights. Criticize this proposal, giving your reasoning.
Answer:
The proposal is bad and has many mistakes in itself.
Explanation:
-Telescopes are preferably to be placed far from cities, ideally a remote place on a mountain or raised land. It should not be erected at sea level.
-It should not be erected at sea level. High humidity present at sea level clouds observations made as infrared observations are not possible at sea level.
-Dome should not be heated at all. Strong air currents are generated during heating which inturn ruins observations. Also, a heated dome will emit infrared radiation which ultimately swamps astronomical signals.
Final answer:
Coastal location and heated dome are unsuitable for infrared telescope due to atmospheric interference, light pollution, and dome-related issues. Consider remote location, dry climate, and remote operation for optimal observations. Consult experts and consider environmental impact.
Explanation:
Criticisms of Building an Infrared Telescope Near the Ocean in a Heated Dome:
While the intention might seem well-meaning, building an infrared telescope near the ocean in a heated dome has several drawbacks:
Unsuitable Location:
Atmospheric interference: Coastal areas have higher humidity and turbulent air, which negatively affects infrared observations due to water vapor absorption and image distortion.
Light pollution: City lights and nearby human activity create significant light pollution, impacting observations of faint infrared sources.
Dome Issues:
Cost: Building and maintaining a heated dome adds significant expense compared to an open-air observatory.
Heat distortion: Heating the dome creates air currents that can distort telescope observations.
Ventilation challenges: Maintaining controlled airflow and humidity within the dome can be complex and costly.
Alternatives:
Remote location: Building the telescope at a high-altitude, dry site with minimal light pollution would be more suitable for infrared observations.
Remote access and automation: Modern telescopes can be operated remotely, eliminating the need for astronomers to be physically present during observations.
Additional considerations:
Expertise: Consult with professional astronomers for advice on telescope placement and operation.
Environmental impact: Consider the potential ecological impact of the telescope and dome on the coastal environment.
Overall, while the concern for astronomer comfort is understandable, a coastal location with a heated dome is not optimal for an infrared telescope. Exploring alternative locations and remote operation technologies would be more effective and cost-effective for achieving high-quality scientific observations.
To practice Problem-Solving Strategy 25.1 Resistor Circuits. Find the currents through and the potential difference across each resistor in the circuit shown on the diagram (Figure 1) . Use the following values: E = 12.0V , R1 = 15.0Ω , R2 = 45.0Ω , R3 = 20.0Ω , and R4 = 25.0Ω .
Part A
Step by step, reduce the circuit to the smallest possible number of equivalent resistors in order to find the equivalent resistance Req of the entire circuit.
Express you answer in ohms to three significant figures.
Part B
Find Ieq, the current through the equivalent resistor.
Express your answer in amperes to three significant figures.
Answer:
I₁ = 0.32 A
I₂ = 0.16 A
I₃ = 0.16 A
I₄ = 0.16 A
Explanation:
Part A
The equivalent resistance of the circuit is
Req = R₁ + (R₂||(R₃ + R₄))
Req = 15 + (45||(20 + 25))
Req = 15 + (45||45) = 15 + ((45×45)/(45+45)) = 15 + 22.5 = 37.5 Ω
Part B
From Ohm's law,
V = IR
Ieq = V/(Req) = 12/(37.5) = 0.32 A
Part C
Current through R₁ is the same as Ieq as R₁ is directly in series with the voltage source.
I₁ = 0.32 A
Then, this current flows through the (R₂||(R₃ + R₄)) loop too as the entire loop is in series with R₁
This current is them split into two branches of R₂ and (R₃ + R₄), since these two branches have equal resistances (45 Ω and 45 Ω), 0.32 A is split equally between the R₂ and (R₃ + R₄) branch.
Current through R₂ (using current divider)
I₂ = (45/90) × 0.32 = 0.16 A
Current through (R₃ + R₄) = 0.16 A too.
And because the two resistors are in series, the same current flows through them.
I₃ = I₄ = 0.16 A
First, we resolve the parallel circuits using the formula for resistors in parallel to get the equivalent resistance, Rp. We then substitute this into the series circuits to get the total equivalent resistance, Req. Using Ohm's Law, we can calculate the equivalent current, Ieq.
Explanation:To solve this, we first need to determine the equivalent resistance of the whole circuit. As it is a combination of parallel and series circuits, we have to start by replacing the resistors in parallel. Using the formula for resistors in parallel, 1/R = 1/R2 + 1/R3, the parallel combination of R2 and R3 can be replaced by one resistor with resistance Rp = 1/ ((1/R2) + (1/R3)). This gives Rp = 13.333Ω. Now we have a simple series circuit with R1, Rp, and R4. The equivalent resistance of the whole circuit is then Req = R1 + Rp + R4 = 15.0Ω + 13.333Ω + 25.0Ω = 53.333Ω.
In part B, we apply Ohm's Law: I = E/R to get the equivalent current through the circuit. Using our values for E (12V) and Req (53.333Ω), we find Ieq = 0.225A.
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A blood-flow meter emits a 1.1-MHz ultrasound pulse to measure the speed of blood moving directly away from the meter. The meter’s sensor detects the pulse reflected back from the blood at a frequency 21 Hz lower than the emitted frequency. Take the speed of sound in the tissues to be 1475 m/s.
At what speed, in centimeters per second, is the blood moving?
Answer:
V = 2.8cm/s
Explanation:
Please see attachment below.
This problem involves the concept of doppler effect.
Answer:
The speed at which the blood is flowing in cm/s = 1.4 cm/s
Explanation:
emitted frequency ( f ) = 1.1 * 10^6 Hz
detected frequency ( F ) = 21 Hz
speed of sound in tissues ( c ) = 1475 m/s
speed ( V ) = ?
To calculate for speed of blood flowing we apply the detected frequency formula :
F = [tex]\frac{2fV}{c}[/tex]
21 = ( 2* 1100000* V ) / 1475
therefore V = (21 * 1475) / (2 * 1100000)
V = 30975 / 2200000
V = 0.0140 m/s = 1.4 cm/sec
A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 18.0 V . Assume that V = 0 at an infinite distance from the sphere.
What is the electric potential at the center of the sphere?
Express your answer with the appropriate units.
Answer:
V_inside = 36 V
Explanation:
Given
We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.
Required
We are asked to calculate the potential at the centre of the sphere
Solution
The potential energy due to the sphere is given by equation
V = (1/4*π*∈o) × (q/r) (1)
Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V
V ∝ 1/r
The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next
V_1/V_2=r_2/r_1
V_inside/V_outside = r/R
V_inside = (r/R)*V_outside (2)
Now we can plug our values for r, R and V_outside into equation (2) to get V_inside
V_inside = (1.2 m )/(0.600)*18
= 36 V
V_inside = 36 V
Answer:
36 V
Explanation:
The solid conducting sphere is a positive charge
and has radius R₁ = 0.6m
at a point R₂ = 1.20 m, the electric potential V = 18.0 V
V, electric potential = K q/R where k = 1/4 πε₀
V is inversely proportional to R
V₁ = electric potential at the center
V₂ = electric potential at 1.2 m
then
V₁ /V₂ = R₂ / R₁
V₁ = V₂ ( R₂ / R₁) = 18.0 V ( 1.2 / 0.6 ) = 36 V
The top of a swimming pool is at ground level. If the pool is 3.00 m deep, how far below ground level does the bottom of the pool appear to be located for the following conditions? (The index of refraction of water is 1.333.)
(a) The pool is completely filled with water.
______m below ground level
(b) The pool is filled halfway with water.
______m below ground level
Answer:
a) 2.25 m
b) 2.625 m
Explanation:
Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.
Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.
In terms of real and apparent depth, the refractive index is given by
η = (real depth)/(apparent depth)
a) Real depth = 3.00 m
Apparent depth = ?
Refractive index, η = 1.333
1.333 = 3/(apparent depth)
Apparent depth = 3/1.3333 = 2.25 m.
Hence the bottom of the pool appears to be 2.25 m below the ground level.
b) Real depth = 1.5 m
Apparent depth = ?
Refractive index, η = 1.333
1.3333 = 1.5/(apparent depth)
Apparent depth = 1.5/1.3333 = 1.125 m
But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.
So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level
The apparent depth of a swimming pool is measured by considering the water's refractive index. With the pool completely filled, the bottom appears to be 2.25m deep. When halfway filled, the pool appears to be 2.625m deep.
Explanation:When light travels from a medium with a high refractive index to one with a lower refractive index, the light is refracted, or bent, making objects appear closer than they actually are. We can calculate this apparent depth by using the formula d' = d / n, where d' is the apparent depth, d is the actual depth, and n is the refractive index.
(a) If the pool is completely filled with water, for a person looking from above ground, the bottom of the pool appears to be closer than it actually is. Substituting the given values into the formula, we get the apparent depth: d' = 3.00m / 1.333 = 2.25 m below ground level.
(b) If the pool is halfway filled with water, the apparent depth of the water is calculated in the same way. However, the depth beneath the water is below the refractive index border and is not subject to refraction. Therefore, the apparent total depth of the partially filled pool is the sum of the actual depth of the air part (1.50m) and the apparent depth of the water part (1.50m / 1.333 = 1.125m). This gives us 2.625m.
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A racquet ball with mass m = 0.238 kg is moving toward the wall at v = 12.4 m/s and at an angle of θ = 31° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.078 s. 1)What is the magnitude of the initial momentum of the racquet ball?
Answer:
[tex]||\vec p || = 2.951\,\frac{kg\cdot m}{s}[/tex]
Explanation:
The initial momentum of the racquet ball is:
[tex]||\vec p || = (0.238\,kg)\cdot (12.4\,\frac{m}{s} )[/tex]
[tex]||\vec p || = 2.951\,\frac{kg\cdot m}{s}[/tex]
A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.
When the beads reach the ends of the rod, the angular speed of the system is approximately 5.76 rad/s.
Explanation:To find the angular speed of the system at the instant the beads reach the ends of the rod, we can use the principle of conservation of angular momentum. Initially, the angular momentum of the system is given by Li = Irodωi + 2(Ibead)ωi, where Irod is the moment of inertia of the rod about its center, Ibead is the moment of inertia of each bead about the center, and ωi is the initial angular speed of the system. When the beads reach the ends of the rod, the moment of inertia of the system changes, but the angular momentum remains constant. So, we have Li = Irodωf + 2(Ibead)ωf, where ωf is the final angular speed of the system. We can rearrange this equation to solve for ωf. Given the values of Irod, Ibead, and ωi, we can substitute them into the equation to find ωf.
Using the given values: mass of the rod = 190 g = 0.19 kg, length of the rod = 43 cm = 0.43 m, mass of each bead = 38 g = 0.038 kg, distance of the beads from the center = 10 cm = 0.1 m, and initial angular speed = 12 rad/s, we can calculate the moments of inertia as follows:
Irod = (1/12)mrodLrod2 = (1/12)(0.19)(0.43)2 = 0.002196 kg.m2
Ibead = mbeadR2 = (0.038)(0.1)2 = 0.000038 kg.m2
Now, substituting the values into the equation Li = Irodωi + 2(Ibead)ωi, we have (0.002196)(12) + 2(0.000038)(12) = (0.002196 + 2(0.000038))ωf. Solving for ωf, we get ωf ≈ 5.76 rad/s.
A 5 meter ladder leans against a wall. The bottom of the ladder is 1 meter from the wall at time t = 0 and slides away from the wall at 1 meters per second. Find the velocity of the top of the ladder at time t = 2.
Answer:
V = –0.89m/s
Explanation:
Please see attachments below.
You are to connect resistors R1 and R2, with R1 > R2, to a battery, first individually, then in series, and then in parallel. Rank those arrangements according to the amount of current through the battery, greatest first.
Answer:
Parallel, R2, R1, series
Explanation:
From Ohm's law, V=IR hence making current the subject of the formula then [tex]I=\frac {V}{R}[/tex]
Since R1>R2 eg 4>2 then current for individual connection of R2 will be greater than that for R1 for example, assume V is 8 then if R2 we will have 8/2=4 A but for R1 we shall have 8/4=2 A.
When in parallel, the equivalent resistance will be given by [tex]\frac {1}{R1}+\frac {1}{R2}[/tex] for example here it will be 1/4+1/2=3/4. Still taking V of 8 then I= 8/(3/4)=10.667 A
When in series connection, equivalent resistance is given by adding R1 and R2 hence using the same figures we shall have 4+2=6 hence I=8/6=1.33A
We can conclude that the arrangement of current from greatest will be parallel, R2, R1, series
According to the amount of current, the arrangement will be "Parallel, R2, R1, Series".
Parallel and Series connectionBy using Ohm's law
→ Voltage (V) = Current (I) × Resistance (R)
or,
→ I = [tex]\frac{V}{R}[/tex]
Since R1 > R2
Let, Current (V) = 8 then the resistance will be:
R2 = [tex]\frac{8}{2}[/tex] = 4 A
and,
R1 = [tex]\frac{8}{2}[/tex] = 4 A
When the resistance is in parallel connection,
= [tex]\frac{1}{R1} + \frac{1}{R2}[/tex]
= [tex]\frac{1}{4} +\frac{1}{2}[/tex]
By taking L.C.M, we get
= [tex]\frac{1+2}{4}[/tex] = [tex]\frac{3}{4}[/tex] and,
The current be:
→ I = [tex]\frac{8}{\frac{3}{4} }[/tex] = 10.667 A
When the resistance is in series connection,
= R1 + R2
= 4 + 2
= 6 and,
The current be:
→ I = [tex]\frac{8}{6}[/tex] = 1.33 A
Thus the response above is appropriate.
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a proton is fired with a speed of 200,000m/s from the midpoint of the capacitor toward the positive plate. This speed is insufficient to reach the positive plate. What is the proton's speed as it collides with the negative plate
Answer: The proton speed = 3 × 10^5m/s
Explanation: The electric P.E change the proton if It can reach the positive plate.
The workdone
I have attached an image of the diagram showing the nature of this motion
Answer:
Protons speed = 2.96 x 10^(5) m/s
Explanation:
A) At closest point of approach to the positive plate, the proton came to rest momentarily.
Thus;
Loss in Kinetic Energy = Gain in Electric potential energy
Hence;
(1/2)(mv^(2)) = eΔV
So, ΔV = (mv^(2))/(2e)
Mass of proton = 1.673 × 10-27 kilograms
Proton elementary charge(e) = 1.6 x 10^(-19) coulumbs
And from the question v = 200,000 m/s
So, ΔV = [1.673 × 10^(-27) x 200000^(2)] / (2 x 1.6 x 10^(-19)) = 209 V
This is less than 250V which is half of the charge at the positive plate shown in the diagram.
Therefore, the speed is insufficient to reach the positive plate from P to Q.
B) Gain in KE = qΔV
Thus; 1/2mvf^(2) - 1/2mvi^(2) = eΔV
Where, vf is final velocity and vi is initial velocity.
So simplifying, we get;
vf^(2) - vi^(2) = (2eΔV)/m
So, vf = √[(2eΔV)/m) + (vi^(2))
= √[(2 x 1.6 x 10^(-19) x 250)/(1.673 × 10^(-27)) + (200,000^(2))
= 2.96 x 10^(5) m/s
Suppose an Atwood machine has a mass of m1 = 6.0 kg and another mass of m2 = 2.0 kg hanging on opposite sides of the pulley. Assume the pulley is massless and frictionless, and the cord is ideal. Determine the magnitude of the acceleration of the two objects and the tension in the cord.
Answer:
Acceleration=[tex]4.9 /s^2[/tex]
Tension=29.4 N
Explanation:
We are given that
[tex]m_1=6 kg[/tex]
[tex]m_2=2 kg[/tex]
We have to find the magnitude of the acceleration of the two objects and the tension in the cord.
Tension, [tex]T=m_1(a+g)[/tex]
[tex]m_2g-T=m_2a[/tex]
Substitute the values
[tex]m_2g-m_1(a+g)=m_2a[/tex]
[tex]m_2g-m_1a-m_1g=m_2a[/tex]
[tex]g(m_2-m_1)=m_2a+m_1a=a(m_1+m_2)[/tex]
[tex]a=\frac{(m_2-m_1)g}{m_1+m_2}[/tex]
Substitute the values
[tex]a=\frac{(2-6)\times 9.8}{2+6}=-4.9m/s^2[/tex]
Where [tex]g=9.8m/s^2[/tex]
Hence, the magnitude of the acceleration of the two objects =[tex]4.9 m/s^2[/tex]
Substitute the values of a
[tex]T=m_1(a+g)=6(-4.9+9.8)=29.4 N[/tex]
A magnetic field has a magnitude of 1.2 \times 10^{-3} T, and an electric field has a magnitude of 4.6 \times 10^{3}N/C. Both fields point in the same direction. A positive 1.8 \mu C charge moves at a speed of 3.1 \times 10^6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.
Answer: F = 113.4.[tex]10^{-3}[/tex]N
Explanation: Net Force is the total forces acting in an object. In this case, there are two forces acting on the charge: one due to magnetic field (Fm) and another due to electric field (Fe). So, net force is
F = Fe + Fm
Force due to electric field
To determine this force:
Fe = q.E, where q is the charge and E is electric field.
Calculating:
Fe = q.E
Fe = 1.8.[tex]10^{-6}[/tex].4.6.[tex]10^{3}[/tex]
Fe = 8.28.[tex]10^{-3}[/tex]N
Force due to magnetic field: It can only happens when the charge is in movement, so
Fm = q.(v×B), where v represents velocity and B is magnetic field
The cross product indicates that force is perpendicular to the velocity and the field.
Calculating:
Fm = q.v.B.senθ
As θ=90°,
Fm = q.v.B
Fm = 1.8.[tex]10^{-6}[/tex].3.1.[tex]10^{6}[/tex].1.2.[tex]10^{-3}[/tex]
Fm = 6.696.[tex]10^{-3}[/tex]N
F, Fm and Fe make a triangle. So, using Pythagorean theorem:
F = [tex]\sqrt{Fe^{2} + Fm^{2} }[/tex]
F = [tex]\sqrt{(8.28.10^{-3} )^{2} +(6.696.10^{-3} )^{2} }[/tex]
F = 113.4.[tex]10^{-3}[/tex]N
The net force acting on the charge is F = 113.4.[tex]10^{-3}[/tex]N
A circular coil that has N = 230 turns and a radius of r = 10.0 cm lies in a magnetic field that has a magnitude of B 0 = 0.0650 T directed perpendicular to the plane of the coil. What is the magnitude of the magnetic flux Φ B through the coil?
Answer:
Φ = 0.469 Wb
Explanation:
Given,
N = 230 turns
Radius, r = 10 cm
Magnetic field, B = 0.0650 T
Magnetic flux = ?
now,
Φ = NBA
Φ = 230 x 0.0650 x π x r²
Φ = 230 x 0.0650 x π x 0.1²
Φ = 0.469 Wb
Hence, the magnetic flux is equal to Φ = 0.469 Wb
The magnitude of the magnetic flux ΦB through the coil is approximately 0.0496 T·m². when A circular coil that has N = 230 turns and a radius of r = 10.0 cm.
Given:
N = 230 turns
r = 10.0 cm = 0.1 m
B₀ = 0.0650 T
The magnetic flux through a coil can be calculated using the formula:
ΦB = B₀ × A × N,
The area of the coil, we can use the formula for the area of a circle:
A = π × r²,
Let's calculate the magnetic flux:
A = π × r² = 3.14159 × (0.1)² = 0.0314159 m²
ΦB = B₀ × A × N = 0.0650 × 0.0314159 × 230 = 0.04958735 T·m²
Therefore, the magnitude of the magnetic flux ΦB through the coil is approximately 0.0496 T·m².
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Use the exact values you enter to make later calculations.You measure the potential difference across a capacitor at different times while it's charging and record the following results.Voltage (V) Time (s)1.4790 0.015843.0000 0.036894.5210 0.066516.0210 0.11700
The final voltage the capacitor reaches after you go get some coffee is 6.500 V.
(a) Determine the time constant from the slope.
Answer:
0.0800 is time constant for slope.
Explanation:
See attached pictures for explanation.
To find the time constant from the slope, use the formula RC = -1/slope, where R is the resistance and C is the capacitance. Calculate the slope by taking the ratio of voltage change to time change between two data points.
Explanation:To determine the time constant from the slope, we can use the formula:
RC = -1/slope
where R is the resistance in the circuit and C is the capacitance of the capacitor.
In this case, since we are only given the voltage and time data, we need to find the slope by taking the ratio of the change in voltage to the change in time between any two points.
Let's take the first and second data points:
Slope = (V2 - V1) / (t2 - t1)
= (3.0000 V - 1.4790 V) / (0.036894 s - 0.015843 s)
Now, calculate the slope:
Slope = (1.5210 V) / (0.021051 s)
Slope ≈ 72.27 V/s
Once we have the slope, we can plug it into the formula RC = -1/slope to find the time constant:
RC = -1 / (72.27 V/s)
Calculate RC:
RC ≈ -0.0138 s/V
So, the time constant (τ) is approximately 0.0138 seconds per volt (s/V). This value represents the product of resistance and capacitance in the circuit.
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In a typical Van de Graaff linear accelerator, protons are accelerated through a potential difference of 20 MV. What is their kinetic energy if they started from rest? Give your answer in (a) eV, (b) keV,(c) MeV, (d) GeV, and (e) joules.
Answer:
a) 2 x10^7 eV
b) 2 x10^4 keV
c) 20 MeV
d) 0.02 Gev
e) 3.2 x 10^-12J
Explanation:
The potential difference = 20 x 10^6 V
The charge on the proton = 1.6 x10^-19
The work done to move the proton will be basically the proton will acquire if it accelerates.
Kinetic energy gained = ΔVq = 20 x10^6 x 1.6 x 10^-19
=3.2 x 10^-12J or 2 x10^7 eV
2 x10^7 eV = 2 x10^4 keV = 20 MeV = 0.02 Gev
Explanation:
Below is an attachment containing the solution.
A 4.9 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F with arrow(x) = (2.6 − x^2) N, , where x is in meters and the initial position of the block is x=0(a) What is the kinetic energy of the block as it passes through x = 2.1 m?
(b) What is the maximum kinetic energy of the block between x = 0 and x = 2.1 m?
Answer with Explanation:
We are given that
Mass of block=m=4.9 kg
Initial velocity, u=0
[tex]F=(2.6-x^2) i N[/tex]
Initial position, x=0
a.We have to find the kinetic energy of the blocks as it passes through x=2.1 m
Work done=Kinetic energy=[tex]\int_{0}^{2.1}(2.6-x^2) dx[/tex]
Kinetic energy of the block=[tex][2.6 x-\frac{x^3}{3}]^{2.1}_{0}[/tex]
Kinetic energy of the block=[tex]2.6\times 2.1-\frac{(2.1)^3}{3}-0=2.373 J[/tex]
Kinetic energy of the block=2.373 J
b.Initial kinetic energy of block=[tex]K_i=\frac{1}{2}(4.9)(0)=0[/tex]
According to work energy theorem
[tex]W=K_f-K_i[/tex]
[tex]2.373 =k_f-0[/tex]
[tex]k_f=2.373 J[/tex]
Hence, the maximum kinetic energy of the block =2.373 J
"A 0.15 kg ball moving at 40 m/s is struck by a bat. The bat reverses the ball's direction and gives it a speed of 50 m/s. What average force does the bat apply to the ball if they are in contact for 6.0 ×10 -3 s?"
Final answer:
The average force exerted by the bat on the ball is 2250 N.
Explanation:
To calculate the average force exerted by the bat on the ball, we can use the impulse-momentum principle. The impulse experienced by the ball is equal to the change in its momentum. We can calculate the initial momentum by multiplying the mass of the ball by its initial velocity, and similarly, we can calculate the final momentum using the mass and final velocity. By subtracting the initial momentum from the final momentum, we get the change in momentum. Finally, dividing the change in momentum by the time of contact gives us the average force.
Using the given values, the initial momentum of the ball is (0.15 kg) × (-40 m/s) = -6 kg·m/s, and the final momentum is (0.15 kg) × (50 m/s) = 7.5 kg·m/s. The change in momentum is 7.5 kg·m/s - (-6 kg·m/s) = 13.5 kg·m/s. Dividing this by the time of contact, 6.0 × 10^-3 s, gives us an average force of 2250 N.
31. Three long, straight, parallel wires all lie in the yz plane and each carries a current of 20 A in the positive z direction. The two outer wires are each 4.0 cm from the center wire. What is the magnitude of the magnetic force on a 50-cm length of either of the outer wires
Final answer:
The magnitude of the magnetic force on a wire can be calculated using the formula F = B * I * L. In this case, the magnetic field created by the two outer wires can be calculated using Ampere's Law. Plugging in the values will give you the magnitude of the magnetic force on the wire.
Explanation:
The magnitude of the magnetic force on a wire can be calculated using the formula:
F = B * I * L
Where F is the magnetic force, B is the magnetic field, I is the current, and L is the length of the wire.
In this case, the magnetic field created by the two outer wires can be calculated using Ampere's Law:
B = (μ0 * I) / (2π * r)
Where μ0 is the permeability of free space, I is the current, and r is the distance from the wire.
Plugging in the values, the magnetic force per meter on either of the outer wires is:
F = (2 * 10-7 Tm/A * 20 A * 0.5 m) / (2π * 0.04 m)
Calculating this will give you the magnitude of the magnetic force on the wire.
A spring with a force constant of 5400 N/m and a rest length of 3.5 m is used in a catapult. When compressed to 1.0 m, it is used to launch a 48 kg rock. However, there is an error in the release mechanism, so the rock gets launched almost straight up. How high does it go (in m)? (Assume the rock is launched from ground height.) m
Answer:
5.51 m
Explanation:
From the question,
The energy used to stretch the spring = the potential energy of the rock.
(1/2)ke² = mgh ................. Equation 1
Where k = spring constant, e = extension/compression, m = mass of the rock, g = acceleration due to gravity, h = height of the rock above the ground
make h the subject of the equation.
h = ke²/2mg ....................equation 2
Given: k = 5400 N/m, e = 1 m, m = 48 kg.
Constant: g = 9.8 m/s²
Substitute into equation 2
h = 5400(1²)/(2×48×9.8)
h = 5400/940.8
h = 5.51 m.
Hence the height of the rock = 5.51 m
You create a plot of voltage (in V) vs. time (in s) for an RC circuit as the capacitor is charging, where V=V_{0} \cdot \left(1- e^{ \frac{-\left(t\right)}{RC} } \right). You curve fit the data using the inverse exponent function Y=A \cdot \left(1- e^{-\left(Cx\right)} \right)+B and LoggerPro gives the following values for A, B, and C. A = 4.211 ± 0.4211 B = 0.1699 ± 0.007211 C = 1.901 ± 0.2051 What is the time constant for and its uncertainty?
Answer:
The time constant and its uncertainty is t ± Δt = 0.526 ± 0.057 s
Explanation:
If we make a comparison we have to:
y = A*(1-e^-(C*x)) + B
If the time remains constant we have to:
t = R*C = 1/C
In this way we calculate the time constant and its uncertainty. this will be equal to:
t ± Δt = (1/1.901) ± (0.2051/1.901)*(1/1.901) = 0.526 ± 0.057 s
Three charged particles form a triangle: particle 1 with charge Q1 = 80.0 nC is at xy coordinates (0, 3.00 mm), particle 2 with charge Q2 is at (0, −3.00 mm), and particle 3 with charge q = 18.0 nC is at (4.00 mm, 0). In unit-vector notation, what is the electrostatic force on particle 3 due to the other two particles if Q2 is equal to (a) 80.0 nC and (b) −80.0 nC?
The net electrostatic force on particle 3 due to the other two particles depends on the charges of these particles. When Q2 = 80.0 nC, the forces from the other two particles cancel out and when Q2 = -80.0 nC, the forces add up.
Explanation:The electrostatic force on a charge due to other charges can be determined using Coulomb’s law. For the setup in the question, the deletion force on particle 3 because of particles 1 and 2 can be obtained by vectorially adding the forces it experiences due to each of these particles separately. When Q2 = -80.0 nC, the forces that particle 3 experiences due to particle 1 and particle 2 are in the same direction this time, therefore they add up to give the net force on particle 3. We can determine the exact value by substituting the given values in the equation for Coulomb's Law.
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If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 56.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answer:
Change in momentum =15.01kgm/s
Impulse applied by bat = 15.01Ns
Explanation:
Please see attachment below.
When a car drives through the Earth's magnetic field, an emf is induced in its vertical 60-cm-long radio antenna. Part A If the Earth's field (5.0×10−5 T) points north with a dip angle of 38∘, which direction(s) will the car be moving to produce the maximum emf induced in the antenna?
Answer:
Explanation:
The magnetic field is in north south direction . In order to cut lines of forces to the maximum extent , car has to move in east- west direction ie towards east or towards west to produce maximum emf.
emf produced = B L V
where B is horizontal component of earth magnetic field
L is length of rod
v is velocity of car carrying antenna rod .
The car should be moving east or west to produce the maximum emf induced in the antenna.
Explanation:When a car drives through the Earth's magnetic field, the maximum emf induced in its vertical radio antenna occurs when the car is moving in a direction perpendicular to the magnetic field. In this case, the Earth's magnetic field points north with a dip angle of 38°. To produce the maximum emf, the car should be moving east or west.
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A current of 8 A exists in a copper (Cu) wire which has a diameter of 5 mm. What is the current density? Each atom of copper contributes one conduction electron, and the average thermal speed r k T m of an electron is 2.4 × 106 m/s . The mass
Explanation:
Below is an attachment containing the solution.
A 0.60-kg particle has a speed of 2.0 m/s at point A and a kinetic energy of 7.5 J at point B. What is (a) its kinetic energy at A? (b) Its speed at point B ? (c) The total work done on the particle as it moves from A to B ?
Answer:
Explanation:
note:
solution is attached due to error in mathematical equation. please find the attachment
Answer:
(a) 1.2 J
(b) 5 m/s
(c) 6.3 J
Explanation:
(a) Kinetic energy at A
Ek = 1/2mv².................. Equation 1
Where Ek = Kinetic energy at A, m = mass of the particle, v = velocity at A.
Given: m = 0.6 kg, v = 2.0 m/s
Substitute into equation 1
Ek = 1/2(0.6)(2²)
Ek = 0.6(2)
Ek = 1.2 J.
(b)
Speed at point B
Ek' = 1/2mv'²............... Equation 2
Make v' the subject of the equation,
v' = √(2Ek'/m).................. Equation 3
Where, Ek' = kinetic energy at B, v' = velocity at B.
Given: Ek' = 7.5 J, m = 0.6 kg,
Substitute into equation 3
v' = √[(2×7.5)/0.6]
v' =√(15/0.6)
v' = √25
v' = 5 m/s.
(c)
Wt = Δ kinetic energy from A to B
Where Wt = total work done as the particle moves from A to B.
Wt = 1/2m(v'²-v²)
Wt = 1/2(0.6)(5²-2²)
Wt = 0.3(25-4)
Wt = 0.3(21)
Wt = 6.3 J
A typical wall outlet voltage in the United States is 120 volts. Personal MP3 players require much smaller voltages, typically 487.0 mV 487.0 mV . If the number of turns in the primary coil is 2464 2464 , calculate the number of turns on the secondary coil of the adapter transformer.
Answer:
Number of turns on the secondary coil of the adapter transformer is 10.
Explanation:
For a transformer,
[tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]
where [tex]V_{s}[/tex] is the voltage induced in the secondary coil
[tex]V_{p}[/tex] is the voltage in the primary coil
[tex]N_{s}[/tex] is the number of turns of secondary coil
[tex]N_{p}[/tex] is the number of turns of primary coil
From the given question,
[tex]\frac{487*10^{-3} }{120}[/tex] = [tex]\frac{N_{s} }{2464}[/tex]
⇒ [tex]N_{s}[/tex] = [tex]\frac{2462*487*10^{-3} }{120}[/tex]
= 9.999733
∴ [tex]N_{s}[/tex] = 10 turns
Answer:
607,145 turns
Explanation:
Output voltage, that is secondary voltage,Es = 120 volts
Input voltage, that is primary voltage, Ep = 487/1000 = 0.487 volts
Number of turns in secondary = Ns
Number of turns in primary, Np = 2464
∴ Es/Ep = Ns/Np
Ns = Es * Np/Ep = 1`20 X 2464/0.487 = 607,145 turns ( Step up transformer)
a carbon steel ball with a 30mm diameter is pressed against a flat carbon steel plate with a force of 20n. calculate the diameter of the circular contact area and the maximum pressure that occurs at the center of the contact area
Final answer:
The diameter of the circular contact area is 30 mm and the maximum pressure at the center of the contact area is approximately 0.14 N/mm^2.
Explanation:
To calculate the diameter of the circular contact area, we need to determine the radius of the contact area first. The radius can be found by dividing the diameter of the ball by 2. In this case, the radius is 15 mm.
The area of the circular contact area can be calculated using the formula A = πr^2, where A is the area and r is the radius. The maximum pressure at the center of the contact area can be found by dividing the force applied on the plate by the area of the contact area.
Using these formulas, the diameter of the circular contact area is 30 mm and the maximum pressure at the center of the contact area is approximately 0.14 N/mm^2.
An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will be
A) three times as large as the initial value.
B) less than three times as large as the initial value.
C) more than three times as large as the initial value.
D) equal to the initial value.
Answer:
A three times as large as the initial value