Explanation:
It is given that the number of electrons passing through the cross-sectional area in 1 s is [tex]3.4 \times 10^{18}[/tex]. Also, we know that charge on an electron is [tex]-1.60 \times 10^{-19} C[/tex], then negative charge crossing to the left per second is as follows.
I- = [tex]3.4 \times 10^{18} electrons \times -1.6 x 10^{-19} C/electrons[/tex]
I- = 0.544 A
As it is given that the number of protons crossing per second is [tex]1.4 \times 10^{18}[/tex], as the charge on the proton is [tex]+1.60 \times 10^{-19} C[/tex], then positive charge crossing to the right per second is calculated as follows.
I+ = [tex]1.4 \times 10^{18} electrons \times 1.6 \times 10^{-19} electrons/C[/tex]
I+ = 0.224 A
I = l I+ l + l I- l
So, I = 0.544 + 0.224
= 0.768 A
Thus, we can conclude that the current in given hydrogen discharge tube is 0.768 A.
A steel ball is whirled on the end of a chain in a horizontal circle of radius R with a constant period T. If the radius of the circle is then reduced to 0.75R, while the period remains T, what happens to the centripetal acceleration of the ball?
When the radius of the circle is reduced while the period remains constant, the centripetal acceleration of the ball increases.
Explanation:When the radius of the circle is reduced to 0.75R while the period remains T, the centripetal acceleration of the ball increases.
Centripetal acceleration is given by the formula ac = v^2 / r, where v is the tangential velocity and r is the radius of the circle. Since the period T remains constant and the radius is reduced, the velocity of the ball must increase in order to cover the smaller circumference in the same amount of time, resulting in an increase in centripetal acceleration.
A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.
Answer:
(a) I_A=1/12ML²
(b) I_B=1/3ML²
Explanation:
We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².
(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².
(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:
[tex]d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L[/tex]
Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:
[tex]x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L[/tex]
Finally, using the Parallel Axis Theorem, we calculate I_B:
[tex]I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}[/tex]
A) Moment of inertia about an axis passing through the point where the two segments meet : [tex]I_{A} = \frac{1}{12} ML^{2}[/tex]
B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends : [tex]Ix_{} = \frac{1}{3} ML^{2}[/tex]
A) The moment of inertia about an axis passing through the point where the two segments meet is [tex]I_{A} = \frac{1}{12} ML^{2}[/tex] given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.
B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends
First step: determine the distance between the ends ( d )
After applying Pythagoras theorem
d = [tex]\frac{\sqrt{2} }{2} L[/tex]
Next step : determine distance between the two axis ( x )
After applying Pythagoras theorem
x = [tex]\frac{\sqrt{2} }{4} L[/tex]
Final step : Calculate the value of Iₓ
applying Parallel Axis Theorem
Iₓ = Iₐ + Mx²
= [tex]\frac{1}{12} ML^{2}[/tex] + [tex]\frac{1}{4} ML^{2}[/tex]
∴ [tex]Ix_{} = \frac{1}{3} ML^{2}[/tex]
Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet : [tex]I_{A} = \frac{1}{12} ML^{2}[/tex], Moment of inertia passing through the point where the midpoint of the line connects its two ends : [tex]Ix_{} = \frac{1}{3} ML^{2}[/tex]
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In the rough approximation that the density of a planet is uniform throughout its interior, the gravitational field strength (force per unit mass) inside the planet at a distance from the center is , where is the radius of the planet. (For the Earth, at least, this is only a rough approximation, because the outer layers of rock have lower density than the inner core of molten iron). 1. Using the uniform-density approximation, calculate the amount of energy required to move a mass m from the center of the Earth to the surface. 2. For comparison, how much energy would be required to move the mass from the surface of the planet to a very large distance away? 3. Imagine that a small hole is drilled through the center of the Earth from one side to the other. Determine the speed of an object of mass m, dropped into this hole, when it reaches the center of the planet.
Answer:
The answers to the questions are;
1. The amount of energy required to move a mass m from the center of the Earth to the surface is 0.5·m·g·R
2. The amount of energy required to move the mass from the surface of the planet to a very large distance away m·g·R.
3. The speed of an object of mass m, dropped into this hole, when it reaches the center of the planet is 9682.41783 m/s.
Explanation:
We note that the Work done W by the force F on the mass to move a small distance is given by
F×dr
The sum of such work to move the body to a required location is
W =[tex]\int\limits {F} \, dr[/tex]
F = mg' = m[tex]\frac{gr}{R}[/tex]
We integrate from 0 to R (the center to the Earth surface)
Therefore W = [tex]\int\limits^R_0 {m\frac{gr}{R}} \, dr[/tex]
Which gives W = [tex]\frac{mg}{R} [\frac{r^2}{2} ]^R_0 = \frac{1}{2}mgR[/tex]
2. To find the work done we have to integrate from the surface to infinity as follows W = [tex]\int\limits^{inf}_R {m\frac{gr}{R}} \, dr[/tex] = [tex]\frac{mg}{R} [\frac{r^2}{2} ]^{inf}_R = mgR[/tex]
The energy required to move the object to a large distance is equal to twice the energy reqired to move the object to the surface.
3 We note that the acceleration due to gravity at the surface is g and reduces to zero at the center of the Earth
v² = u² + 2·g·s
Radius of the Earth = 6371 km
From surface to half radius we have
v₁² = 2×9.81×6371/2×1000 = 62499460.04
v₁ = 7905.66 m/s
From the half the radius of the earth to the Earth center =
v₂² = 7905.66² + 2×9.81/2×6371/2×1000 = 93749215.04
v₂ = 9682.41783 m/s
The speed of an object of mass m, dropped into this hole, when it reaches the center of the planet. is 9682.41783 m/s
1. Energy to Move from Center to Surface: Using the uniform-density approximation, the energy required to move a mass m from Earth's center to the surface is [tex]\(-\frac{GMm}{R}\)[/tex]. 2. Energy to Move from Surface to Infinity: Moving the mass from the surface to infinity requires zero energy. 3. Speed at Earth's Center: The speed of an object dropped through a hole from the surface to the Earth's center is [tex]\(\sqrt{\frac{2GM}{R}}\)[/tex].
1. Energy to Move Mass from Center to Surface:
The gravitational potential energy U is given by:
[tex]\[ U = -\frac{GMm}{r} \][/tex]
where:
- G is the gravitational constant,
- M is the mass of the planet,
- m is the mass being moved,
- r is the distance from the center.
For this scenario, r is the radius of the planet R.
[tex]\[ U_{\text{center to surface}} = -\frac{GMm}{R} \][/tex]
2. Energy to Move Mass from Surface to Infinity:
When moving the mass to a very large distance away [tex](\( \infty \))[/tex], the potential energy becomes zero.
[tex]\[ U_{\text{surface to infinity}} = 0 \][/tex]
3. Speed of Object Dropped Through Earth's Center:[tex]\[ U_{\text{surface}} = K_{\text{center}} \]\[ -\frac{GMm}{R} = \frac{1}{2}mv^2 \][/tex]converted into kinetic energy at the center.
Solve for v:
[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]
Given:
- M is the mass of the Earth,
- m is the mass being moved.
These expressions provide the required energy calculations and the speed of the object dropped through the Earth's center.
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A 25-kg iron block initially at 350oC is quenched in an insulated tank that contains 100 kg of water at 18oC. Assuming the water that vaporizes during the process condenses back in the tank, determine the total entropy change during this process.
Answer:
The value of total entropy change during the process
[tex]dS = 0.608 \frac{KJ}{K}[/tex]
Explanation:
mass of iron [tex]m_{iron}[/tex] = 25 kg
Initial temperature of iron [tex]T_{1}[/tex] = 350°c = 623 K
Mass of water [tex]m_{w}[/tex] = 100 kg
Initial temperature of water [tex]T_{2}[/tex] = 180°c = 453 K
When iron block is quenched inside the water the final temperature of both iron & water becomes equal. this is = [tex]T_{f}[/tex]
Thus heat lost by the iron block = heat gain by the water
⇒ [tex]m_{iron}[/tex] [tex]C_{iron}[/tex] ( [tex]T_{1}[/tex] - [tex]T_{f}[/tex] ) = [tex]m_{w}[/tex] [tex]C_{w}[/tex] ( [tex]T_{f}[/tex] - [tex]T_{2}[/tex] )
⇒ 25 × 0.448 × ( [tex]T_{1}[/tex] - [tex]T_{f}[/tex] ) = 100 × 4.2 × ( [tex]T_{f}[/tex] - [tex]T_{2}[/tex] )
⇒ [tex]( T_{1} - T_{f} ) = 37.5 ( T_{f} - T_{2} )[/tex]
⇒ [tex]( 623 - T_{f} ) = 37.5 ( T_{f} - 453 )[/tex]
⇒ [tex]( 623 - T_{f} ) = 37.5 T_{f} - 16987.5[/tex]
⇒ [tex]38.5 T_{f} = 17610.5[/tex]
⇒ [tex]T_{f} = 457.41 K[/tex]
This is the final temperature after quenching.
The total entropy change is given by,
[tex]dS = m_{iron}\ C_{iron} \ ln \frac{T_{f} }{T_{1} } + m_{w}\ C_{w} \ ln \frac{T_{f} }{T_{2} }[/tex]
Put all the values in above formula,
[tex]dS =[/tex] 25 × 0.448 × [tex]ln \frac{457.41}{623}[/tex] + 100 × 4.2 × [tex]ln \frac {457.41}{453}[/tex]
[tex]dS =[/tex] - 3.46 + 4.06
[tex]dS = 0.608 \frac{KJ}{K}[/tex]
This is the value of total entropy change.
. A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall
Explanation:
Below is an attachment containing the solution.
With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you release the ball 1.4 mm above the ground? Solve this problem using energy.
Answer:
The initial velocity is 0.5114 m/s or 511.4 mm/s
Explanation:
Let the initial velocity be 'v'.
Given:
Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]
Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]
Final height of the ball (h₂) = 15 mm = 0.015 m
Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.
Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.
Change in kinetic energy is given as:
[tex]\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity[/tex]
As it just touches the 15 mm high roof, the final velocity will be zero. So,
[tex]v_f=0\ m/s[/tex].
Now, the change in kinetic energy is equal to:
[tex]\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2[/tex]
Change in gravitational potential energy = Final PE - Initial PE
So,
[tex]\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J[/tex] [ g = 9.8 m/s²]
Now, Change in KE = Change in PE
[tex]0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s[/tex]
Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s
An airplane needs to reach a velocity of 199.0 km/h to take off. On a 2000-m runway, what is the minimum acceleration necessary for the plane to take flight? Assume the plane begins at rest at one end of the runway.
Answer:
[tex]0.76m/s^2[/tex]
Explanation:
We are given that
Final velocity, v=199 km/h=[tex]199\times \frac{5}{18}=55.3m/s[/tex]
1km/h=[tex]\frac{5}{18}m/s[/tex]
Initial velocity, u=0
S=2000 m
We know that
[tex]v^2-u^2=2as[/tex]
Using the formula
[tex](55.3)^2=2a(2000)[/tex]
[tex]a=\frac{(55.3)^2}{2\times 2000}[/tex]
[tex]a=0.76m/s^2[/tex]
Hence, the minimum acceleration necessary for the plane to take flight=[tex]0.76m/s^2[/tex]
Explain the differences between the geocentric theory of the universe and the heliocentric theory
Geocentric Theory
In astronomy, the geocentric model is a superseded description of the Universe with Earth at the center. Under the geocentric model, the Sun, Moon, stars, and planets all orbited Earth
Heliocentric Theory
Heliocentrism is the astronomical model in which the Earth and planets revolve around the Sun at the center of the Solar System. Historically, heliocentrism was opposed to geocentrism, which placed the Earth at the center
G-Theory is the earth is the center of the universe.
H-Theory is the sun is the center of the universe.
When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is 4.00 atm, is opened, what will be the final pressure in the two bulbs
Answer:
[tex]P_{C} = 3.2\, atm[/tex]
Explanation:
Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:
Bulb A (2 L, 2 atm) - Before opening:
[tex]P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T[/tex]
Bulb B (3 L, 4 atm) - Before opening:
[tex]P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T[/tex]
Bulbs A & B (5 L) - After opening:
[tex]P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T[/tex]
After some algebraic manipulation, a formula for final pressure is derived:
[tex]P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}[/tex]
And final pressure is obtained:
[tex]P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}[/tex]
[tex]P_{C} = 3.2\, atm[/tex]
How fast (in rpm) must a centrifuge rotate if a particle 7.50 cm from the axis of rotation is to experience an acceleration of 119000 g's? If the answer has 4 digits or more, enter it without commas, e.g. 13500.
Explanation:
Below is an attachment containing the solution.
Suppose you move along a wire at the same speed as the drift speed of the electrons in the wire. Do you now measure a magnetic field of zero?
Answer:
False. Field is non-zero
Explanation:
If you were moving along with the electrons, they would appear stationary to you. You would measure a current of zero. However, the fixed positive charges in the wire seem to move backwards relative to you, creating the equivalent current as if you weren't moving. You would measure the same field, but the field would be caused by the 'backward' motion of positive particles.
Moving at the same speed as the drift speed of electrons does not result in a zero magnetic field. The movement of electrons in a wire creates a magnetic field, and this field would still be present even if you were moving at the same speed as the electrons.
Explanation:When moving along a wire at the same speed as the drift speed of the electrons, you will not measure a magnetic field of zero. The drift speed of electrons refers to the average velocity at which the electrons move in a conductor when an electric field is applied. This speed is generally very slow, but it does not mean that there is no magnetic field.
The movement of electrons in a wire creates a magnetic field around the wire, even if the drift speed is small. This is because the electric current generated by the movement of electrons is what produces the magnetic field.
So, even if you were to move at the same speed as the drift speed of the electrons in the wire, you would still measure a non-zero magnetic field because the movement of electrons in the wire is what generates the magnetic field.
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An elevator cab that weighs 27.8 kN moves upward. What is the tension in the cable if the cab's speed is (a) increasing at a rate of 2 1.22 / msand (b) decreasing at a rate of 2 1.22 / ms?
Answer:
a)T = 8.63 × 10 ⁴ N, b)T = -3.239 × 10 ⁴ N
Explanation:
Given:
W = 27.8 KN = 27.8 × 10 ³ N,
For upward motion: Fnet is upward, Tension T is upward and weight W is downward so
a) a=21.22 m/s² ( not written clearly the unit. if it is acceleration?) then
Fnet = T - W
⇒ T = F + W = ma + W
T = (W/g)a + W (W=mg ⇒m=W/g)
T = (27.8 × 10 ³ N / 9.8 ) 21.22 m/s² + 27.8 × 10 ³ N
T = 86,263.265 N
T = 8.63 × 10 ⁴ N
b) For Declaration in upward direction a = -21.22 m/s²
Fnet = T - W
⇒ T = F + W = ma + W
T = (W/g)a + W
T = (27.8 × 10 ³ N / 9.8 ) (-21.22 m/s²) + 27.8 × 10 ³ N
T = -32395.5 N
T = -3.239 × 10 ⁴ N
as Tension can not be negative I hope the value of acceleration and deceleration is correct.
1. At the synaptic terminal, voltage-gated ______________ channels open, thereby stimulating the synaptic vesicles to release their neurotransmitters by exocytosis.
Ion
Explanation:
At the synaptic terminal, voltage-gated ion channels open, thereby stimulating the synaptic vesicles to release the neurotransmitters by exocytosis.
These ion channels are the signaling molecules in neurons. They are the transmembrane proteins that form ion channels. The membrane potential changes the conformation of the channel proteins that regulates their opening and closing. These channels play an important role in neurotransmitter release in presynaptic nerve endings.
For example - Ca²⁺ gated ion channel.
Suppose you watch a leaf bobbing up and down as ripples pass it by in a pond. You notice that it does two full up and down bobs each second. What is true of the ripples on the pond?
Answer:
The frequency of the ripples is 2Hz, and their period is 0.5 seconds.
Explanation:
Since the ripples on the pond are making the leaf oscillate up and down at a rate of two times per second, we can calculate the period T and the frequency f of the ripples on the pon[tex]T=\frac{1}{0.5Hz} =[/tex]d.
The frequency, by definition, is the number of waves per unit of time. In this case, we have two waves per second, so the frequency is 2s⁻¹, or 2Hz.
The period is the inverse of the frequency, so
[tex]T=\frac{1}{2Hz} =0.5s[/tex]
Then, the period is equal to 0.5 seconds.
The New England Merchants Bank Building in Boston is 152 mm high. On windy days it sways with a frequency of 0.20 HzHz , and the acceleration of the top of the building can reach 2.5 %% of the free-fall acceleration, enough to cause discomfort for occupants.What is the total distance, side to side, that the top of the building moves during such an oscillation?
Answer:
The total distance, side to side, that the top of the building moves during such an oscillation = 31 cm
Explanation:
Let the total side to side motion be 2A. Where A is maximum acceleration.
Now, we know know that equation for maximum acceleration is;
A = α(max) / [(2πf)^(2)]
So 2A = 2[α(max) / [(2πf)^(2)] ]
α(max) = (0.025 x 9.81) while frequency(f) from the question is 0.2Hz.
Therefore 2A = 2 [(0.025 x 9.81) / [((2π(0.2)) ^(2)] ] = 2( 0.245 / 1.58) = 0.31m or 31cm
When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb
Answer:
343/1500
Explanation:
Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).
From the question,
P' = mg×v................. Equation 1
Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.
Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²
Substitute into equation 1
P' = 700(2.5)(9.8)
P' = 17150 W.
If the full power generated by the engine = 75000 W
The fraction of the engine power used to make the climb = 17150/75000
= 343/1500
What happens to the direction of the magnetic field about an electric current when the direction of the current is reversed?
Answer:
The direction of the magnetic field is also reversed.
Explanation:
The direction of the magnetic field is also reversed when viewed form the same side if the direction of current is reversed. The direction of the magnetic field with respect to the direction of electric current is determined by the Maxwell's right-hand thumb rule.According to this rule we place our palm with the thumb pointing the direction of current flow and curling our finger in an action of gripping the wire. This position the the direction of curled fingers represents the direction of magnetic field.A student with a mass of 80.0 kg runs up three flights of stairs in 12.0 sec. The student has gone a vertical distance of 8.0 m. Determine the amount of work done by the student to elevate his body to this height. Determine the power consumed by the student. Assume that his speed is constant.
Explanation:
Work = Force x Displacement
Force = Weight of student
Weight = Mass x Acceleration due to gravity
Mass, m = 80 kg
Acceleration due to gravity, g = 9.81 m/s²
Weight = 80 x 9.81 = 784.8 N
Displacement = 8 m
Work = 784.8 x 8 = 6278.4 J
The amount of work done by the student to elevate his body to this height is 6278.4 J
Power is the ratio of work to time taken
[tex]P=\frac{W}{t}\\\\P=\frac{6278.4}{12}\\\\P=523.2W[/tex]
Power is 523.2 Watts
The power consumed by the student is 523.2 watts.
The calculation is as follows:[tex]Work = Force \times Displacement[/tex]
Force = Weight of student
[tex]Weight = Mass \times Acceleration\ due\ to\ gravity[/tex]
Mass, m = 80 kg
Acceleration due to gravity, g = 9.81 m/s²
[tex]Weight = 80 \times 9.81 = 784.8 N[/tex]
Displacement = 8 m
[tex]Work = 784.8 \times 8 = 6278.4 J[/tex]
So,
Power consumed should be
[tex]= 6278.4 \div 12[/tex]
= 523.2
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The Nichrome wire is replaced by a wire of the same length and diameter, and same mobile electron density but with electron mobility 4 times as large as that of Nichrome. Now what is the electric field inside the wire?
Answer:
The electric field inside the wire will remain the same or constant, while the drift velocity will by a factor of four.
Explanation:
Electron mobility, μ = [tex]\frac{v_d}{E}[/tex]
where
[tex]v_d[/tex] = Drift velocity
E = Electric field
Given that the electric field strength = 1.48 V/m,
Therefore since the electric potential depends on the length of the wire and the attached potential difference, then when the electron mobility is increased 4 times the Electric field E will be the same but the drift velocity will increase four times. That is
4·μ = [tex]\frac{4*v_d}{E}[/tex]
Answer:
Explanation:
Usually, the electron drift velocity in a material is directly proportional to the electric field, which means that the electron mobility is a constant (independent of electric field).
μ × E = Vd
Where,
D = electric field
Vd = drift velocity
μ = electron mobility
μ2 = μ × 4
μ/Vd1 = 4 × μ/Vd2
Vd2 = 4 × Vd1
The electric field is the same but drift velocity increases.
Kirchhoff’s loop rule for circuit analysis is an expression of__________?
Answer:
Conservation of energy
Explanation:
Gustav Kirchhoff, a German physicist proposed a rule called Kirchhoff's loop rule also known as Kirchhoff's second rule which states that for any closed loop, the potential difference including the voltage supply is zero. That means energy supplied by a voltage source must absorbed for balance by other components in the loop
This fulfils the law of conservation of energy
A centrifuge at a museum is used to separate seeds of different sizes. The average rotational acceleration of the centrifuge according to a sign is 30 rad/s2rad/s2. Part A If starting at rest, what is the rotational velocity of the centrifuge after 10 ss?
Answer:
[tex]\omega = 300 rad/s[/tex]
Explanation:
Given,
rotational acceleration = 30 rad/s²
initial angular speed = 0 m/s
time, t = 10 s
Final angular speed = ?
Using equation of rotation motion
[tex]\omega = \omega_o + \alpha t[/tex]
[tex]\omega =0+30\times 10[/tex]
[tex]\omega = 300 rad/s[/tex]
Rotational velocity after 10 s = 300 rad/s.
An object of mass 0.77 kg is initially at rest. When a force acts on it for 2.9 ms it acquires a speed of 16.2 m/s. Find the magnitude (in N) of the average force acting on the object during the 2.9 ms time interval. (Enter a number.)
Answer: 4.3KN
Explanation:f=m(v-u)/t
m=0.77kg
t=0.0029s
s=16.2m/s
F= 0.77*16.2/0.0029
F=12.474/0.0029
F= 4301.38N
F=4.3KN
To find the magnitude of the average force acting on an object, one can use the derived form of the second law of motion, F = mΔv/Δt. Placing the given values into the equation, we calculate the force to be approximately 4307 N.
Explanation:To solve this question, we can use the equation F = mΔv/Δt which is derived from the second law of motion (Force = mass × acceleration), where F is the average force, m is the mass of the object, Δv is the change in velocity, and Δt is the time interval.
Substituting the given values, m = 0.77 kg, Δv = 16.2 m/s (the final velocity) - 0 m/s (the initial velocity) = 16.2 m/s and Δt = 2.9 ms = 2.9 × 10⁻³ s (as time should be converted to seconds).
Therefore, F = (0.77 kg × 16.2 m/s) / 2.9 × 10⁻³ s = 4306.9 N. Therefore, the magnitude of the average force acting on the object during the 2.9 ms time interval is approximately 4307 N.
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Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 5.5 mi2/hr. How rapidly is radius of the spill increasing when the area is 6 mi2?
Explanation:
The area of a circle can be represented by A = π r² I
Differentiating both sides w.r.t time
[tex]\frac{dA}{dt}[/tex] = 2π r [tex]\frac{dr}{dt}[/tex] II
Dividing II by I , we have
[tex]\frac{dA}{A}[/tex] = 2 x [tex]\frac{dr}{r}[/tex]
substituting the values
[tex]\frac{dr}{r}[/tex] = [tex]\frac{5.5}{12}[/tex] = 0.46 mi per unit radius
or dr = 1.4 x 0.46 = 0.64 mi/hr
here 1.4 mi is the radius , when area of circle is 6 mi²
A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 2.0 cm. What will be the compression if the same block collides with the spring at a speed of 2v?
Answer:
4.0 cm
Explanation:
For the compression of the spring, the kinetic energy of the mass equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² ⇒ x = (√m/k)v
Since m and k are constant since its the same spring x ∝ v
If our speed is now v₁ = 2v, our compression is x₁
x₁ = (√m/k)v₁ = (√m/k)2v = 2(√m/k)v = 2x
x₁ = 2x
Since x = 2.0 cm, our compression for speed = 2v is
x₁ = 2(2.0) = 4.0 cm
If the same block collides with the spring at a speed of 2v, the compression will be 4.0cm.
Given the data in the question;
Compression; [tex]x_1 = 2.0cm[/tex]Velocity 1; [tex]v_1 = v[/tex]Velocity 2; [tex]v_2 = 2v[/tex]Using conservation of energy:
Kinetic energy of the mass = Elastic potential energy of the spring
We have:
[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2\\\\kx^2 = mv^2[/tex]
"v" is directly proportional to "x"
Hence,
[tex]\frac{x_1}{x_2} = \frac{v_1}{v_2}[/tex]
We substitute in our given values
[tex]\frac{2.0cm}{x_2} = \frac{v}{2v}\\\\x_2 = \frac{v(2.0cm*2)}{v} \\\\x_2 = (2.0cm*2)\\\\x_2 = 4.0cm[/tex]
Therefore, if the same block collides with the spring at a speed of 2v, the compression will be 4.0cm.
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A 497−g piece of copper tubing is heated to 89.5°C and placed in an insulated vessel containing 159 g of water at 22.8°C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/°C, what is the final temperature of the system (c of copper = 0.387 J/g·°C)?
Final answer:
To calculate the initial temperature of the copper piece, use the principle of energy conservation and the equations for heat gained and heat lost. The final temperature of the system is approximately 24.8 °C.
Explanation:
To calculate the initial temperature of the copper piece, we can use the principle of energy conservation. The heat gained by the water is equal to the heat lost by the copper. The heat gained by the water can be calculated using the formula:
Q = m * c * ΔT
Where Q is the heat gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The heat lost by the copper can be calculated using the formula:
Q = m * c * ΔT
Where Q is the heat lost, m is the mass of the copper, c is the specific heat capacity of copper, and ΔT is the change in temperature. Since the final temperature is known, we can rearrange the formulas to solve for the initial temperature of the copper:
Initial temperature of copper = (heat gained by water / (m * c)) + final temperature
Substituting the given values into the formulas, we get:
Heat gained by water = (159 g) * (4.18 J/g·°C) * (39.9 °C - 22.8 °C)
Heat lost by copper = (497 g) * (0.387 J/g·°C) * (final temperature - 89.5 °C)
Setting the two equations equal, we can solve for the final temperature:
(159 g) * (4.18 J/g·°C) * (39.9 °C - 22.8 °C) = (497 g) * (0.387 J/g·°C) * (final temperature - 89.5 °C)
Solving the equation, the final temperature of the system is approximately 24.8 °C.
A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C exists along the +x axis. A magnetic field also exists, and its x and y components are Bx = +1.9 T and By = +1.9 T. Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is (a) stationary, (b) moving along the +x axis at a speed of 345 m/s, and (c) moving along the +z axis at a speed of 345 m/s.
a)
[tex]F_{E_x}=1.19\cdot 10^{-3}N[/tex] (+x axis)
[tex]F_{B_x}=0[/tex]
[tex]F_{B_y}=0[/tex]
b)
[tex]F_{E_x}=1.19\cdot 10^{-3} N[/tex] (+x axis)
[tex]F_{B_x}=0[/tex]
[tex]F_{B_y}=3.21\cdot 10^{-3}N[/tex] (+z axis)
c)
[tex]F_{E_x}=1.19\cdot 10^{-3} N[/tex] (+x axis)
[tex]F_{B_x}=3.21\cdot 10^{-3} N[/tex] (+y axis)
[tex]F_{B_y}=3.21\cdot 10^{-3}N[/tex] (-x axis)
Explanation:
a)
The electric force exerted on a charged particle is given by
[tex]F=qE[/tex]
where
q is the charge
E is the electric field
For a positive charge, the direction of the force is the same as the electric field.
In this problem:
[tex]q=+4.9\mu C=+4.9\cdot 10^{-6}C[/tex] is the charge
[tex]E_x=+242 N/C[/tex] is the electric field, along the x-direction
So the electric force (along the x-direction) is:
[tex]F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N[/tex]
towards positive x-direction.
The magnetic force instead is given by
[tex]F=qvB sin \theta[/tex]
where
q is the charge
v is the velocity of the charge
B is the magnetic field
[tex]\theta[/tex] is the angle between the directions of v and B
Here the charge is stationary: this means [tex]v=0[/tex], therefore the magnetic force due to each component of the magnetic field is zero.
b)
In this case, the particle is moving along the +x axis.
The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,
[tex]F_{E_x}=1.19\cdot 10^{-3} N[/tex] (towards positive x-direction)
Concerning the magnetic force, we have to analyze the two different fields:
- [tex]B_x[/tex]: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, [tex]\theta=0^{\circ}[/tex], so the force due to this field is zero.
[tex]- B_y[/tex]: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, [tex]\theta=90^{\circ}[/tex]. Therefore, [tex]\theta=90^{\circ}[/tex], so the force due to this field is:
[tex]F_{B_y}=qvB_y[/tex]
where:
[tex]q=+4.9\cdot 10^{-6}C[/tex] is the charge
[tex]v=345 m/s[/tex] is the velocity
[tex]B_y = +1.9 T[/tex] is the magnetic field
Substituting,
[tex]F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N[/tex]
And the direction of this force can be found using the right-hand rule:
- Index finger: direction of the velocity (+x axis)
- Middle finger: direction of the magnetic field (+y axis)
- Thumb: direction of the force (+z axis)
c)
As in part b), the electric force has not change, since it does not depend on the veocity of the particle:
[tex]F_{E_x}=1.19\cdot 10^{-3}N[/tex] (+x axis)
For the field [tex]B_x[/tex], the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is
[tex]F_{B_x}=qvB_x[/tex]
And by substituting,
[tex]F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N[/tex]
And by using the right-hand rule:
- Index finger: velocity (+z axis)
- Middle finger: magnetic field (+x axis)
- Thumb: force (+y axis)
For the field [tex]B_y[/tex], the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is
[tex]F_{B_y}=qvB_y[/tex]
And by substituting,
[tex]F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N[/tex]
And by using the right-hand rule:
- Index finger: velocity (+z axis)
- Middle finger: magnetic field (+y axis)
- Thumb: force (-y axis)
The force on a charged particle in an electric field is given by Felectric = qE, and in a magnetic field while moving, it is Fmagnetic = qv × B, with the right-hand rule determining the direction of Fmagnetic. A stationary particle only experiences Felectric. When moving, it may experience both forces, depending on the motion's relationship to the field's direction.
Explanation:The force on a charged particle in an electric field is given by Felectric = qE, where q is the charge and E is the electric field. If the particle is stationary, only the electric force acts on it. When moving in a magnetic field, a magnetic force Fmagnetic = qv × B also acts on it, where v is the velocity and B is the magnetic field. The direction of this force is perpendicular to both v and B as per the right-hand rule.
For a particle with charge q = +4.9 μC:
(a) When stationary, the force due to the electric field is F = qEx, only in the direction of E.(b) When moving along the +x axis, the magnetic force is zero as v and B are parallel.(c) When moving along the +z axis, both electric and magnetic forces act, and they can be calculated using the given formulas.
The astronomer who discovered the dwarf planet Eris suggests there might be another object far beyond the Kupier belt. If this Planet X exists, it would be about 10 times the mass of Earth and 2-3 times the size of Earth, putting it in the ice giant category, and have an orbit with a semi-major axis of 700 AU. You can read more about this object on NASA's page. If this object exists, what would we classify it as?
These objects would be classified as extreme trans Neptunian object (ETNO).
Explanation:
ETNO’s are the objects lying beyond the planet Neptune and orbiting the Sun. They follow a highly eccentric path which is tilted. ETNO has been grouped into three major according to their respective perihelia.
Within this region (beyond Neptune’s orbit), a hypothetical planet has been discovered. It was discovered following its gravitational effect on the other objects of Kuiper Belt (region beyond the orbit of the Neptune- the last planet of our Solar system)
The Planet is assumed to be around 2 times the Earth’s size and around 10 times heavier than Earth.
Planet X, if it exists, would likely be classified as a planet rather than a dwarf planet because it is much larger and more massive, suggesting it could clear its orbital neighborhood, which is one of the criteria that differentiate planets from dwarf planets.
The hypothetical object proposed by the astronomer beyond the Kuiper Belt, nicknamed Planet X, would be classified differently from Eris and other dwarf planets like Pluto, Makemake, and Haumea. Since it is speculated to be about 10 times the mass of Earth and 2-3 times its size, placing it within the ice giant category, it would resemble Uranus or Neptune rather than the smaller dwarf planets in the solar system.
Dwarf planets are generally smaller bodies that, while orbiting the Sun and having sufficient mass for their self-gravity to overcome rigid body forces, have not cleared their neighboring region of other objects. In contrast, Planet X, being significantly larger and more massive, would likely be considered a full-fledged planet if its existence were confirmed, primarily because of its mass, size, and potential to clear its orbit, aligning closer with the current criteria for a planet.
A bugle can be thought of as an open pipe. If a bugle were straightened out, it would be 2.65 mlong.a.) If the speed of sound is 343m/????, find the lowest frequency that is resonant for a bugle (ignoring end corrections)b.) Find the next two resonant frequencies for the bugle.
Answer:
(a). The lowest frequency is 64.7 Hz.
(b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.
Explanation:
Given that,
Length = 2.65 m
Speed of sound = 343 m
We need to calculate the wavelength
Using formula of wavelength
[tex]\lambda=2l[/tex]
Put the value into the formula
[tex]\lambda=2\times2.65[/tex]
[tex]\lambda=5.3\ m[/tex]
(a). We need to calculate the lowest frequency
Using formula of frequency
[tex]f_{1}=\dfrac{v}{\lambda_{1}}[/tex]
Put the value into the formula
[tex]f_{1}=\dfrac{343}{5.3}[/tex]
[tex]f_{1}=64.7\ Hz[/tex]
(b). We need to calculate the next two resonant frequencies for the bugle
Using formula of resonant frequency
[tex]f_{2}=\dfrac{v}{\lambda_{2}}[/tex]
[tex]f_{2}=\dfrac{v}{l}[/tex]
Put the value into the formula
[tex]f_{2}=\dfrac{343}{2.65}[/tex]
[tex]f_{2}=129.4\ Hz[/tex]
For third frequency,
[tex]f_{3}=\dfrac{v}{\lambda_{3}}[/tex]
[tex]f_{3}=\dfrac{3v}{2l}[/tex]
Put the value into the formula
[tex]f_{3}=\dfrac{3\times343}{2\times2.65}[/tex]
[tex]f_{3}=194.2\ Hz[/tex]
Hence, (a). The lowest frequency is 64.7 Hz.
(b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.
The next two resonant frequencies for the bugle are 64.72Hz and 323.6Hz
The lowest frequency is expressed as:
f = v/2l
[tex]f_0=\frac{343}{2(2.65)} f_0=\frac{343}{5.3}\\f_0= 64.72 Hz[/tex]
Since the bugle is an open pipe the next two resonant frequencies are;
[tex]f_2 =3f_0=3(64.72) = 194.15Hz\\\\f_3 = 5f_0 = 5(64.72) =323.6Hz \\[/tex]
Hence the next two resonant frequencies for the bugle are 64.72Hz and 323.6Hz
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The drag is proportional to the square of the speed of the boat, in the form Fd=bv2 where b= 0.5 N⋅ s2/m2. What is the acceleration of the boat just after the rain starts? Take the positive x axis along the direction of motion.
The first part of the question is missing and it is:
A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0 kg/hr.
Answer:
a(o) = 18.03 x 10^(-3) m/s^(-2)
Explanation:
First of all, if we make a momentum balance in the direction of motion (the x-direction), we'll discover that the change of the momentum will be equal to the forces acting on the boat.
Hence, there is only the drag force, which acts against the direction of motion as:
d(m·v)/dt = - k·v² (since f_d = - k·v²)
where k = 0.5 Ns²/m²
Now let's simplify the time derivative on the Left, by applying product rule of differentiation and we obtain:
v·dm/dt + m·dv/dt = - k·v²
where dm/dt = 10kg/hr (this is the change of the mass of the boat)
Furthermore, acceleration is the time derivative of time velocity, Thus;
a = dv/dt = - (k·v² + v·dm/dt) / m
Now, for the moment,when the rain starts; since we know all the values on the right hand side, let's solve for the acceleration ;
a(o) = - (ko·vo² + vo·dm/dt) / mo
= -[ (0.5(3)²) + (3x10/3600)/250
(dt = 3600secs because 1hr = 3600 secs)
So a(o) = 18.03 x 10^(-3) m/s^(-2)
A football punter accelerates a football from rest to a speed of 15 m/s during the time in which his toe is in contact with the ball (about 0.15 s). If the football has a mass of 0.44 kg, what average force does the punter exert on the ball?
Answer:
Force, F = 44 N
Explanation:
Given that,
Initial speed of the football, u = 0
Final speed, v = 15 m/s
The time of contact of the ball, t = 0.15 s
The mass of football, m = 0.44 kg
We need to find the average force exerted on the ball. It is given by the formula as :
[tex]F=ma\\\\F=\dfrac{mv}{t}\\\\F=\dfrac{0.44\times 15}{0.15}\\\\F=44\ N[/tex]
So, the average force exerted on the ball is 44 N. Hence, this is the required solution.