Explanation:
A trapeze is rotating with 1 rotation per second .
Thus its angular velocity ω = 2π n
here n is the number of rotations per second
Thus ω = 2π b because n = 1 in this case
Suppose the moment of inertia of his is = I
Then angular momentum L₁ = I ω = 2 I π
In the second case , the moment of inertia becomes = 0.4 I
Let his angular velocity is ω₀
Thus angular momentum L₂ = 0.4 I ω₀
Because no external torque is applied , therefore angular momentum will remain constant .
Thus L₁ = L₂
Therefore 2 I π = 0.4 I x 2 n₀ π
here n₀ is the number of rotations per second
n₀ = [tex]\frac{1}{0.4}[/tex] = [tex]\frac{5}{2}[/tex] = 2.5
Assuming that only air resistance and gravity act on a falling object, we can find that the velocity of the object, v, must obey the differential equation dv m mg bv dt . Here, m is the mass of the object, g is the acceleration due to gravity, and b > 0 is a constant. Consider an object that has a mass of 100 kilograms and an initial velocity of 10 m/sec (that is, v(0) = 10). If we take g to be 9.8 m/sec2 and b to be 5 kg/sec, find a formula for the velocity of the object at time t. Further, find the terminal (or limiting) velocity of the object. Circle your velocity formula and the terminal velocity.
Answer:
v = 196 - 186*e^( - 0.05*t )
v-terminal = 196 m/s
Explanation:
Given:
- The differential equation for falling object velocity v in gravity with air resistance is given by:
m*dv/dt = m*g - b*v
- The initial conditions and constants are as follows:
v(0) = 10 , m = 100 kg , b = 5 kg/s , g = 9.8 m/s^2
Find:
- Find a formula for the velocity of the object at time t. Further, find the terminal (or limiting) velocity of the object. Circle your velocity formula and the terminal velocity.
Solution:
- Rewrite the differential equation in te form:
dv/dt + (b/m)*v = g
- The integration factor function P(t) = b/m. The integrating factor u(t) is:
u(t) = e^∫P(t).dt
u(t) = e^∫(b/m).dt
u(t) = e^[(b/m).t]
- Solve the differential equation after expressing in form:
v.u(t) = ∫u(t).g.dt
v.e^[(b/m).t] = g*∫e^[(b/m).t].dt
v.e^[(b/m).t] = g*m*e^[(b/m).t] / b + C
v = g*m/b + C*e^[-(b/m).t]
- Apply the initial conditions v(0) = 10 m/s and evaluate C:
10 = 9.8*100/5 + C*e^[-(b/m).0]
10 = 9.8*100/5 + C
C = -186
- The final ODE solution is:
v = 196 - 186*e^( - 0.05*t )
- The Terminal velocity vt can be expressed by a limiting value for v(t), where t ->∞.
vt = Lim t ->∞ ( v(t) )
vt = Lim t ->∞ ( 196 - 186*e^( - 0.05*t ) )
vt = 196 - 0 = 196 m/s
The goal of this experiment is to investigate quantitatively how the magnetic force that a bar magnet and a current-carrying coil exert on each other depends on the distance between them.
Available equipment: Bar magnet, coil with connecting wires, power supply, digital balance, ring stand, clamp, meter stick, tape
Can you design an experiment with that info.
Answer:
Answer explained below
Explanation:
magnetic force of charged particle placed in uniform magnetic field is given by
F = iLB sin theta
where i is current
L is length
B = magnetic field
theta is the angle between L and B
so we can design an experiment with the given apparatus
as we have device for measuring length ( meter stick and scale)
magnitude of magnetic field from bar magnetic using magnetic moments principle
Finally from conducting wires and power, we can measure current i using Ammeter
Steam enters a well-insulated turbine operating at steady state at 4 MPa with a specific enthalpy of 3015.4 kJ/kg and a velocity of 10 m/s. The steam expands to the turbine exit where the pressure is 0.07 MPa, specific enthalpy is 2400 kJ/kg, and the velocity is 90 m/s. The mass flow rate is 30 kg/s. Neglecting potential energy effects, determine the power developed by the turbine, in kW.
Answer:
power developed by the turbine = 18342 kW
Explanation:
given data
pressure = 4 MPa
specific enthalpy h1 = 3015.4 kJ/kg
velocity v1 = 10 m/s
pressure = 0.07 MPa
specific enthalpy h2 = 2400 kJ/kg
velocity v2 = 90 m/s
mass flow rate = 30 kg/s
solution
first we apply here thermodynamic equation that is express as energy equation is
[tex]h1 + \frac{v1}{2} + q = h2 + \frac{v2}{2} + w[/tex] .......................1
we know turbine is insulated so q is 0
put here value we get
[tex]3015.4 \times 1000 + \frac{10^2}{2} = 2400 \times 1000 + \frac{90^2}{2} + w[/tex]
w = 611400 J/kg = 611.4 kJ/kg
and now we get power developed by the turbine W is
W = mass flow rate × w ................2
put here value
W = 30 × 611.4
W = 18342 kW
power developed by the turbine = 18342 kW
We would like to use the relation V(t)=I(t)R to find the voltage and current in the circuit as functions of time. To do so, we use the fact that current can be expressed in terms of the voltage. This will produce a differential equation relating the voltage V(t) to its derivative. Rewrite the right-hand side of this relation, replacing I(t) with an expression involving the time derivative of the voltage. Express your answer in terms of dV(t)/dt and quantities given in the problem introduction.
Answer:
Explanation:
The explanation or solution is given in the attach document
Final answer:
To find the voltage and current as functions of time in a circuit, replace current I(t) with C*dV(t)/dt, and substitute this into Ohm's law V(t)=I(t)R to obtain a differential equation V(t) = (C*dV(t)/dt)*R.
Explanation:
To rewrite the relation V(t)=I(t)R by replacing I(t) with an expression involving the time derivative of the voltage, first understand that the current I(t) is the rate of change of charge Q with respect to time, that is I(t) = dQ/dt. However, since Q is related to the voltage across a capacitor by the equation Q=CV(t), where C is the capacitance, the current can also be expressed in terms of the voltage as I(t) = C*(dV(t)/dt). Replacing this into the original equation gives us V(t) = (C * dV(t)/dt) * R, which is a first-order differential equation relating the voltage to its time derivative.
An avant-garde composer wants to use the Doppler effect in his new opera. As the soprano sings, he wants a large bat to fly toward her from the back of the stage. The bat will be outfitted with a microphone to pick up the singer's voice and a loudspeaker to rebroadcast the sound toward the audience. The composer wants the sound the audience hears from the bat to be, in musical terms, one half-step higher in frequency than the note they are hearing from the singer. Two notes a half-step apart have a frequency ratio of 2⁽¹/¹²⁾ = 1.059.
With what speed must the bat fly toward the singer?
Answer:
v’= 9.74 m / s
Explanation:
The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.
Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer
f₁ ’= f₀ (v + v₀)/v
Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest
f₂’= f₁’ v/(v - vs)
Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’
v’= vo = vs
Let's replace
f₂’= f₀ (v + v’)/v v/(v -v ’)
f₂’= f₀ (v + v’) / (v -v ’)
(v –v’ ) f₂’ / f₀ = v + v ’
v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)
v’ (1 + 1.059) = 340 (1.059 - 1)
v’= 20.06 / 2.059
v’= 9.74 m / s
Delicate measurements indicate that the Earth has an electric field surrounding it, similar to that around a positively charged sphere. Its magnitude at the surface of the Earth is about 100 N/C. What charge would an oil drop of mass 2.0 x 10 15 kg have to have, in order to remain suspended by the Earth’s electric field? Give your answer in Coulombs ?
Answer:
q = 1.96 10⁴ C
Explanation:
The elective force is given by
[tex]F_{e}[/tex] = q E
Where E is the electric field and q the charge.
Let's use Newton's law of equilibrium for the case of the suspended drop
F_{e} –W = 0
F_{e} = W
q E = m g
q = m g / E
Let's calculate
q = 2.0 10⁵ 9.8 / 100
q = 1.96 10⁴ C
To maintain suspension in the Earth’s electric field, an oil drop with a mass of 2.0 × 10^-15 kg requires a charge of 1.96 × 10^-16 Coulombs, calculated by equating the electric force with the gravitational force.
Explanation:To determine the charge needed for an oil drop of mass 2.0 × 10-15 kg to remain suspended by the Earth’s electric field of 100 N/C, we can apply the equilibrium condition between the electric force and the gravitational force. The electric force (Felectric) is equal to the charge (q) multiplied by the electric field (E), so Felectric = q × E. The gravitational force (Fgravity) is the mass (m) multiplied by the acceleration due to gravity (g), which is approximately 9.8 m/s2.
For the oil drop to remain suspended, these two forces must be equal: q × E = m × g, which gives us q = (m × g) / E. Using the provided values, the charge q is calculated as follows:
q = (2.0 × 10-15 kg × 9.8 m/s2) / 100 N/C
q = (2.0 × 10-15 × 9.8) / 100
q = 1.96 × 10-16 C
Therefore, the oil drop must have a charge of 1.96 × 10-16 Coulombs to remain suspended in the Earth’s electric field.
Daisy walks across a force platform, and forces exerted by her foot during a step are recorded. The peak velocity reaction force is 1200 (this force acts upward on Daisy). At the same instant, the frictional force is 200N(this force acts forward on Daisy).(a) how large is the resultant of these forces(b) What is the direction of the resultant forces?
Answer:
(a). The resultant of these forces is 1216.55 N.
(b). The direction of the resultant forces is 80.53°.
Explanation:
Given that,
First force = 1200 N
Second force = 200 N
(a). We need to calculate the resultant of these forces
Using cosine law
[tex]F=\sqrt{F_{1}^2+F_{2}^2+2F_{1}F_{2}\cos\theta}[/tex]
Put the value into the formula
[tex]F=\sqrt{1200^2+200^2+2\times1200\times200\cos90}[/tex]
[tex]F=\sqrt{1200^2+200^2}[/tex]
[tex]F= 1216.55\ N[/tex]
The resultant of these forces is 1216.55 N.
(b). We need to calculate the direction of the resultant forces
Using formula of direction
[tex]\tan\alpha=\dfrac{F_{1}}{F_{2}}[/tex]
Put the value into the formula
[tex]\alpha=\tan^{-1}(\dfrac{1200}{200})[/tex]
[tex]\alpha=80.53^{\circ}[/tex]
Hence, (a). The resultant of these forces is 1216.55 N.
(b). The direction of the resultant forces is 80.53°.
Answer:
a) [tex]F_r=1216.55\ N[/tex]
b) [tex]\theta=80.54^{\circ}[/tex]
Explanation:
Given:
force acting upward on the, [tex]F_y=1200\ N[/tex]force acting forward on daisy, [tex]F_x=200\ N[/tex]a)
Now the resultant of these forces:
Since the forces are mutually perpendicular,
[tex]F_r=\sqrt{F_x^2+F_y^2}[/tex]
[tex]F_r=\sqrt{200^2+1200^2}[/tex]
[tex]F_r=1216.55\ N[/tex]
b)
The direction of this force from the positive x-direction:
[tex]\tan\theta=\frac{F_y}{F_x}[/tex]
[tex]\tan\theta=\frac{1200}{200}[/tex]
[tex]\theta=80.54^{\circ}[/tex]
An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.60 A in the -c-direction. At an instant when the electron is at point (0, 0.200 m, 0) and the electron's velocity is v(5.00 What is the force that the wire exerts on the electron? Enter the z, y, and z components of the force separated by commas. 104 m/s)^-(3.00 x 104 m/s)3.
Answer:
The force that the wire exerts on the electron is [tex]-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Explanation:
Given that,
Current = 8.60 A
Velocity of electron [tex]v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s[/tex]
Position of electron = (0,0.200,0)
We need to calculate the magnetic field
Using formula of magnetic field
[tex]B=\dfrac{\mu I}{2\pi d}(-k)[/tex]
Put the value into the formula
[tex]B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}[/tex]
[tex]B=0.0000086\ T[/tex]
[tex]B=-8.6\times10^{-6}k\ T[/tex]
We need to calculate the force that the wire exerts on the electron
Using formula of force
[tex]F=q(\vec{v}\times\vec{B}[/tex]
[tex]F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )[/tex]
[tex]F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k[/tex]
[tex]F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Hence, The force that the wire exerts on the electron is [tex]-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Part A) Components of the Force The force components on the electron are: [tex]\[F_x = -8.26 \times 10^{-20} \, \text{N}, \quad F_y = -1.38 \times 10^{-19} \, \text{N}, \quad F_z = 0 \, \text{N}\][/tex]
Part B) Magnitude of the Force The magnitude of the force is:[tex]\[F \approx 1.60 \times 10^{-19} \, \text{N}\][/tex]
Part A: Calculate the force components
The force on a moving charge in a magnetic field is given by the Lorentz force equation:
[tex]\[\vec{F} = q \vec{v} \times \vec{B}\][/tex]
First, we need to find the magnetic field [tex]\(\vec{B}\)[/tex] produced by the wire at the position of the electron. The magnetic field due to a long, straight current-carrying wire is given by:
[tex]\[B = \frac{\mu_0 I}{2 \pi r}\][/tex]
where:
- [tex]\(\mu_0 = 4 \pi \times 10^{-7} \, \text{T} \cdot \text{m/A}\)[/tex] (the permeability of free space)
- [tex]\(I = 8.60 \, \text{A}\)[/tex] (the current through the wire)
- [tex]\(r = 0.200 \, \text{m}\)[/tex] (the distance from the wire to the electron)
Calculating [tex]\(B\)[/tex]:
[tex]\[B = \frac{4 \pi \times 10^{-7} \times 8.60}{2 \pi \times 0.200} = \frac{4 \times 10^{-7} \times 8.60}{0.200} = \frac{3.44 \times 10^{-6}}{0.200} = 1.72 \times 10^{-5} \, \text{T}\][/tex]
The direction of [tex]\(\vec{B}\)[/tex] follows the right-hand rule. Since the current flows in the [tex]\(-x\)[/tex]-direction, at the point [tex]\((0, 0.200, 0)\)[/tex], the magnetic field [tex]\(\vec{B}\)[/tex] is directed into the page (negative [tex]\(z\)[/tex]-direction):
[tex]\[\vec{B} = -1.72 \times 10^{-5} \hat{k} \, \text{T}\][/tex]
Now we use the Lorentz force equation with:
[tex]\[q = -1.60 \times 10^{-19} \, \text{C} \quad (\text{charge of an electron})\][/tex]
[tex]\[\vec{v} = (5.00 \times 10^4 \hat{i} - 3.00 \times 10^4 \hat{j}) \, \text{m/s}\][/tex]
[tex]\[\vec{B} = -1.72 \times 10^{-5} \hat{k} \, \text{T}\][/tex]
The cross product [tex]\(\vec{v} \times \vec{B}\)[/tex]:
[tex]\[\vec{v} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\5.00 \times 10^4 & -3.00 \times 10^4 & 0 \\0 & 0 & -1.72 \times 10^{-5}\end{vmatrix}= \hat{i}( (-3.00 \times 10^4)(-1.72 \times 10^{-5}) - 0) - \hat{j}( (5.00 \times 10^4)(-1.72 \times 10^{-5}) - 0)\][/tex]
[tex]\[= \hat{i}( 5.16 \times 10^{-1}) - \hat{j}( -8.60 \times 10^{-1})\][/tex]
[tex]\[= 0.516 \hat{i} + 0.860 \hat{j} \, \text{N/C}\][/tex]
Now, multiply by the charge of the electron:
[tex]\[\vec{F} = q \vec{v} \times \vec{B} = -1.60 \times 10^{-19} (0.516 \hat{i} + 0.860 \hat{j})\][/tex]
[tex]\[\vec{F} = -0.516 \times 1.60 \times 10^{-19} \hat{i} - 0.860 \times 1.60 \times 10^{-19} \hat{j}\][/tex]
[tex]\[\vec{F} = -8.26 \times 10^{-20} \hat{i} - 1.376 \times 10^{-19} \hat{j} \, \text{N}\][/tex]
So, the components of the force are:
[tex]\[F_x = -8.26 \times 10^{-20} \, \text{N}, \quad F_y = -1.376 \times 10^{-19} \, \text{N}, \quad F_z = 0 \, \text{N}\][/tex]
Part B: Calculate the magnitude of the force
The magnitude of the force is given by:
[tex]\[F = \sqrt{F_x^2 + F_y^2 + F_z^2}\][/tex]
[tex]\[F = \sqrt{(-8.26 \times 10^{-20})^2 + (-1.376 \times 10^{-19})^2}\][/tex]
[tex]\[F = \sqrt{(6.82 \times 10^{-39}) + (1.89 \times 10^{-38})}\][/tex]
[tex]\[F = \sqrt{2.57 \times 10^{-38}}\][/tex]
[tex]\[F \approx 1.60 \times 10^{-19} \, \text{N}\][/tex]
So, the magnitude of the force is approximately [tex]\(1.60 \times 10^{-19} \, \text{N}\).[/tex]
The complete question is attached here:
An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.60 A in the -z-direction. At an instant when the electron is at point (0, 0.200 m, 0) and the electron's velocity is (5.00 x 104 m/s) -(3.00 x 104 m/s).
Part A:What is the force that the wire exerts on the electron?
Part B:Calculate the magnitude of this force.
SPEAR is a storage ring at the Stanford Linear Accelerator which has a circulating beam of electrons that are moving at nearly the speed of light (2.998 108 m/s). If a similar ring is about 68.0 m in diameter and has a 0.37 A beam, how many electrons are in the beam
Answer:
The no. of electron in the beam = [tex]1.64\times10^{12}[/tex]
Explanation:
Given :
The diameter of circular ring = 68 m.
The current flowing in the beam = 0.37 A
Speed of light = [tex]3\times10^{8} ms^{-1}[/tex]
We know that the current is equal to the charge per unit time.
⇒ [tex]I = \frac{Q}{t}[/tex]
∴ [tex]Q=It[/tex]
Here given in the question, electron revolving in a circle with the diameter
[tex]d = 68[/tex]m
⇒ Time take to complete one round [tex](t) =[/tex] [tex]\frac{\pi d }{v}[/tex]
∴ [tex]Q = \frac{I\pi d }{v}[/tex]
[tex]Q = \frac{0.37 \times 3.14 \times 68}{3 \times 10^{8} }[/tex]
[tex]Q = 26.33 \times 10^{-8}[/tex]
Now, for finding the no. of electron we have to divide [tex]Q[/tex] to the charge of the electron [tex]q = 1.6 \times 10^{-19}[/tex]
∴ [tex]n[/tex] = [tex]\frac{26.33 \times 10^{-8} }{1.6 \times 10^{-19} }[/tex]
[tex]n = 1.64 \times 10^{12}[/tex]
Thus, the no. of electron in the beam is [tex]1.64 \times 10^{12}[/tex].
Final answer:
To calculate the number of electrons in the beam, first determine the total charge passing a point in one second using the current, then divide by the charge of one electron. Using the given current of 0.37 A, this method reveals the total number of electrons in the beam.
Explanation:
To determine the number of electrons in a beam with a current of 0.37 A, the given data of the SPEAR storage ring can be used. First, recall that the charge of one electron is approximately ‑1.602 × 10⁻¹⁹ C (coulombs).
Current (I) is defined as the rate of charge (Q) flow through a given point, over time (t), expressed as I = Q/t. Therefore, the total charge in the beam can be calculated for one second as the product of the current and time.
To find the number of electrons (N), the total charge is divided by the charge of one electron, mathematically represented as N = Q / e. Substituting the given value of 0.37 A for I and using 1 second for t, the calculation would be N = (0.37 C/s) / (1.602 × 10⁻¹⁹ C/electron). This gives the total number of electrons circulating in the beam.
What is an ideal diode? a. The ideal diode acts as an open circuit for forward currents and as a short circuit with reverse voltage applied. b. The ideal diode acts as an open circuit regardless of forward voltage or reverse voltage applied. c. The ideal diode acts as a short circuit regardless of forward voltage or reverse voltage applied. d. The ideal diode acts as a short circuit for forward currents and as an open circuit with reverse voltage applie
Answer:
d. The ideal diode acts as a short circuit for forward currents and as an open circuit with reverse voltage applied.
Explanation:
Ideal diode acts like an ideal conductor. In case of forward voltage it acts like an ideal conductor. However when it is reverse biased then it behaves like an ideal insulator. You can understand it bu considering a switch. When the voltage is forward then ideal diode acts like a closed switch. When the voltage is reverse biased then ideal diode behaves like an open switch.
That is why we can say that the ideal diode acts as a short circuit (higher conduction) for forward currents and as an open circuit ( zero conduction) with reverse voltage applied.
Jeff was riding in a car. Which change demonstrates the GREATEST effect of inertia?
Answer: The car makes a sudden stop.
Explanation:
Manuel is coasting on his bike. Because he is not pedaling, his bike will come to a stop. Which of these will cause Manuel's bike to stop? - the force of friction.
Hope this helps! enjoy your day.
A mass is hanging on the end of a spring and oscillating up and down. There are three forces acting on the mass, the force of the spring, the force of gravity and the force of air resistance. Which of these forces can be associated with potential energy function? a. The spring force only. b. Gravity only. c. Air resistance only. d. The spring force and gravity. e. The spring force and air resistance. f. Gravity and air resistance. g. All three forces.
Answer:
d. The spring force and gravity.
Explanation:
The forces that can be associated with a potential energy function are only the conservative forces. These are the forces whose work done on an object by the force does not depend on the path taken, but only on the initial and final position of the object.
The only two conservative forces in this problem are:
- Gravity
- The spring force
While the air resistance is a non-conservative force.
The potential energy associated with the gravitational force is:
[tex]U=mgh[/tex]
where
m is the mass of the object
g is the acceleration due to gravity
h is the position of the object with respect to a reference level (e.g. the ground)
The potential energy associated to the spring force is
[tex]U=\frac{1}{2}kx^2[/tex]
where:
k is the spring constant
x is the elongation of the spring with respect to the equilibrium position
A motor has an armature resistance of 3.75 Ω . Part A If it draws 9.10 A when running at full speed and connected to a 120-V line, how large is the back emf?
Back emf is 85.9 V.
Explanation:
Given-
Resistance, R = 3.75Ω
Current, I = 9.1 A
Supply Voltage, V = 120 V
Back emf = ?
Assumption - There is no effects of inductance.
A motor will have a back emf that opposes the supply voltage, as the motor speeds up the back emf increases and has the effect that the difference between the supply voltage and the back emf is what causes the current to flow through the armature resistance.
So if 9.1 A flows through the resistance of 3.75Ω then by Ohms law,
The voltage across the resistance would be
v = I x R
= 9.1 x 3.75
= 34.125 volts
We know,
supply voltage = back emf + voltage across the resistance
By plugging in the values,
120 V = back emf + 34.125 V
Back emf = 120 - 34.125
= 85.9 Volts
Therefore, back emf is 85.9 V.
KATZPSEF1 30.P.041. My Notes Ask Your Teacher A charged particle enters a region of space with a uniform magnetic field B = 1.98 i T. At a particular instant in time, it has velocity v = (1.46 ✕ 106 i + 2.42 ✕ 106 j) m/s. Based on the observed acceleration, you determine that the force acting on the particle at this instant is F = 1.60 k N. What are the following?(a) the sign of the charged particle
(b) the magnitude of the charged particle
Answer: The charge on the particle is positive
While the magnitude = 0.00028C
Explanation:
Please find the attached file for the solution
A rectangular block of copper has sides of length 15 cm, 26 cm, and 43 cm. If the block is connected to a 5.0 V source across two of its opposite faces, find the following. (a) What is the maximum current the block can carry?
Answer:
the case is the one with the greatest current, L=15 cm , i = 2.19 10⁸ A
Explanation:
Ohm's law is
V = i R
Resistance is
R = ρ L / A
Where L is the length of the electrons pass and A the area perpendicular to the current
i = V / R
i = V (A / ρ L)
i = V / ρ (A / L)
We can calculate the relationship between the area and the length to know in which direction the maximum currents
Case 1
L = 0.15 m
A = 0.26 0.43 = 0.1118 m2
A / L = 0.1118 / 0.15
A / L = 0.7453 m
Case 2
L = 0.26 m
A = 0.15 0.43 = 0.0645 m2
A / L = 0.248 m
Case 3
L = 0.43 m
A = 0.15 0.26 = 0.039 m2
A / L = 0.0907 m
We can see that the case is the one with the greatest current, L=15 cm
Let's calculate the current
i = 5 / 1.7 10⁻⁸ (0.7453)
i = 2.19 10⁸ A
Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.40×105 V/m . When the space is filled with dielectric, the electric field is E= 2.20×105 V/m . Part A What is the charge density on each surface of the dielectric?
Answer:
[tex]\sigma_i=1.06*10^{-6}C[/tex]
Explanation:
When the space is filled with dielectric, an induced opposite sign charge appears on each surface of the dielectric. This induced charge has a charge density related to the charge density on the electrodes as follows:
[tex]\sigma_i=\sigma(1-\frac{E}{E_0})[/tex]
Where E is the eletric field with dielectric and [tex]E_0[/tex] is the electric filed without it. Recall that [tex]\sigma[/tex] is given by:
[tex]\sigma=\epsilon_0E_0[/tex]
Replacing this and solving:
[tex]\sigma_i=\epsilon_0E_0(1-\frac{E}{E_0})\\\sigma_i=8.85*10^{-12}\frac{C^2}{N\cdot m^2}*3.40*10^5\frac{V}{m}*(1-\frac{2.20*10^5\frac{V}{m}}{3.40*10^5\frac{V}{m}})\\\sigma_i=1.06*10^{-6}C[/tex]
You throw a rock straight up from the edge of a cliff. It leaves your hand at time t = 0 moving at 13.0 m/s. Air resistance can be neglected. Find both times at which the rock is 4.00 m above where it left your hand. Enter your answers in ascending order separated by a comma. Express your answer in seconds.
Answer:
0.36s, 2.3s
Explanation:
Let gravitational acceleration g = 9.81 m/s2. And let the throwing point as the ground 0 for the upward motion. The equation of motion for the rock leaving your hand can be written as the following:
[tex]s = v_0t + gt^2/2[/tex]
where s = 4 m is the position at 4m above your hand. [tex]v_0 = 13 m/s[/tex] is the initial speed of the rock when it leaves your hand. g = -9.81m/s2 is the deceleration because it's in the downward direction. And t it the time(s) it take to get to 4m, which we are looking for
[tex]4 = 13t - 9.81t^2/2[/tex]
[tex]4.905 t^2 - 13t + 4 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{13\pm \sqrt{(-13)^2 - 4*(4.905)*(4)}}{2*(4.905)}[/tex]
[tex]t= \frac{13\pm9.51}{9.81}[/tex]
t = 2.3 or t = 0.36
Explanation:
Below is an attachment containing the solution.
a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical and horizontal velocity components.
Answer:
Vx = 10.9 m/s , Vy = 15.6 m/s
Explanation:
Given velocity V= 19 m/s
the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°
θ = 55°
to Find Vx = ? and Vy= ?
Vx = V cos θ
Vx = 19 m/s × cos 55°
Vx = 10.9 m/s
Vx = V sin θ
Vy = 19 m/s × sin 55°
Vy = 15.6 m/s
15.56m/s and 10.90m/s respectively
Explanation:The vertical and horizontal components of a given vector, say A, are given by
[tex]A_{Y}[/tex] = A sin θ ----------------(i)
[tex]A_{X}[/tex] = A cos θ ----------------(ii)
Where;
[tex]A_{Y}[/tex] is the vertical component of the vector A
[tex]A_{X}[/tex] is the horizontal component of the vector A
A is the magnitude of the vector A
θ is the angle the vector makes with the positive x-axis (horizontal direction).
Now, from the question;
The vector is the velocity of the 1-kg discus. Lets call it vector V
The magnitude of the velocity vector V = V = 19m/s
The angle that the vector makes with the positive x-axis = θ
To calculate θ;
Notice that the velocity vector makes an angle of 35° from the vertical direction rather than the horizontal direction.
Therefore, to get the horizontal direction of the velocity vector, we subtract 35° from 90° as follows;
θ = 90° - 35° = 55°
Now, the vertical and horizontal components of the velocity vector, V, are given by
[tex]V_{Y}[/tex] = V sin θ --------------------(iii)
[tex]V_{X}[/tex] = V cos θ ------------------------(iv)
Substitute all the necessary values into equations(iii) and (iv) as follows;
[tex]V_{Y}[/tex] = 19 sin 55° = 19m/s x 0.8192 = 15.56m/s
[tex]V_{X}[/tex] = 19 cos 55° = 19m/s x 0.5736 = 10.90m/s
Therefore, the vertical and horizontal velocity components are respectively 15.56m/s and 10.90m/s.
In another experiment, the intensity of the incident light and the temperature of the metal are held constant. Assuming that the initial light incident on the metal surface causes electrons to be ejected from the metal, what happens if the frequency of the incident light is increased?
Answer:
Explanation:
Answer:
Explanation:
The intensity and the temperature of metal is constant so the number of photo electrons remains constant. As the number of photo electrons remains same so the photo electric remains constant.
As the frequency is increased, the kinetic energy of the photo electrons increases and thus, the speed of photo electrons increases.
When the frequency of the incident light is increased above a certain threshold, electrons will start being ejected from the metal's surface, with their maximum velocity increasing linearly with the frequency.
As the frequency of the incident light is increased beyond a certain threshold frequency, electrons will begin to be ejected from the surface of the metal. However, their kinetic energy will not increase proportionally with the frequency, rather the maximum velocity of the ejected electrons will increase linearly with the frequency of the incident light.
Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to truthfully report the mass). (a) What is the force constant of the spring in such a scale if it the spring stretches 8.00 cm for a 10.0 kg load
Answer:
K= 1226.25 N/m
Explanation:
Given: mass m = 10 kg, Distance x= 8 cm = 0.08 m, g= 9.81 m/s²
By Hook's Law
F=K x
F=W=mg = 10 kg x 9.8 m/s² = 98.1 N
to Find Spring constant k = F/x = 98 N /0.08 m
K= 1226.25 N/m
a straight, 2.5-m wire carries a typical household current of 1.5 a (in one direction) at location where the earth's magnetic field i 0.55 gauss from south to north at a location where the earth's magnetic field is 0.55 gauss from south to north. find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) is the magnetic force ever large enough to cause significant effects under normal household current of 1.5 A
Answer:
a) When the current is from west to east and the magnetic field is from south to north the magnitude of the force is 2.1x10⁻⁴N and the direction is upwards.
b) The current is moving vertically upward, the magnitude of the force is 2.1x10⁻⁴N and the direction is west.
c) The force is 0 because the magnetic field and the direction of the current are in parallel.
d) No, the force is less.
Explanation:
Given:
L = length of the wire = 2.5 m
i = current in wire = 1.5 A
B = magnetic field = 0.55x10⁻⁴T
a) The magnitude of magnetic force is equal to:
[tex]F=BiL=0.55x10^{-5} *1.5*2.5=2.1x10^{-4} N[/tex]
b) The same way to a):
F = 2.1x10⁻⁴N
c) F = 0
The magnetic field and the direction of the current are in parallel.
d) The answer is no, the force is less
You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it isn't so by calculating how much weaker gravity is 400 km above the Earth's surface.
Answer:
It's only 1.11 m/s2 weaker at 400 km above surface of Earth
Explanation:
Let Earth radius be 6371 km, or 6371000 m. At 400km above the Earth surface would be 6371 + 400 = 6771 km, or 6771000 m
We can use Newton's gravitational law to calculate difference in gravitational acceleration between point A (Earth surface) and point B (400km above Earth surface):
[tex]g = G\frac{M}{r^2}[/tex]
where G is the gravitational constant, M is the mass of Earth and r is the distance form the center of Earth to the object
[tex]\frac{g_B}{g_A} = \frac{GM/r^2_B}{GM/r^2_A}[/tex]
[tex]\frac{g_B}{g_A} = \left(\frac{r_A}{r_B}\right)^2 [/tex]
[tex]\frac{g_B}{g_A} = \left(\frac{6371000}{6771000}\right)^2 [/tex]
[tex]\frac{g_B}{g_A} = 0.94^2 = 0.885[/tex]
[tex]g_B = 0.885 g_A[/tex]
So the gravitational acceleration at 400km above surface is only 0.885 the gravitational energy at the surface, or 0.885*9.81 = 8.7 m/s2, a difference of (9.81 - 8.7) = 1.11 m/s2.
A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 562 Hz tone, what is the wavelength of that tone in air at standard conditions?
To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,
Frequency [tex]= f = 562Hz[/tex]
Speed of sound in air [tex]= v = 331m/s[/tex]
The definition of wavelength is,
[tex]\lambda = \frac{v}{f}[/tex]
Here,
v = Velocity
f = Frequency
Replacing,
[tex]\lambda = \frac{331m/s}{562Hz}[/tex]
[tex]\lambda = 0.589m[/tex]
Therefore the wavelength of that tone in air at standard conditions is 0.589m
The wavelength of the tone in air 0.59 Hz
The trombone can produce pitches wavelength ranging from 85 Hz to 660 Hz
The trombone produces a tone of 562 Hz
The tone of air is at standard conditions, hence the velocity of the sound in air is 331 m/s
velocity= frequency/wavelength
331= 562/wavelength
wavelength= 331/562
= 0.59 HZ
Hence the wavelength of the tone is 0.59 Hz
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a solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls without slipping a distance L = 6.0 m down a roof that is inclined at the angle theta = 30degree.
(a) What is the angular speed of the cylinder about its center as it leaves the roof?
(b) The roofs edge is at height H = 5.0 m. How far horizontally from the roof's edge does the cylinder hit the level ground?
Answer:
Explanation:
Acceleration of cylinder
a = g sin 30 / 1+ k² / r² where k is radius of gyration and r is radius of cylinder.
For cylinder k² = (1 / 2) r²
acceleration
= gsin30 / 1.5
= g / 3
= 3.27
v² = u² + 2as
= 2 x 3.27 x 6
v = 6.26 m /s
v = angular velocity x radius
6.26 = angular velocity x .10
angular velocity = 62.6 rad / s
b ) vertical component of velocity
= 6.26 sin 30
= 3.13 m /s
h = ut + 1/2 g t²
5 = 3.13 t + .5 t²
.5 t²+ 3.13 t- 5 = 0
t = 1.32 s
horizontal distance covered
= 6.26 cos 30 x 1.32
= 7.15 m
The conservation of energy and kinematics allows to find the results for the questions about the movement of the cylinder on the ceiling and when falling are:
a) The angular velocity is w = 6.26 rad / s
b) the distance to the ground is: x = 7,476 m
Given parameters
Cylinder radius r = 10 cm = 0.10 m Mass m = 12 kg Distance L = 6.0 m Roof angle θ = 30º Ceiling height H = 5.0 mTo find
a) The angular velocity.
b) Horizontal distance.
Mechanical energy is the sum of kinetic energy and potential energies. If there is no friction, it remains constant at all points.
Linear and rotational kinematics study the motion of bodies with linear and rotational motions.
a) Let's write the mechanical energy at the points of interest.
Starting point. When it comes out of the top
Em₀ = U = m g h
Final point. On the edge of the roof.
[tex]Em_f[/tex] = K = ½ mv² + ½ I w²
Since the cylinder does not slide, friction is zero and energy is conserved.
Em₀ = [tex]Em_f[/tex]
mg h = ½ m v² + ½ I w²
The moment of inertia of the cylinder is;
I = ½ m r²
Linear and angular variables are related.
v = w r
let's substitute.
m g h = ½ m (wr) ² + ½ (½ m r²) w²
gh = ½ w² r² (1 + ½) = ½ w² r² [tex]\frac{3}{2}[/tex]
w² = [tex]\frac{4}{3 } \ \frac{gh}{r^2}[/tex]
Let's use trigonometry to find the height of the ceiling.
sin θ = h / L
h = L sin θ
We substitute.
w= [tex]\sqrt{ \frac{4}{3} \ \frac{g \ L sin \theta }{r^2} }[/tex]
Let's calculate.
w = [tex]\sqrt{\frac{4}{3} \frac{9.8 \ 6.0 sin 30}{0.10^2} }[/tex]
Let's calculate
w = Ra 4/3 9.8 6.0 sin 30 / 0.10²
w = 62.6 rad / s
b) For this part we can use the projectile launch expressions.
Let's find the time it takes to get to the floor.
y = y₀ + go t - ½ g t²
The initial height is y₀ = H, when it reaches the ground its height is y = 0 and let's use trigonometry for the vertical initial velocity.
sin 30 = [tex]\frac{v_{oy}}{v_o}[/tex]I go / v
[tex]v_{oy}[/tex] = v sin 30 = wr sin 30
[tex]v_{oy}[/tex] = 62.6 0.1 sin 30
[tex]v_{oy}[/tex] = 3.13 m / s
0 = H + voy t - ½ g t²
0 = 5 + 3.13 t - ½ 9.8 t³
t² - 0.6387 t - 1.02 = 0
We solve the quadratic equation.
t =[tex]\frac{0.6387 \pm \sqrt{0.6387^2 - 4 \ 1.02} }{2}[/tex]
t = [tex]\frac{0.6378 \pm 2.118}{2}[/tex]
t₁ = 1.379 s
t₂ = -0, 7 s
The time o must be a positive quantity, therefore the correct answer is:
t = 1.379 s
We look for the horizontal distance.
x = v₀ₓ t
vₓ = v cos θ
v = wr
Let's substitute.
x = wr cos t
Let's calculate.
x = 62.6 0.1 cos 30 1.379
x = 7.476 m
In conclusion using the conservation of energy and kinematics we can find the results for the questions about the movement of the cylinder on the ceiling and when falling are:
a) The angular velocity is w = 6.26 rad / s
b) the distance to the ground is: x = 7,476 m
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A constant voltage of 8.00 V has been observed over a certain time interval across a 2.30 H inductor. The current through the inductor, measured as 1.00 A at the beginning of the time interval, was observed to increase at a constant rate to a value of 8.00 A at the end of the time interval. How long was this time interval
Answer:2.01 s
Explanation:
Given
Applied voltage [tex]V=8\ V[/tex]
Inductance [tex]L=2.3\ H[/tex]
Change in Current [tex]\Delta i=8-1=7\ A[/tex]
Induced EMF is given by
[tex]V=L\times \dfrac{\Delta A}{\Delta t}[/tex]
[tex]8=2.3\times \dfrac{7}{\Delta t}[/tex]
[tex]\Delta t=\dfrac{2.3\times 7}{8}[/tex]
[tex]\Delta t=2.0125\ s[/tex]
The emission spectrum of an incandescent light bulb is continuous rather than discrete. This is true even through the filament is made of tungsten, an element. Why do you think this is
Answer:
This is the property of metals like W (Tungsten) to produce a continuous spectrum.
Explanation:
The molecules of a gas are highly differentiated and single atoms absorb photons instead of bulk like metals and only give line spectrum for example line spectrum given by hydrogen etc. On the other hand, metals like Tungsten gets very hot and emission of every possible wavelength of light is observed. This emission is continuous because the wavelengths of light in the spectrum has no breakages.
Two stones are launched from the top of a tall building. One stone is thrown in a direction 15.0 ∘∘ above the horizontal with a speed of 20.0 m/sm/s ; the other is thrown in a direction 15.0 ∘∘ below the horizontal with the same speed.
Which stone spends more time in the air? (Neglet air resistance)
a. The stone thrown upward spends more time in the air.
b. The stone thrown downward spends more time in the air.
c. Both stones spend the same amount of time in the air.
Answer:a
Explanation:
Given
First stone is thrown [tex]15^{\circ}[/tex] above the horizontal with some speed let say u
Second stone is thrown at [tex]15^{\circ}[/tex] below the horizontal with speed u
For a height h of building
For first stone (motion in vertical direction)
using
[tex]v^2-u^2=2ah [/tex]
where v=final velocity
u=initial velocity
a=acceleration
h=displacement
[tex]h=u\sin 15(t_1)-\frac{1}{2}gt_1^2---1[/tex]
For second stone
[tex]h=(-u\sin 15)(t_2)-\frac{1}{2}gt_2^2----2[/tex]
Equating 1 and 2
[tex]u\sin 15(t_1+t_2)-\frac{1}{2}g(t_1-t_2)(t_1+t_2)=0[/tex]
[tex](t_1+t_2)[u\sin 15-4.9(t_1-t_2)]=0[/tex]
as [tex]t_1+t_2[/tex] cannot be zero
so [tex]t_1-t_2=1.05\ s[/tex]
[tex]t_1=t_2+1.056[/tex]
therefore time taken by first stone(thrown upward) will be more.
Answer:
a. The stone thrown upward spends more time in the air.
Explanation:
Given:
projection of first stone, [tex]\theta_1=15^{\circ}[/tex] above the horizontal
initial velocity of projectiles, [tex]u_1=u_2=20\ m.s^{-1}[/tex]
projection of second stone,[tex]\theta_2=15^{\circ}[/tex] below the horizontal
The stone thrown upward will spend more time in the air because it travels more distance than the one thrown downwards.
The stone thrown upwards faces deceleration due to the gravity because it goes opposite to the gravity initially, then reaches a velocity zero for a moment and then falls freely from a greater height.
While the second stone posses an initial velocity downward in the direction of the gravity and which further increases its velocity and it travels a short distance.
A block with mass m is pulled horizontally with a force F_pull leading to an acceleration a along a rough, flat surface.
Find the coefficient of kinetic friction between the block and the surface.
Answer:
[tex]\mu_k=\frac{a}{g}[/tex]
Explanation:
The force of kinetic friction on the block is defined as:
[tex]F_k=\mu_kN[/tex]
Where [tex]\mu_k[/tex] is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:
[tex]N=mg\\F_k=F=ma[/tex]
Replacing this in the first equation and solving for [tex]\mu_k[/tex]:
[tex]ma=\mu_k(mg)\\\mu_k=\frac{a}{g}[/tex]
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.
Explanation:
Boyle law states that the pressure is inversely proportional to volume if we kept the temperature constant.
[tex]P_{1} V_{1} = P_{2} V_{2}[/tex] (1)
here [tex]P_{1}[/tex] and [tex]V_{1}[/tex] are the initial pressure and volume.
[tex]P_{2}[/tex] and [tex]V_{2}[/tex] are the final pressure and volume respectively.
now
[tex]P_{1} = p\\V_{1} = v\\P_{2} = ?\\V_{2} = \frac{v}{3}[/tex]
by putting these values in equation 1.
hence it is proved that " if the volume is decreased to one third of its original value then the pressure will be three times larger then its initial value".
From the top of a cliff overlooking a lake, a person throws two stones. The two stones have identical initial speeds of v0 = 13.4 m/s and are thrown at an angle θ = 30.1°, one below the horizontal and one above the horizontal. What is the distance between the points where the stones strike the ground?
Answer:
X = 15.88 m
Explanation:
Given:
Initial Velocity V₀ = 13.4 m/s
θ = 30.1 °
g = 9.8 m/s²
To Find horizontal distance let "X" we have to time t first.
so from motion 2nd equation at Height h = 0
h = V₀y t + 1/2 (-g) t ² (ay = -g)
0 = 13.4 sin 30.1° t - 0.5 x 9.81 x t² (V₀y = V₀ Sin θ)
⇒ t = 1.37 s
Now For Horizontal distance X, ax =0m/s²
X = V₀x t + 1/2 (ax) t ²
X = 13.4 m × cos 30.1° x 1.37 s + 0
X = 15.88 m
To find the distance between the points where the stones strike the ground, analyze the motion of each stone separately and find the horizontal displacements.
Explanation:To find the distance between the points where the stones strike the ground, we can analyze the motion of each stone separately. For the stone thrown below the horizontal, we can use the equations of motion in the x and y directions to find the time of flight and the horizontal displacement. For the stone thrown above the horizontal, we can use the same approach. Finally, we can subtract the two horizontal displacements to find the distance between the points where the stones strike the ground.
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