The unknown capacitance C connected parallelly with a 30-μF capacitor initially charged at 80 V is calculated to be 90 μF. After being connected, the total potential difference across both capacitors reduces to 20 V.
Explanation:The question deals with capacitors and how they function in a circuit. A capacitor is an electronic component that stores electrical energy and releases it in the circuit when necessary. When a charged 30-μF capacitor is connected across an initially uncharged capacitor, the voltage or potential difference across the connected capacitors will equalize. In this case, the final potential difference is noted as 20 V.
We know that for capacitors connected in parallel, the total charge stored is the sum of the charges stored in each capacitor, i.e., Q = Q₁ + Q₂, where Q is the total charge, and Q₁ and Q₂ are the charges stored in capacitor 1 (the 30-μF capacitor) and 2 (unknown capacitance C), respectively.
By using the formula Q = CV (charge = capacitance × voltage), where C is the capacitance and V is the voltage, we understand that Q₁ (charge on the 30 uF capacitor) is initially 30 μF × 80 V = 2400 μC (micro coulombs). After connecting the uncharged capacitor, the voltage drops to 20V, thus the final charge on the 30 uF capacitor becomes 30 μF × 20 V = 600 μC. The remaining charge must then be stored in the previously uncharged capacitor.
So, the charge on the unknown capacitor C would be the difference i.e., 2400 μC - 600 μC = 1800 μC. Now using the Q = CV formula, we can calculate the unknown capacitance C. V here is 20V as capacitors connected in parallel have the same voltage. So, C = Q / V = 1800 μC / 20 V = 90 μF. Thus, the unknown capacitance is 90 μF.
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(a) If TH = 1100 K and TC = 400 K, what is the thermal efficiency?
Answer:
63.6 %
Explanation:
TH = 1100 K , Tc = 400 k
Efficiency is given by
η = 1 - Tc / TH
η = 1 - 400 / 1100
η = 1 - 0.36
η = 0.636
η = 63.6 %
When ________ is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system? pressure and volume temperature temperature and volume pressure volume
Answer:
temperature
Explanation:
When temperature is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system.
When pressure is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system
What is enthalpy?Enthalpy, the sum of the internal energy and the product of the pressure and volume of a thermodynamic system. Enthalpy is an energy-like property or state function—it has the dimensions of energy (and is thus measured in units of joules or ergs),
and its value is determined entirely by the temperature, pressure, and composition of the system and not by its history. In symbols, the enthalpy, H, equals the sum of the internal energy, E, and the product of the pressure, P, and volume, V, of the system: H = E + PV.
According to the law of energy conservation, the change in internal energy is equal to the heat transferred to, less the work done by, the system.
If the only work done is a change of volume at constant pressure, the enthalpy change is exactly equal to the heat transferred to the system
Hence when pressure is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system
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What was the speed of a space shuttle that orbited Earth at an altitude of 1482 km?
Answer:
v = 7121.3 m/s
Explanation:
As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed
so here we will have
[tex]\frac{mv^2}{r} = \frac{GMm}{r^2}[/tex]
here we know that
r = orbital radius = 6370 km + 1482 km
[tex]r = 7.852 \times 10^6 m[/tex]
also we know that
[tex]M = 5.97 \times 10^{24} kg[/tex]
now we will have
[tex]v^2 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.852 \times 10^6}[/tex]
[tex]v^2 = 5.07 \times 10^7[/tex]
[tex]v = 7121.3 m/s[/tex]
The speed of the space shuttle that orbited the earth at an altitude of 1482km will be [tex]V=7121.3\dfrac{m}{s}[/tex]
What will be the speed of a space shuttle that orbited Earth at an altitude of 1482 km?
As we know that the centripetal force for the space shuttle is due to the gravitational force of the earth due to which it will rotate in a circular path with constant speed
so here we will have
[tex]\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}[/tex]
[tex]V^2=\dfrac{GM}{r}[/tex]
here we know that
[tex]\rm r= orbital \ radius =6370+1482=7852 \ km[/tex]
[tex]r=7.852\times10^6\ m[/tex]
mass of earth [tex]M=5.97\times10^{24}\ kg[/tex]
Gravitational constant [tex]G=6.67\times10^{-11}[/tex]
By putting all the values we get
[tex]V^2=\dfrac{(6.67\times10^{-11} )(5.97\times10^{24})}{7.852\times10^{6}}[/tex]
[tex]V^2=5.07\times10^7[/tex]
[tex]V=7121.3 \dfrac{m}{s}[/tex]
Thus the speed of the space shuttle that orbited the earth at an altitude of 1482km will be [tex]V=7121.3\dfrac{m}{s}[/tex]
what Is the gravitational force between a roast beef sandwich (m = .05 kg) and a hungry tarantula (m = 87 grams) if they are 17 cm apart?
Answer:
Gravitational force, [tex]F=1.003\times 10^{-11}\ N[/tex]
Explanation:
It is given that,
Mass of roast beef sandwich, m₁ = 0.05 kg
Mass of hungry tarantula, m₂ = 87 gm = 0.087 kg
Distance between two objects, d = 17 cm = 0.17 m
We need to find the gravitational force between them. It is given by :
[tex]F=G\dfrac{m_1m_2}{d^2}[/tex]
[tex]F=6.67\times 10^{-11}\times \dfrac{0.05\ kg\times 0.087\ kg}{(0.17\ m)^2}[/tex]
[tex]F=1.003\times 10^{-11}\ N[/tex]
So, the gravitational force between two objects is [tex]F=1.003\times 10^{-11}\ N[/tex]. Hence, this is the required solution.
A piece of aluminum has a volume of 4.89 x 10-3 m3. The coefficient of volume expansion for aluminum is = 69 x 10-6(C°)-1. The temperature of this object is raised from 19.1 to 357 oC. How much work is done by the expanding aluminum if the air pressure is 1.01 x 105 Pa?
Answer:
11.515 Joule
Explanation:
Volume of aluminium = V = 4.89×10⁻³ m³
Coefficient of volume expansion for aluminum = α = 69×10⁻⁶ /°C
Initial temperature = 19.1°C
Final temperature = 357°C
Pressure of air = 1.01×10⁵ Pa
Change in temperature = ΔT= 357-19.1 = 337.9 °C
Change in volume
ΔV = αVΔT
⇒ΔV = 69×10⁻⁶×4.89×10⁻³×337.9
⇒ΔV = 114010.839×10⁻⁹ m³
Work done
W = PΔV
⇒W = 1.01×10⁵×114010.839×10⁻⁹
⇒W = 11.515 J
∴ Work is done by the expanding aluminum is 11.515 Joule
The heating of the filament is what causes the light production (photon emission), and heating is caused by the current in the light bulb. Why isn’t it a good idea to drive light bulbs using constant current sources rather than constant voltage? (Hint: think about the answer to Question 4).
Answer:
explained
Explanation:
Yes, the heating of filament is what causes the light production (photon emission), and this heating is caused because of current in the light bulb
(H= i^2*R*t i=current, H= heat, t= time and R= resistance).But using constant current source is not a good idea because in constant current source resistance is very low that can cause short circuit and ultimately fusing it. Whereas in constant voltage source current adjusts itself and prevents fusing because of high resistance in the circuit.
Final answer:
Using a constant current source for light bulbs is not ideal because the filament's resistance changes as it heats up, which would lead to inefficient operation and potential damage. A constant voltage source is preferred as it adapts better to the changing resistance, ensuring a stable operation.
Explanation:
The heating of the filament in light bulbs, which leads to light production, is inherently linked to the current flowing through the filament. Using a constant current source instead of a constant voltage source is not advisable due to the changing resistance of the filament. As the filament heats up, its resistance increases significantly, which, according to Ohm's law (V=IR), would necessitate a higher voltage to maintain the constant current. Initially, when the light bulb is turned on and the filament is at a lower temperature, the required voltage to maintain a constant current would be lower. As the temperature and resistance rise, so would the required voltage to maintain that current. This varying voltage could lead to inefficient operation and potentially damage the bulb or reduce its lifespan. The constant voltage approach is preferred as it naturally adapts to the changing resistance of the filament, providing a more stable and predictable operation of the bulb.
Torque is dependent on the angle between the force applied and the length of the level arm. When is the torque maximum? Not enough information. θ = 90 θ = 0 θ = 45
Answer:
90 degree
Explanation:
According to the formula of torque
torque = force x displacement x Sine of angle between force and displacement
So, for the maximum torque, the value of Sin theta should be maximum.
the maximum value of Sin theta is 1.
that means the value of theta is 90 degree.
Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the magnetic field at a point in the plane of the wires and 10 cm from each wire is 4.0 μT. What is the larger of the two currents?
Answer:
3A is the larger of the two currents.
Explanation:
Let the currents in the two wires be I₁ and I₂
given:
Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T
Distance, R = 10cm = 0.1m
Ratio of the current = I₁ : I₂ = 3 : 1
Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as
[tex]B = \frac{\mu_oI}{2\pi R}[/tex]
Where [tex]\mu_o[/tex] is the magnitude constant = 4π×10⁻⁷ H/m
Thus, the magnitude of a magnetic field due to I₁ will be
[tex]B_1 = \frac{\mu_oI_1}{2\pi R}[/tex]
[tex]B_2 = \frac{\mu_oI_2}{2\pi R}[/tex]
given,
B = B₁ - B₂ (since both the currents are in the same direction and parallel)
substituting the values of B, B₁ and B₂
we get
4.0×10⁻⁶T = [tex]\frac{\mu_oI_1}{2\pi R}[/tex] - [tex]\frac{\mu_oI_2}{2\pi R}[/tex]
or
4.0×10⁻⁶T = [tex]\frac{\mu_o}{2\pi R}\times (I_1-I_2 )[/tex]
also
[tex]\frac{I_1}{I_2} = \frac{3}{1}[/tex]
⇒[tex]I_1 = 3\times I_2[/tex]
substituting the values in the above equation we get
4.0×10⁻⁶T = [tex]\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)[/tex]
⇒[tex]I_2 = 1A[/tex]
also
[tex]I_1 = 3\times I_2[/tex]
⇒[tex]I_1 = 3\times 1A[/tex]
⇒[tex]I_1 = 3A[/tex]
Hence, the larger of the two currents is 3A
A flywheel in the form of a uniformly thick disk of radius 1.23 m has a mass of 93.6 kg and spins counterclockwise at 369 rpm . Calculate the constant torque required to stop it in 2.25 min .
Answer:
20.26 Nm
Explanation:
r = 1.23 m , m = 93.6 kg, w = 0, f0 = 369 rpm = 369 / 60 = 6.15 rps
w0 = 2 x 3.14 x 6.15 = 38.622 rad/s
t = 2.25 min = 2.25 x 60 = 135 second
Moment of inertia = 1/2 m r^2 = 0.5 x 93.6 x 1.23 x 1.23 = 70.8 kg m^2
use first equation of motion for rotational motion
w = w0 + α t
0 = 38.622 - α x 135
α = 0.286 rad/s^2
torque = moment of inertia x angular acceleration
Torque = 70.8 x 0.286 = 20.26 Nm
How much energy is dissipated as heat during a two-minute time interval by a 1.5- kΩ resistor which has a constant 20- V potential difference across its leads?
Answer:
32 J
Explanation:
Power is V²/R = 20²/1500 = 4/15 . . . watts
In 120 seconds, the energy is ...
(4/15 J/s)×(120 s) = 32 J
In a 2-minute period 32 joules are dissipated as heat.
We calculate the power dissipation rate of the 1.5-kΩ resistor under a constant potential difference of 20- V, which is 0.27 Watts or 0.27 Joules per second. Multiplying this rate by the time interval of two minutes (or 120 seconds), we find the total energy dissipated as heat, which is 32.4 Joules.
Explanation:To understand the amount of energy dissipated during the given time interval, we first need to determine the power of the 1.5-kΩ resistor. This can be done using the equation P = V²/R, where V is the voltage, and R is the resistance. For this case, we have P = (20 V) ² / (1.5 kΩ) = 0.27 Watts.
Since the power dissipation rate is 0.27 Watts, which is equivalent to 0.27 Joules per second, we can now find out the energy dissipated during the two-minute interval using the equation E = Pt. Here, t represents the time in seconds, so t = 2 minutes * 60 seconds/minute = 120 seconds. Therefore, substituting the known values, we get E = (0.27 J/s)(120 s) = 32.4 Joules.
So, this is the total energy dissipated as heat by the 1.5- kΩ resistor under a constant potential difference of 20- V during the two-minute interval.
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Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 100 cm2, how many turns do you need?
Answer:
The number of turns is 64.
Explanation:
Given that,
Magnetic field = 0.050 T
Area of coil = 100 cm²
Frequency = 60 Hz
Output voltage emf= 12 V
We need to calculate the number of turns
Using formula of induced emf
[tex]emf =NAB\omega[/tex]
[tex]N=\dfrac{emf}{A\times B\times2\pi f}[/tex]
[tex]N=\dfrac{12}{0.01\times0.050\times2\times3.14\times60}[/tex]
[tex]N =63.6 = 64\ turns[/tex]
Hence, The number of turns is 64.
Answer:
You need 63.66 turns.
Explanation:
The number of turns of a magnetic field is given by the following formula:
[tex]N = \frac{V}{S*T*2\pi f}[/tex]
In which N is the number of turns, V is the maximum output voltage, S is the area of the rotating coil, in square meters and T is the measure of the magnetic field and f is the frequency.
In this problem, we have that:
Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. This means that [tex]V = 12[/tex] and [tex]f = 60[/tex].
A uniform magnetic field of 0.050 T is available. This means that [tex]T = 0.050[/tex].
If the area of the rotating coil is 100 cm2, how many turns do you need?
This means that [tex]S = 0.01[/tex]m². So:
[tex]N = \frac{V}{S*T*2\pi f}[/tex]
[tex]N = \frac{12}{0.01*0.05*120\pi}[/tex]
[tex]N = 63.66[/tex]
You need 63.66 turns.
A resistor dissipates 0.25 W when current of 20 mA passes through it. Part A How much current would be needed for the resistor to dissipate 0.50 W? How much current would be needed for the resistor to dissipate 0.50 ? 56 mA 28 mA 40 mA 80 mA
Power dissipated by a resistor = (current)² x (resistance).
0.25 W = (0.02 A)² x (resistance)
This resistor = (0.25 W) / (0.02 A)²
This resistor = 625 ohms
If we want it to dissipate 0.5 W, then
0.5 W = (current)² x (625 ohms)
Current = √(0.5/625)
Current = √(0.0008)
Current = 28.28 mA
============================
A slightly easier way:
Since the power is = I²R, it grows in proportion to (current)² .
We want to double the power dissipated, so we only need to increase the current by the factor of √2 .
(20 mA) x (√2) = 28.28 mA
Final answer:
Using the power dissipation formula, we calculate the resistance as 625 ohms. To dissipate 0.50 W, a current of sqrt(0.50 W / 625 ohms) is required, which equals 28 mA.
Explanation:
To determine how much current would be needed for a resistor to dissipate 0.50 W, we'll use the formula for electric power dissipation in a resistor which is P = I^2 R, where P is the power in watts, I is the current in amperes, and R is the resistance in ohms. In the given example, a resistor dissipates 0.25 W when a 20 mA (or 0.02 A) current passes through it. With P = I^2 R, we can derive R as R = P / I^2 which gives a resistance value through this initial condition.
The first step is to calculate the resistance of the resistor using the initial power dissipation value:
R = 0.25 W / (0.02 A)^2 = 0.25 / 0.0004 = 625 ohms.
Now, to find the current for 0.50 W power dissipation, we use
I = sqrt(P / R) = sqrt(0.50 W / 625 ohms) = sqrt(0.0008) = 0.0283 A or 28.3 mA.
Therefore, 28 mA of current is needed for the resistor to dissipate 0.50 W.
A powerful grasshopper launches itself at an angle of 45° above the horizontal and rises to a maximum height of 1.01 m during the leap. With what speed v did it leave the ground? Neglect air resistance. Hint: Use conservation of energy. A. 6.29 m/s B. 7.15 m/s C. 5.98 m/s D. 6.72 m/s E. 5.37 m/s
Answer:
6.29 m/s option (A)
Explanation:
theta = 45 degree, H = 1.01 m
let v be the launch speed
Use the formula for the maximum height for the projectile
H = v^2 Sin^θ / 2g
1.01 = v^2 x Sin^2(45) / (2 x 9.8)
1.01 = 0.0255 v^2
v^2 = 39.59
v = 6.29 m/s
The initial velocity of the grasshopper is 6.29 m/s.
Initial velocity of the grasshopper
The Initial velocity of the grasshopper is calculated from the following kinematic equation.
[tex]H = \frac{v_0^2 sin^2 \theta}{2g}[/tex]
where;
H is the maximum heightv is the initial velocity[tex]v_0^2 = \frac{2gH}{sin^2\theta} \\\\v_0^2 = \frac{2 \times 9.8 \times 1.01 }{(sin45)^2} \\\\v_0^2 = 39.6\\\\v_0 = 6.29 \ m/s[/tex]
Thus, the initial velocity of the grasshopper is 6.29 m/s.
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A thin, circular disc is made of lead and has a radius of 0.250 cm at 20.0 °C. Determine the change in the area of the circle if the temperature is increased to 800.0 °C. The coefficient of linear thermal expansion for lead is 29.0 x 10^-6/C°.
Answer:
The change in the area of the circle is [tex]8.88\times10^{-7}\ m^2[/tex]
Explanation:
Given that,
Radius = 0.250 cm
Temperature = 20.0°C
Final temperature =800.0°C
Coefficient of linear thermal expansion for lead[tex]\alpha =29.0\times10^{-6}\ /\°C [/tex]
We calculate the change in temperature,
[tex]\Delta T=800.0-20.0=780^{\circ}[/tex]
Now, We calculate the area of the disc
[tex]A = \pi r^2[/tex]
Put the value into the formula
[tex]A=3.14\times(2.5\times10^{-3})^2[/tex]
[tex]A =1.9625\times10^{-5}\ m^2[/tex]
We need to calculate the areal expansion
[tex]\Delta A=2\alpha\times A\times\Delta T[/tex]
[tex]\Delta A=2\times29.0\times10^{-6}\times1.9625\times10^{-5}\times780[/tex]
[tex]\Delta A=8.88\times10^{-7}\ m^2[/tex]
Hence, The change in the area of the circle is [tex]8.88\times10^{-7}\ m^2[/tex]
Final answer:
The change in area of the lead circular disc can be calculated using the formulas for linear thermal expansion. The change in radius is calculated using the coefficient of linear expansion and the change in temperature. The change in area is then calculated using the change in radius and the original radius.
Explanation:
To determine the change in area of the circular disc made of lead, we need to calculate the change in its radius using the coefficient of linear thermal expansion. The formula for linear thermal expansion is given by AL = a * L * AT, where AL is the change in length, a is the coefficient of linear expansion, L is the original length, and AT is the change in temperature.
In this case, we are interested in the change in radius, so we can use the formula AR = a * R * AT, where AR is the change in radius and R is the original radius.
Substituting the given values, we have:
AR = (29.0 × 10^-6/°C) * (0.250 cm) * (800.0 °C - 20.0 °C)
AR = 0.00145 cm
The change in area can be calculated using the formula AE = 2π * R * AR. Substituting the values, we have:
AE = 2π * (0.250 cm) * (0.00145 cm)
AE = 2.31 x 10^-3 cm²
Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 61.3 N, Jill pulls with 83.9 N in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 137 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Find the magnitude of the net force the people exert on the donkey.
Answer:
[tex]F_{net} = 220.8 N[/tex]
Explanation:
It is pulled by three forces as given below
1. Jack pulls directly ahead of the donkey with a force of 61.3 N,
2. Jill pulls with 83.9 N in a direction 45° to the left, and
3. Jane pulls in a direction 45° to the right with 137 N.
Now net force directly in forward direction given as
[tex]F_x = 61.3 N + 83.9 cos45 + 137cos45[/tex]
[tex]F_x = 217.5 N[/tex]
Now similarly in perpendicular to this we have
[tex]F_y = 137 sin45 - 83 sin45 [/tex]
[tex]F_y = 38.2 N[/tex]
Now net force is given by them
[tex]F_{net} = \sqrt{F_x^2 + F_y^2}[/tex]
[tex]F_{net} = \sqrt{217.5^2 + 38.2^2}[/tex]
[tex]F_{net} = 220.8 N[/tex]
A motorboat is moving at 4.0 m/s when it begins to accelerate at 1.0 m/s2. To the nearest tenth of a second, how long does it take for the boat to reach a speed of 17.0 m/s?
Answer:
13 sec
Explanation:
Hello
by definition the acceleration is a vector derived magnitude that indicates the speed variation per unit of time
[tex]a=\frac{V_{f}-V_{i}}{t_{f}-t_{i}}\\a(t_{f}-t_{i})={V_{f}-V_{i} }}\\t_{f}-t_{i}=\frac{V_{f}-V_{i}}{a} \\\\t_{f}=\frac{V_{f}-V_{i}}{a}+t_{i} \\\\[/tex]
Let
Vi=4.0 m/s
a=1.0 m/s2
Vf=17.0 m/s
ti= (0)sec
t2= unknown
[tex]t_{f}=\frac{V_{f}-V_{i} }{a}+t_{i}\\t_{f}=\frac{17-4 }{1}+0\\t_{f}=\ 13\ sec\\[/tex]
Answer: 13 sec
I hope it helps
A whistle of frequency 589 Hz moves in a circle of radius 54.6 cm at an angular speed of 16.1 rad/s. What are (a) the lowest and (b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle? (Take the speed of sound in air to be 343 m/s.)
Answer:
[tex]f_{min} = 574.3 Hz[/tex]
[tex]f_{max} = 604.5 Hz[/tex]
Explanation:
As per Doppler's effect of sound we know that when source and observer moves relative to each other then the frequency of sound observed by the observer is different from real frequency of sound.
As the source is moving here in this case so the frequency is given as
[tex]f = f_o\frac{v}{v\pm v_s}[/tex]
part a)
for lowest frequency we will have
[tex]f_{min} = 589(\frac{343}{343 + R\omega})[/tex]
here we know that
R = 54.6 cm
[tex]\omega = 16.1 rad/s[/tex]
now we have
[tex]f_{min} = 589(\frac{343}{343 + 0.546(16.1)})[/tex]
[tex]f_{min} = 574.3 Hz[/tex]
part b)
for maximum frequency we will have
[tex]f_{max} = 589(\frac{343}{343 - R\omega})[/tex]
here we know that
R = 54.6 cm
[tex]\omega = 16.1 rad/s[/tex]
now we have
[tex]f_{max} = 589(\frac{343}{343 - 0.546(16.1)})[/tex]
[tex]f_{max} = 604.5 Hz[/tex]
A motorcycle moving at 13.2 m/s increases in speed to 25.7 m/s over a period of 8.6 s. If the motorcycle's mass is 352 kg, what net force in Newtons acts on the motorcycle?
Answer:
510.4 N
Explanation:
u = 13.2 m /s, v = 25.7 m/s, t = 8.6 s, m = 352 kg
Use first equation of motion
v = u + a t
a = (25.7 - 13.2) / 8.6 = 1.45 m/s^2
Use Newton's second law
F = m a = 352 x 1.45 = 510.4 N
Frequency and velocity of a particle in simple harmonic motion Problem in that the amplitude is 0.24 in. and the maximum acceleration is 225 ft/s maximum velocity of the particle ist/s, and the frequency of its motion s z Print 20r
Answer:
velocity maximum = 21.6 ft /s
frequency = 16.88 Hz
Explanation:
Amplitude, A = 0.24 in = 0.02 ft
maximum acceleration, a = 225 ft/s\
The formula for maximum acceleration is
a = ω² A
225 = ω² x 0.02
ω² = 11250
ω = 106.06 rad/s
Maximum velocity, v = ω A
v = 106.06 x 0.02 = 2.1 ft/s
Let f be the frequency
ω = 2 x 3.14 x f
f = 106.06 / (2 x 3.14) = 16.88 Hz
velocity maximum = 21.6 ft /s
frequency = 16.88 Hz
Amplitude, A = 0.24 in = 0.02 ft
maximum acceleration, a = 225 ft/s
The formula for maximum acceleration is
a = ω² A
225 = ω² x 0.02
ω² = 11250
ω = 106.06 rad/s
Maximum velocity, v = ω A
v = 106.06 x 0.02 = 2.1 ft/s
Let f be the frequency
ω = 2 x 3.14 x f
f = 106.06 / (2 x 3.14) = 16.88 Hz
When 7.9×1014 Hz light shines on a plate of unknown material, it is determined that the stopping potential is 1.1 V. Determine the work function (in eV) & the cutoff frequency. Wo = eV fo =
Answer:
W = 2.158 eV
fo = 5.23 x 10^14 Hz
Explanation:
f = 7.9 x 10^14 Hz, Vo = 1.1 V
Let W be the work function.
Use the Einstein equation
Energy = W + eVo
hf = W + eVo
where, h is the Plank's constant and e be the electronic charge.
W = hf - eVo
W = (6.6 x 10^-34 x 7.9 x 10^14) - (1.6 x 10^-19 x 1.1)
W = 5.214 x 10^-19 - 1.76 x 10^-19 = 3.454 x 10^-19 J
W = (3.454 x 10^-19) / (1.6 x 10^-19) = 2.158 eV
Let fo be the cut off frequency
W = h fo
fo = W / h = (3.454 x 10^-19) / (6.6 x 0^-34) = 5.23 x 10^14 Hz
A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage?
The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite each other. Between the electrodes but not near their edges, the electric field strength is 2.5 × 106 N/C. What is the magnitude of the charge on each electrode? (ε0 = 8.85 × 10-12 C2/N ∙ m2)
Answer:
1.77 x 10^-8 C
Explanation:
Let the surface charge density of each of the plate is σ.
A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2
d = 2 mm
E = 2.5 x 10^6 N/C
ε0 = 8.85 × 10-12 C2/N ∙ m2
Electric filed between the plates (two oppositively charged)
E = σ / ε0
σ = ε0 x E
σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2
The surface charge density of each plate is ± σ / 2
So, the surface charge density on each = ± 22.125 x 10^-6 / 2
= ± 11.0625 x 10^-6 C/m^2
Charge on each plate = Surface charge density on each plate x area of each plate
Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C
Two particles, one with charge -5.45 uC and one with charge 3.55 uC, are 4.34 cm apart. What is the magnitude of the force that one particle exerts on the other?
Answer:
[tex]F = 92.45 N[/tex]
Explanation:
As we know that the force between two charge particles is given by
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we know that
[tex]q_1 = 3.55 \mu C[/tex]
[tex]q_2 = 5.45 \mu C[/tex]
now the distance between the two charges is
r = 4.34 cm
now from the formula of electrostatic force we will have
[tex]F = \frac{(9\times 10^9)(3.55 \mu C)(5.45 \mu C)}{0.0434^2}[/tex]
[tex]F = 92.45 N[/tex]
A battery with an emf of 12 V and an internal resistance of 1 Ω is used to charge a battery with an emfof 10 V and an internal resistance of 1 Ω. The current in the circuit is : A) 2 A B) 1 A C) 4A D) 11A E) 22 A
Answer:
Option B
Explanation:
The net emf in the circuit
E = 12 - 10 = 2 V
Total effective resistance,
r = 1 + 1 = 2 ohm
By using Ohm's law
E = I × R
I = 2 / 2 = 1 A
To find the current in the circuit, use Ohm's Law and solve two equations.
Explanation:The current in the circuit can be found using Ohm's Law: V = IR, where V is the voltage, I is the current, and R is the resistance.
In this case, the emf of the first battery is 12 V with an internal resistance of 1 Ω. Let's assume the current in the circuit is I1. So, the terminal voltage across the first battery can be calculated using Ohm's Law: 12 = I1 * (1 + 1).
Similarly, for the second battery with emf of 10 V and internal resistance of 1 Ω, the terminal voltage can be calculated as 10 = (I1 - I) * (1 + 1), where I is the current flowing in the second battery.
Now, we need to solve these two equations to find the value of I, which is the current in the circuit.
A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold welghs 198 N and is 3.2 m long. what is the tension in each rope whien the 6oo-N worker stands 1.12 m from one end? smaller tension arger tension
Answer:
[tex]T_2 = 309 N[/tex]
[tex]T_1 = 489 N[/tex]
Explanation:
As we know that total tension in both the ropes is counter balancing the weight of scaffold and worker both
so here we will have
[tex]T_1 + T_2 = (m + M)g[/tex]
now we have
[tex]T_1 + T_2 = 198 + 600 = 798 N[/tex]
now we also know that net torque due to both tension force in the string with respect to the position of worker must be zero so that platform will remain in equilibrium and horizontal in position
so here we will have
[tex]T_1(1.12) + (198)(1.6 - 1.12) = T_2(3.2 - 1.12)[/tex]
[tex]T_1+ 84.86 = 1.86 T_2[/tex]
now from above two equations we will have
[tex](1.86 T_2 - 84.86) + T_2 = 798 [/tex]
[tex]T_2 = 309 N[/tex]
also we have
[tex]T_1 = 489 N[/tex]
Final answer:
To calculate the tension in each rope supporting a scaffold with a window washer, the principles of torques and equilibrium are applied. By considering the system's equilibrium, equations based on the torques created by the worker's weight, the scaffold's weight, and the tensions in the ropes can be set up and solved.
Explanation:
To solve for the tension in each rope when a window washer stands on a scaffold, we use the concept of torques and equilibrium. The entire system is in equilibrium, meaning the sum of torques around any pivot point is zero. Considering the window washer weighing 600 N and standing 1.12 m from one end of a 3.2 m scaffold that weighs 198 N, we choose the pivot point at one end of the scaffold for easier calculations.
Let T1 be the tension in the rope closest to the worker and T2 be the tension in the other rope. The force due to the worker creates a clockwise torque, and the force due to the scaffold creates a clockwise torque as well, whereas the tensions in the ropes create counterclockwise torques. The torques are calculated by multiplying the force by the distance from the pivot point.
To find T1 and T2, we set up the equilibrium condition for torques and solve the equations considering the weight of the worker and the scaffold acting at their respective centers of mass. The specific calculations require numerical values that depend on the distances given and the gravitational force.
Ultimately, by solving these equations, we find the tensions T1 and T2 that support the system in equilibrium.
One billiard ball is shot east at 2.00 m/s. A second, identical billiard ball is shot west at 1.00 m/s. The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90° and sending it north at 1.41 m/s. What are the speed and direction of the first ball after the collision?
Answer:
Velocity is 1.73 m/s along 54.65° south of east.
Explanation:
Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.
Here momentum is conserved.
Initial momentum = Final momentum
Initial momentum = m x 2i + m x (-1)i = m i
Final momentum = m x v + m x 1.41 j = mv + 1.41 m j
Comparing
mi = mv + 1.41 m j
v = i - 1.41 j
Magnitude of velocity
[tex]v=\sqrt{1^2+(-1.41)^2}=1.73m/s[/tex]
Direction,
[tex]\theta =tan^{-1}\left ( \frac{-1.41}{1}\right )=-54.65^0[/tex]
Velocity is 1.73 m/s along 54.65° south of east.
Using the law of conservation of momentum, one can deduce the speed and direction of the first ball after the collision. It's found to be traveling east at 1 m/s.
Explanation:The scenario described is an example of a two-dimensional collision. In such collision, the law of conservation of momentum applies both in the east-west direction (x-axis) and the north-south direction (y-axis).
With the given velocities before the collision, the total momentum in the x-axis before the collision is: momentum_east = mass*velocity = m*2.00 m/s, and momentum_west = m*(-1.00) m/s. Therefore, total momentum in the x-axis before the collision = m*2.00 m/s + m*(-1.00) m/s = m m/s.
After the collision, the first ball keeps moving in the east-west direction (since the second ball is deflected north), but we don't know its velocity, let's call it v1. Applying the conservation of momentum after the collision in the x-direction, we get: total momentum = m*v1 + 0 (since the second ball no longer moves in the east-west direction) = m m/s. From this, we can solve for v1 and find that v1 = 1 m/s east. Thus, the first billiard ball is traveling east at 1 m/s after the collision.
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The crankshaft in a race car goes from rest to 3000 rpm in 3.0 s . (a) What is the crankshaft's angular acceleration?
(b) How many revolutions does it make while reaching 3000 rpm ?
Answer:
75 rotations
Explanation:
f0 = 0, f = 3000 rpm = 50 rps, t = 3 s
(a) use first equation of motion for rotational motion
w = w0 + α t
2 x 3.14 x 50 = 0 + α x 3
α = 104.67 rad/s^2
(b) Let θ be the angular displacement
use second equation of motion for rotational motion
θ = w0 t + 1/2 α t^2
θ = 0 + 0.5 x 104.67 x 3 x 3
θ = 471.015 rad
The angle turn in one rotation is 2 π radian.
Number of rotation = 471.015 / (2 x 3.14) = 75 rotations
You are in downtown Chicago (where streets run N-S and E-W). You started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station. Length of one block is 100 [m]. What is the magnitude and direction of your displacement from the start position?
Answer:
Displacement is 565.69 m at 45° west of north
Explanation:
Let north represent positive y axis and east represent positive x axis.
We have journey started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station.
3 blocks toward north = 300 j m
4 blocks toward west = -400 i m
1 blocks toward north = 100 j m
Total displacement = -400 i + 400 j m
Magnitude
[tex]s=\sqrt{(-400)^2+400^2}=565.69m[/tex]
Direction,
[tex]\theta =tan^{-1}\left ( \frac{400}{-400}\right )=135^0[/tex]
Direction is 45° west of north.
Displacement is 565.69 m at 45° west of north
How much heat would be needed to completely evaporate 31.5 g of boiling water at a temperature of 100 "C? Express your answer in units of joules.
The heat needed to evaporate 31.5 g of boiling water is approximately 7,150 J.
Explanation:To calculate the heat needed to completely evaporate water, we use the equation Q = m × ΔHvap, where Q is the heat, m is the mass, and ΔHvap is the heat of vaporization.
The heat of vaporization for water is approximately 40.7 kJ/mol. To find the heat for 31.5 g of water, we need to convert the mass to moles by dividing by the molar mass of water (18.015 g/mol).
Using the equation, Q = (31.5 g ÷ 18.015 g/mol) × (40.7 kJ/mol × 1000 J/kJ), the answer is approximately 7,150 J.
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Final answer:
To evaporate 31.5 g of boiling water at 100°C, 70,875 joules of heat are needed. This is determined by multiplying the mass by the heat of vaporization for water, which is 2,250 J/g.
Explanation:
To calculate the amount of heat required to completely evaporate 31.5 g of boiling water at 100°C, the heat of vaporization of water is needed. The heat of vaporization of water is 2,250 J/g. To find the total heat we will multiply the mass of the water by the heat of vaporization:
Heat required = mass × heat of vaporization
Heat required = 31.5 g × 2,250 J/g
Heat required = 70,875 J
Therefore, 70,875 joules of heat would be needed to completely evaporate 31.5 g of boiling water at 100°C.
For fully developed laminar pipe flow in a circular pipe, the velocity profile is u(r) = 2(1-r2 /R2 ) in m/s, where R is the inner radius of the pipe. Assuming the pipe diameter is 4 cm, find the maximum and average velocities in the pipe as well as the volume flow rate
Answer:
a) [tex]v_{max} = 2\ \textup{m/s}[/tex]
b) [tex]v_{avg} = 1\ \textup{m/s}[/tex]
c) Q = 1.256 × 10⁻³ m³/s
Explanation:
Given:
The velocity profile as:
[tex]u(r) = 2(1-\frac{r^2}{R^2} )[/tex]
Now, the maximum velocity of the flow is obtained at the center of the pipe
i.e r = 0
thus,
[tex]v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )[/tex]
or
[tex]v_{max} = 2\ \textup{m/s}[/tex]
Now,
[tex]v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}[/tex]
or
[tex]v_{avg} = \frac{2}}{2}\ \textup{m/s}[/tex]
or
[tex]v_{avg} = 1\ \textup{m/s}[/tex]
Now, the flow rate is given as:
Q = Area of cross-section of pipe × [tex]v_{avg}[/tex]
or
Q = [tex]\frac{\pi D^2}{4}\times v_{avg}[/tex]
or
Q = [tex]\frac{\pi 0.04^2}{4}\times 1[/tex]
or
Q = 1.256 × 10⁻³ m³/s
For fully developed laminar pipe flow in a circular pipe, the maximum velocity is 2 m/s, the average velocity is (4/3) m/s, and the volume flow rate is (16/3)π cm^3/s. The velocity profile equation and the formulas for maximum velocity, average velocity, and volume flow rate are explained in detail.
Explanation:For fully developed laminar pipe flow in a circular pipe, the velocity profile is given by the equation u(r) = 2(1-r^2/R^2) in m/s, where R is the inner radius of the pipe. To find the maximum velocity, we need to substitute r = R into the equation. The maximum velocity u_max is then equal to 2(1-1^2/R^2), which simplifies to 2 m/s.
The average velocity can be found by integrating the velocity profile equation over the entire cross-sectional area of the pipe. The cross-sectional area is given by A = πR^2, so the average velocity v_avg is equal to (1/A) times the integral of 2(1-r^2/R^2) over the range r = 0 to r = R. Simplifying the integral and dividing by A, we get v_avg = (4/3) m/s.
The volume flow rate Q can be calculated by multiplying the cross-sectional area A by the average velocity v_avg. Using the given diameter of the pipe, we can find the radius R = 2 cm. Substituting the values into the equation Q = (πR^2)(4/3), we get Q = (16/3)π cm^3/s.
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