Answer: q = 7.06 × 10^-21 C
the magnitude of the charges that make up the dipole is 7.06 × 10^-21 C
Explanation:
Given:
Torque, τ = 6.60×10^−26N⋅m
Angle made by p with a uniform electric field, θ = 90° (perpendicular)
Electric field, E = 8.50×10^4N/C
Length between dipole r = 1.10×10^−10 m
Torque acting on the dipole is given by the relation,
τ = pE sinθ....1
But,
p = qr .....2
Substituting equation 1 to 2
τ= qrEsinθ ....3
Making q the subject of formula
q = τ/rEsinθ .....4
Where;
q = magnitude of the charges that make up the dipole.
Substituting the given values into equation 4:
q = 6.60×10^−26N⋅m/(1.10×10^−10 m × 8.50×10^4N/C × sin90°)
q = 0.70588 × 10^-20 C
q = 7.06 × 10^-21 C
The magnitude of the charge making up the electric dipole, given a torque of 6.60×10−26N⋅m when the dipole moment is perpendicular to an electric field of 8.50×104N/C, and a separation of 1.10×10−10m between the charges, is approximately 7.05×10−21 C.
Explanation:The torque (τ) on a dipole in a uniform electric field is given by the equation τ = pEsinθ, where p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field. In this case, the dipole moment is perpendicular to the electric field, so θ= 90 degrees, and sinθ= 1. The dipole moment, p, is the product of the magnitude of the charge (q) and the separation (d) between the charges, so p = qd.
Given τ = 6.60×10−26N⋅m, E = 8.50×104N/C, and d = 1.10×10−10m, we can first use the torque equation to find the dipole moment: p = τ / E = 6.60×10−26N⋅m / 8.50×104N/C = 7.76×10−31 C⋅m. Then, use p = qd to calculate the charge: q = p / d = 7.76×10−31 C⋅m / 1.10×10−10m = 7.05×10−21 C.
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By what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on each object is 5.5 N
Answer:
Distance between both the object will be 404.51 m
Explanation:
We have given charge on two objects [tex]q_1=q_2=1C[/tex]
Coulomb force between the two objects is given F = 5.5 N
We have to fond the distance between both the object so that force between them is 5.5 N
According to coulomb law force between two charge particle is [tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}[/tex]
So [tex]5.5=\frac{9\times 10^9\times 1\times 1}{r^2}[/tex]
[tex]r^2=16.36\times 10^8[/tex]
r = 404.51 m
So distance between both object will be 404.51 m
A raft is made of 14 logs lashed together. Each log is 42 cm in diameter and a length of 6.4 m. 42% of the log volume is above the water when no one is on the raft. Determine the following: the specific gravity of the logs.
Answer:
Explanation:
Given
No of logs [tex]n=14[/tex]
diameter of log [tex]d=42\ cm[/tex]
Length of log [tex]L=6.4\ m[/tex]
42 % of log volume(V) is above water when no one is on raft
so 58 % of log volume(V) is submerged in the water
Weight of 14 log
[tex]W=14\times \rho _{log}\times V\times g[/tex]
Buoyancy force on 14 logs [tex]F_b=14\times \rho _{water}\times 0.58V\times g[/tex]
as system is in equilibrium so
[tex]W=F_b[/tex]
[tex]14\times \rho _{log}\times V\times g=14\times \rho _{water}\times 0.58V\times g[/tex]
[tex]\rho _{log}=0.58\rho _{water}[/tex]
[tex]\frac{\rho _{log}}{\rho _{water}}=0.58[/tex]
Specific gravity of log [tex]=0.58[/tex]
What is the magnitude of the net gravitational force on the m1=20kgm1=20kg mass? Assume m2=10kgm2=10kg and m3=10kg.
Answer:
The net force on the 20kg mass m1 is equal to 6.09 x 10^-7 N. This force is due to the sun of the vertical components of the forces F1 and F2 of the masses m2 and m3 respectively on the mass m1.
The law of gravitational force has been applied and the use of the Pythagorean theorem also used.
Explanation:
The full solution can be found in the attachment below.
Thank you for reading this post and I hope it is helpful to you.
The magnitude of the net gravitational force on the masses can be determined using Newton's law when the distance between the two masses is known.
Magnitude of gravitational force
The magnitude of gravitaional force on the two given masses can be determined by applying Newton's law of universal gravitation as shown below;
F = Gm₁m₂/r²
where;
G is universal gravitation constantr is the distance between the two massesm₁ is the first mass m₂ is the second massThus, the magnitude of the net gravitational force on the masses can be determined using Newton's law when the distance between the two masses is known.
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A girl delivering newspapers covers her route by traveling 5.00 blocks west, 5.00 blocks north, and then 7.00 blocks east. What is her final position relative to her starting location?
Answer:
P = 2i + 5j
Therefore she is 2 blocks east and 5 blocks north.
Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks
Angle = taninverse (5/2)
Angle = 68.2°
Explanation:
Given:
Let west be negative and east be positive x axis.
Let north be the positive y axis.
5.00blocks west = -5.00 i
5.00 blocks north = 5.00 j
7.00 blocks east = 7.00i
Addition of the vector form of hee position is;
P = -5i +7i -5j
P = 2i + 5j
Therefore she is 2 blocks east and 5 blocks north.
Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks
Angle = taninverse (5/2)
Angle = 68.2°
Two thin conducting plates, each 24.0 cm on a side, are situated parallel to one another and 3.4 mm apart. If 1012 electrons are moved from one plate to the other, what is the electric field between the plates?
Answer:
E=3.307×10⁻⁴N/C
Explanation:
Given data
Length of the plate side L=24.0 cm =0.24 m
Distance between the plates d= 3.4 mm
Number of electron moves from one plate to others n=1012 electrons
Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²
Electron charge e=-1.6×10⁻¹⁹C
To find
Electric field between the plates
Solution
E=σ/ε₀
[tex]E=\frac{(Q/A)}{E}\\ E=\frac{Q}{EA}\\ E=\frac{ne}{EA}\\ E=\frac{1012*()1.6*10^{-19} }{8.85*10^{-12}(0.24)(0.24)}\\ E=3.307*10^{-4}N/C[/tex]
Two force vectors are perpendicular, that is, the angle between their directions is ninety degrees and they have the same magnitude. If their magnitudes are 655 newtons, then what is the magnitude of their sum?
Answer:
[tex]F=926.31N[/tex]
Explanation:
To calculate the magnitude of the sum of the vectors we must find their components on the x and y axes, in this case, since they are perpendicular, one vector is on the x axis and the other is on the y axis:
[tex]F=\sqrt{F_x^2+F_y^2}\\F_x=A_x+B_x=A\\F_y=A_y+B_y=B\\F=\sqrt{A^2+B^2}\\F=\sqrt{(655N)^2+(655N)^2}\\F=926.31N[/tex]
List and explain briefly similarities and differences between the electric force between two charges and the gravitational force between two masses.
Answer:
Please see below as the answers are self-explanatory.
Explanation:
Similarities1) The resultant force is along the line that joins both charges or both masses (assuming both objects can be represented as points)
2) Both type of forces obey Newton's 3rd law.
3) Both are proportional to the product of the property that is affected by the force (charges and masses)
4) Both obey an inverse - square law (consequence of our universe being three-dimensional)
Differences1) Main difference, is that while the gravitational force is always attractive, the electrostatic force can be attractive or repulsive, as there are two types of charges, which attract each other being of different type, and repel each other if they are of the same type.
2) It is possible, artificially, to block the influence of the electrostatic force, shielding a room, for instance, which is not possible for the gravitational force.
A tuned mass damper for a skyscraper consists of a mass–spring system with spring constant 0.288 M/m. What should be its mass if it’s to oscillate with a period of 5.91 s
Answer:
m = 0.255kg
Explanation:
from the formular of a mass - spring system
T = 2π√m/k
making m as the subject of formular
m = T² k/4π²
T =5.91s
k = 0.288 N/m
m = 10.059/39.489
m = 0.255kg
The velocity of P-waves in the crust is ~ 7 km/s. Of the epicenter of an M 7.6 earthquake occurred 280 km from the closest seismic station, how long does it take the P-wave to arrive at the station
Answer:
t = 40 s
Explanation:
given,
Speed of the P-wave = 7 km/s
distance of the seismic station = 280 Km
time taken by the P-wave = ?
we know,
distance = speed x time
[tex]t = \dfrac{d}{s}[/tex]
[tex]t = \dfrac{280}{7}[/tex]
t = 40 s
time taken by the P-wave to arrive at the station is equal to 40 s.
Final answer:
The P-wave takes approximately 40 seconds to arrive at the closest seismic station.
Explanation:
The P-waves in the crust travel at a velocity of approximately 7 km/s. To determine the time it takes for the P-wave to arrive at a seismic station, we can use the equation:
Time = Distance / Velocity
In this case, the distance to the epicenter of the earthquake is given as 280 km. Plugging this value into the equation, we get:
Time = 280 km / 7 km/s = 40 seconds
So, it takes approximately 40 seconds for the P-wave to arrive at the closest seismic station.
A compressor receives air at 290 K, 100 kPa and a shaft work of 5.5 kW from a gasoline engine. It is to deliver a mass flow rate of 0.01 kg/s air to a pipeline. Assuming a constant-pressure specific heat of Cp = 1.004 kJ/kg-K for the air, determine the maximum possible exit pressure of the compressor
Answer:
[tex]P_2=4091\ KPa[/tex]
Explanation:
Given that
T₁ = 290 K
P₁ = 100 KPa
Power P =5.5 KW
mass flow rate
[tex]\dot{m}= 0.01\ kg/s[/tex]
Lets take the exit temperature = T₂
We know that
[tex]P=\dot{m}\ C_p (T_2-T_1)[/tex]
[tex]5.5=0.01\times 1.005(T_2-290})\\T_2=\dfrac{5.5}{0.01\times 1.005}+290\ K\\\\T_2=837.26\ K[/tex]
If we assume that process inside the compressor is adiabatic then we can say that
[tex]\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{0.285}[/tex]
[tex]\dfrac{837.26}{290}=\left(\dfrac{P_2}{100}\right)^{0.285}\\2.88=\left(\dfrac{P_2}{100}\right)^{0.285}\\[/tex]
[tex]2.88^{\frac{1}{0.285}}=\dfrac{P_2}{100}[/tex]
[tex]P_2=40.91\times 100 \ KPa[/tex]
[tex]P_2=4091\ KPa[/tex]
That is why the exit pressure will be 4091 KPa.
In a Young's double-slit experiment, a set of parallel slits with a separation of 0.142 mm is illuminated by light having a wavelength of 576 nm and the interference pattern observed on a screen 3.50 m from the slits. (a) What is the difference in path lengths from the two slits to the location of a fourth order bright fringe on the screen? μm (b) What is the difference in path lengths from the two slits to the location of the fourth dark fringe on the screen, away from the center of the pattern? μm
Answer:
[tex]1.152\ \mu m[/tex]
[tex]1.44\ \mu m[/tex]
Explanation:
d = Gap between slits = 0.142 mm
[tex]\lambda[/tex] = Wavelength of light = 576 nm
L = Distance between light and screen = 3.5 m
m = Order = 2
Difference in path length is given by
[tex]\delta=dsin\theta=m\lambda\\\Rightarrow \delta=2\times 576\times 10^{-9}\\\Rightarrow \delta=0.000001152\ m\\\Rightarrow \delta=1.152\times 10^{-6}\ m=1.152\ \mu m[/tex]
The difference in path lengths is [tex]1.152\ \mu m[/tex]
For dark fringe the difference in path length is given by
[tex]\delta=(m+\dfrac{1}{2})\lambda\\\Rightarrow \delta=(2+\dfrac{1}{2})\times 576\times 10^{-9}\\\Rightarrow \delta=0.00000144\ m=1.44\ \mu m[/tex]
The difference in path length is [tex]1.44\ \mu m[/tex]
True / False:
O An ammeter must be placed in parallel with a resistor to measure the current through the resistor.
O A voltmeter is used to measure voltage.
O A voltmeter has a small internal resistance.
O An ammeter is used to measure current.
O A voltmeter must be placed in parallel with a resistor to measure the voltage across the resistor.
O An ammeter has a large internal resistance.
Answer: 1. False, an ammeter must be placed in a series connection to experience the same current which it is measuring
2. True, a voltmeter measures voltage (electric potential) in volts
3. False, a voltmeter should have high internal resistance such that it does not significantly alters current in a circuit.
4. True, a meters measure current
5. True, voltmeter would measure electric potential across a resistor
6. False, since they tend to influence the current in a circuit, ideally it should have very low resistance near zero
Explanation:
Lightning As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and a cloud at an altitude of 650 m to be the other plate. Assume the surface area of the cloud to be the same as the area of a square that is 0.70 km on a side. a. What is the capacitance of this capacitor?b. How much charge can the cloud hold before the dielectric strength of the air is exceeded and a spark (lightning) results?
Answer:
[tex]6.67154\times 10^{-9}\ F[/tex]
13.009503 C
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
k = Dielectric constant of air [tex]3\times 10^6\ V/m[/tex]
Side of plate = 0.7 km
A = Area
d = Distance = 650 m
Capacitance is given by
[tex]C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 700^2}{650}\\\Rightarrow C=6.67154\times 10^{-9}\ F[/tex]
The capacitance is [tex]6.67154\times 10^{-9}\ F[/tex]
Electric field is given by
[tex]Q=CV\\\Rightarrow Q=Ckd\\\Rightarrow Q=6.67154\times 10^{-9}\times 3\times 10^6\times 650\\\Rightarrow Q=13.009503\ C[/tex]
The charge on the cloud is 13.009503 C
The capacitance of the cloud-ground system is 6.67 nF, and the cloud can hold up to 13 C of charge before lightning occurs.
To model lightning using a parallel-plate capacitor, assume the ground is one plate and a cloud at 650 m altitude is the other. Let’s calculate the capacitance and the charge the cloud can hold before a lightning strike occurs.
We have the height h = 650 m and the side length L of the cloud's surface area is 0.70 km (700 m), making the area A = 700 m × 700 m = 490,000 m².The formula for capacitance C of a parallel-plate capacitor is:Therefore, the capacitance of the capacitor formed by the cloud and the ground is 6.67 nF, and the cloud can hold up to 13 C of charge before a lightning strike occurs.
In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and particle 1 experiences an attractive force of 3.44 N.
What is the magnitude and sign of q2?
Answer:
Charge of particle 2, [tex]q_2=-7.13\ \mu C[/tex]
Explanation:
Given that,
Charge 1, [tex]q_1=3.11\ \mu C=3.11\times 10^{-6}\ C[/tex]
The distance between charges, r = 0.241 m
Force experienced by particle 1, F = 3.44 N
We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]q_2=\dfrac{Fr^2}{kq_1}[/tex]
[tex]q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}[/tex]
[tex]q_2=7.13\times 10^{-6}\ C[/tex]
or
[tex]q_2=7.13\ \mu C[/tex]
So, the magnitude of electric charge 2 is [tex]q_2=7.13\ \mu C[/tex]. Since, the force is attractive then the magnitude of charge 2 must be negative.
Final answer:
To determine the magnitude and sign of q2, Coulomb's Law is applied, revealing that q2 has a negative charge of approximately -4.48 μC, due to the attractive force observed.
Explanation:
To find the magnitude and sign of q2, we will use Coulomb's Law, which quantifies the electric force between two charged particles. Coulomb's Law is given by the formula F = k |q1*q2| / r^2, where F is the magnitude of the force between the charges, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
Given that q1 = +3.11 μC (or 3.11 x 10^-6 C), r = 0.241 m, and F = 3.44 N, and knowing that the force is attractive (indicating that q2 must have an opposite sign to q1), we can solve for q2 as follows:
F = k |q1*q2| / r^2
3.44 = (8.99 x 10^9) |(3.11 x 10^-6) * q2| / (0.241)^2
Solving for q2, we get q2 ≈ -4.48 x 10^-6 C. The negative sign indicates that q2 is negatively charged, and the magnitude of this charge is 4.48 μC.
Estimate the acceleration you subject yourself to if you walk into a brick wall at normal walking speed. Make a reasonable estimate of your speed and the time it takes you to come to a stop. Explain your answer!
Answer:
Walking into a brick wall at normal walking speed (1.4 m/s), you will come to a complete stop in a short time (0.1 s) and experience much more acceleration (14 m/s²) back the way you came, but because you are softer than the wall, the inelastic collision will probably cause you to bounce back off the wall, changing the actual experienced acceleration you feel.
Explanation:
Assuming normal human walking speeds and time it takes to come to rest.
Normal human walking speed = 5km/h = 1.4 m/s
Time it takes you to come to a complete stop = 0.1 seconds
acceleration = Δ Velocity/ Time
Δ Velocity = final velocity - initial velocity
Δ Velocity = (1.4 - 0)m/s = 1.4 m/s
acceleration = 1.4/0.1
acceleration = 14 m/s²
Walking into a brick wall at normal walking speed, you will experience much more acceleration back the way you came, but because you are softer than the wall, the inelastic collision will probably cause you to bounce back off the wall, changing the actual experienced acceleration you feel.
Acceleration of the subject.
The acceleration is the rate at which the body velocity changes with time. It can be referred to the body the speed and direction. The point of the object moves in the straight line as acceleration is both a magnitude and a direction. Estimated acceleration of the subject and yourself
thus the answer is 1.4 m/s and stops at 0.1s
As per the question of you walking into the brick wall at a normal walking speed than then walking towards the wall at normal walking at speed is 1.4 m, you will completely stop at a short time of 0.1 sec. The experience much more acceleration (14 m/s²) back the way you came, but due to the inelastic collision that probably causes you to bounce back by the wall, which will change the initial experience you felt.Hence answer is 1.4 and 01 sec.Learn more about the acceleration.
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the earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. where did this energy come from>
Explanation:
Earth or any planet are actually born from huge clouds of gas and dust. Their stellar mass are fairly distributed at a radius from the axis of rotation. Gravitational force cause the cloud to come together. Now the whole gathered in smaller area. Now, individual particles come close to the roational axis. Thus, decreasing the moment of inertia of the planet.
As
I=mr^2
reducing r reduces I. However, the angular moment of the system remains always conserved. So, to conserve the angular momentum the angular velocity of the planet increases and so did the otational kinetic energy
The additional rotational kinetic energy of the Earth compared to the initial cloud comes from the gravitational potential energy that was present in the cloud before it collapsed. The conservation of angular momentum dictates that as the cloud shrinks, its rotation speed must increase, leading to an increase in rotational kinetic energy.
The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed because of the conservation of angular momentum. As the cloud collapsed under its own gravity, the rotation rate increased to conserve angular momentum, which is the product of moment of inertia and rotational velocity. Since the moment of inertia decreases as the mass moves closer to the axis of rotation (due to the shrinking size of the collapsing cloud), the rotational velocity must increase to keep the angular momentum constant. This results in an increase in rotational kinetic energy, which is given by the equation [tex]\( KE_{rot} = \frac{1}{2} I \omega^2 \)[/tex], where [tex]\( I \)[/tex] is the moment of inertia and [tex]\( \omega \)[/tex] is the angular velocity.
The gravitational potential energy of the cloud is converted into kinetic energy as the cloud collapses. The initial potential energy is high because the particles in the cloud are far from the center of mass. As the cloud contracts, this potential energy is transformed into kinetic energy, both linear and rotational, due to the conservation of energy. The increase in rotational kinetic energy is a direct consequence of this conversion and the conservation of angular momentum.
In summary, the additional rotational kinetic energy of the Earth compared to the initial cloud comes from the gravitational potential energy that was present in the cloud before it collapsed. The conservation of angular momentum dictates that as the cloud shrinks, its rotation speed must increase, leading to an increase in rotational kinetic energy.
The complete question is:
The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. Where did this energy come from?
Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of 20C
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
[tex]Q =\frac{KA(\delta T)}{L}[/tex]
[tex]Q =\frac{25.87*1*20}{1}[/tex]
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
Answer: Rate of heat transfer q = 517.4W
Explanation:
Thermal conductivity is a material property that describes its ability to conduct heat. Thermal conductivity is the quantity of heat transmitted through a unit thickness of a material, in a direction normal to a surface of unit area,due to a unit temperature gradient under steady state conditions. It can be shown mathematically using the equation below;
Q/t = kA(∆T)/d
q = kA(∆T)/d
q = Q/t = rate of heat transfer
Q = total amount of heat transfer
t = time
k = thermal conductivity of material
A = Area
d= distance
For the case above, the material used is still air.
k for still air at 20°C and 1bara = 0.02587W/mK
A = 1m^2
d = 1mm = 0.001m
∆T = 20°C = 20K
q = 0.02587×1×20/0.001
q = 517.4W
Therefore, the rate of heat conduction through the still air is 517.4W
In testing thousands of different materials for use as lightbulb filaments, Thomas Edison best illustrated a problem-solving approach known as:
Group of answer choices
a. fixation.
b. belief perseverance.
c. trial and error.
d. the confirmation bias.
e. the representativeness heuristic.
Answer:
Trial and error.
Explanation:
This approach of Thomas Edison where in testing of thousand of different material for light bulb filament are used to find the most suitable material is called trial and error approach of problem solving.
Trial and error pursue a method, seeing if it performs, and not trying a new mechanism. This mechanism will be repeated until a solution or success is attained. Assume moving a large object like a couch into your house for example.
Tests reveal that a normal driver takes about 0.75s before he or she can react to a situation to avoid a collision. It takes about 3s for a driver having 0.1% alcohol in his system to do the same.
If such drivers are traveling on a straight road at 30mph (44 )ft/s and their cars can decelerate at 24ft/s^2 , determine the shortest stopping distance d for normal driver from the moment he or she see the pedestrians.
Also, Determine the shortest stopping distance for drunk driver from the moment he or she see the pedestrians.
Answer:
NORMAL DRIVER: d = 73.3 ft
DRUNK DRIVER: d = 172.3
Explanation:
NORMAL DRIVER:
Distance covered in initial 0.75s = 0.75s *44 = 33ft
USING THE THIRD EQUATION OF MOTION
V^2-U^2 = 2as
0-(44)^2 = 2 (-24) s
s = 1936/48 =40.3 ft
d = 33 + 40.3 = 73.3 ft
DRUNK DRIVER:
Distance covered in initial 3s = 3s *44 = 132 ft
USING THE THIRD EQUATION OF MOTION
V^2-U^2 = 2as
0-(44)^2 = 2 (-24) s
s = 1936/48 =40.3 ft
d = 132 + 40.3 = 172.3 ft
A 0.500-kg mass suspended from a spring oscillates with a period of 1.36 s. How much mass must be added to the object to change the period to 2.04 s?
The mass that must be added is 0.628 kg
Explanation:
The period of a mass-spring system is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where
m is the mass
k is the spring constant
For the initial mass-spring system in the problem, we have
m = 0.500 kg
T = 1.36 s
Solving for k, we find the spring constant:
[tex]k=(\frac{2\pi}{T})^2 m = (\frac{2\pi}{1.36})^2 (0.500)=10.7 N/m[/tex]
In the second part, we want the period of the same system to be
T = 2.04 s
Therefore, the mass on the spring in this case must be
[tex]m=(\frac{T}{2\pi})^2 k =(\frac{2.04}{2\pi})^2 (10.7)=1.128 kg[/tex]
Therefore, the mass that must be added is
[tex]\Delta m = 1.128 - 0.500 = 0.628 kg[/tex]
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A force of 10 N is applied to an object. How much work is done by moving the object a distance of 5 m
Answer:
Work done on the object will be 50 J
Explanation:
We have given force force applied on the object F = 10 N
Object move by a distance of 5 m
So d = 5 m
We have to find the work done on the object by the force
Work done is given by [tex]work=force\times distance[/tex]
So work done [tex]=10\times 5=50J[/tex]
So work done on the object will be 50 J
The largest and smallest slits in these experiments are 0.16mm and 0.04mm wide, respectively. The wavelength of the laser light is 650nm. How many wavelengths wide are these slits?
To solve this problem we will convert the values given to international units (meters) and then proceed to calculate the number of wavelengths through the division of the width over the wavelength. This is,
[tex]\lambda = 650 nm =650 * 10^{-9} m[/tex]
[tex]w_1 = 0.16 mm = 0.16 * 10^{-3} m[/tex]
[tex]w_2 = 0.04 mm =0.04 * 10^{-3} m[/tex]
Now the number of wavelengths is the division between the total width over the wavelength therefore
First case,
[tex]\frac{w_1}{\lambda} = \frac{ 0.16 * 10^{-3}}{650 * 10^{-9}} = 4062.5 * 10^6[/tex]
Second case,
[tex]\frac{w_2}{\lambda} = \frac{0.04 * 10^{-3} }{650 * 10^{-9} } = 16250 * 10^6[/tex]
a) What is the resistance of a lightbulb that uses an average power of 25.0 W when connected to a 60.0-Hz power source having a maximum voltage of 170 V?
To solve this problem we will start considering the relationship between rms voltage and the maximum voltage. Once the RMS voltage is obtained, we will proceed to find the system current through the given power and the voltage found. Finally by Ohm's law we will find the resistance of the system
The relation of RMS Voltage is,
[tex]V_{rms} = \frac{1}{\sqrt{2}}*V_{max}[/tex]
Note that this conversion is independent from the Frequency.
[tex]V_{rms} = \frac{1}{\sqrt{2}}*(170)[/tex]
[tex]V_{rms}= 120.20V[/tex]
Now the current is,
[tex]I =\frac{P}{V}[/tex]
[tex]I = \frac{(25 W)}{(120.20V)}[/tex]
[tex]I = 0.2079A[/tex]
By Ohm's Law
[tex]V = IR \rightarrow R = \frac{V}{I}[/tex]
Replacing,
[tex]R = \frac{ (120.20 V)}{ ( 0.2079 A)}[/tex]
[tex]R = 578.16 \Omega[/tex]
Therefore the resistance of this lightbulb is[tex]578.16 \Omega[/tex]
The resistance of the lightbulb is 1156.5 Ω. It can be calculated using Ohm's Law, which states that resistance is equal to voltage divided by current.
Explanation:The resistance of a lightbulb can be found using Ohm's Law, which states that resistance is equal to voltage divided by current. In this case, we need to find the resistance of a lightbulb that uses an average power of 25.0 W when connected to a 60.0-Hz power source with a maximum voltage of 170 V. We can start by finding the current using the formula P = IV, where P is power and I is current. Rearranging the formula gives us I = P/V. Plugging in the given values, we get I = 25.0 W / 170 V = 0.147 A. Now we can use Ohm's Law to find the resistance. R = V/I = 170 V / 0.147 A = 1156.5 Ω.
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Under electrostatic conditions, the electric field just outside the surface of any charged conductor:
A. is equal to the electric field just inside the surface of the conductor.
B. can have nonzero components perpendicular to and parallel to the surface of the conductor, if it is not symmetric.
C. is always zero.
D. is always perpendicular to the surface of the conductor.
E. is always parallel to the surface of the conductor.
Answer:
D. is always perpendicular to the surface of the conductor
Explanation:
1) Answer is (D) option. Electric field just outside surface of charged conductor is normal to conductor at that point.
It can be explained on the basis of the fact that, Electric field inside conductor under static condition is zero. As a result potential difference between any two points with in conductor is zero. So whole of conductor is equipotential body.
Equipotential surface and Electric field lines always cut at 90 degrees to each other. Conductor being equipotential body, Electric field lines starting or terminating at conductor must be normal to surface. Hence electric field just outside conductor is perpendicular or normal to surface.
What equation gives the position at a specific time for an object with constant acceleration? a. x=x0+v0t+1/2}at^2b. x=v0t+at^2c. vf=v0+atd. v^2f=v0^2+2aΔx
The first option is mathematically described as
[tex]x = x_0 +v_0 t +\frac{1}{2} at^2[/tex]
Here,
[tex]x_0 =[/tex]Initial position
[tex]v_0 =[/tex] Initial velocity
t = Time
a = Acceleration
As we can see in this equation, the position of a body is described taking into account its initial point with respect to the reference system, the initial velocity of this body and the acceleration when it is constant. All this depending on time.
The second option despises the initial position, so it does not allow the exact calculation of the position.
The third option does not consider the position, only the speed and acceleration with respect to time
The fourth option considers acceleration and distance but does not take into account the time of the object.
Therefore the correct answer is A.
A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/s 2 . What is their velocity after the collision
To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then
[tex]\text{Initial Momentum} = \text{Final Momentum}[/tex]
[tex]m_1u_1 +m_2u_2 = m_1v_1+m_2v_2[/tex]
Here,
[tex]m_{1,2}[/tex]= Mass of each object
[tex]u_{1,2}[/tex]= Initial velocity of each object
[tex]v_{1,2}[/tex]= Final velocity of each object
When they position the final velocities of the bodies it is the same and the car is stationary then,
[tex]m_2u_2 = (m_1+m_2)v_f[/tex]
Rearranging to find the final velocity
[tex]v_f = \frac{m_2u_2}{ (m_1+m_2)}[/tex]
[tex]v_f = \frac{ 12500*7.8}{ 12500+7430}[/tex]
[tex]v_f = 4.8921ft/s[/tex]
The expression for the impulse received by the first car is
[tex]I = m_1 (v-u)[/tex]
[tex]I = \frac{W}{g} (v-u)[/tex]
Replacing,
[tex]I = \frac{12500}{32.2}(4.89-7.8)[/tex]
[tex]I = -1129.65lb\cdot s[/tex]
The negative sign show the opposite direction.
The largest building in the world by volume is the boeing 747 plant in Everett, Washington. It measures approximately 632 m long, 710 yards wide, and 112 ft high. what is the cubic volume in feet, convert your result from part a to cubic meters.
Answer with Explanation:
We are given that
a.Length building=632 m
1 m=3.2808 feet
632 m=[tex]3.2808\times 632=2073.47 feet[/tex]
Length of building,l=2073.47 feet
Width of building,b=710 yards=[tex]710\times 3=2130 ft[/tex]
1 yard=3 feet
Height of building,h=112 ft
Volume of building=[tex]l\times b\times h[/tex]
Volume of building=[tex]2073.47\times 2130\times 112=494647003.2ft^3[/tex]
Volume of building=494647003.2 cubic feet
b. 1 cubic feet=0.028 cubic meter
Volume of building= [tex]494647003.2\times 0.028[/tex]cubic meters
Volume of building=13850116.0896 cubic meters
A 1.40 kg object is held 1.15 m above a relaxed, massless vertical spring with a force constant of 300 N/m. The object is dropped onto the spring.
How far does the object compress the spring?
Repeat part (a), but now assume that a constant air-resistance force of 0.600 N acts on the object during its motion.
How far does the object compress the spring if the same experiment is performed on the moon, where g = 1.63 m/s2 and air resistance is neglected?
Answer:
a) [tex]\Delta x =0.32433\ m= 324.33\ mm[/tex]
b) [tex]\delta x=0.087996\ m=87.996\ mm[/tex]
c) [tex]\delta x=0.13227\ m=132.27\ mm[/tex]
Explanation:
Given:
mass of the object, [tex]m=1.4\ kg[/tex]height of the object above the spring, [tex]h=1.15\ m[/tex]spring constant, [tex]k=300\ N.m^{-1}[/tex]a)
When the object is dropped onto the spring whole of the gravitational potential energy of the mass is converted into the spring potential energy:
[tex]PE_g=PE_s[/tex]
[tex]m.g.h=\frac{1}{2} \times k.\Delta x^2[/tex]
[tex]1.4\times 9.8\times 1.15=0.5\times 300\times \Delta x^2[/tex]
[tex]\Delta x =0.32433\ m= 324.33\ mm[/tex] is the compression in the spring
b)
When there is a constant air resistance force of 0.6 newton then the apparent weight of the body in the medium will be:
[tex]w'=m.g-0.6[/tex]
[tex]w'=1.4\times 9.8-0.6[/tex]
[tex]w'=1.01\ N[/tex]
Now the associated gravitational potential energy is converted into the spring potential energy:
[tex]PE_g'=PE_s[/tex]
[tex]w'.h=\frac{1}{2} \times k.\delta x^2[/tex]
[tex]1.01\times 1.15=0.5\times 300\times \delta x^2[/tex]
[tex]\delta x=0.087996\ m=87.996\ mm[/tex]
c)
On moon, as per given details:
[tex]m.g'.h=\frac{1}{2} \times k.\delta x^2[/tex]
[tex]1.4\times 1.63\times 1.15=0.5\times 300\times \delta x^2[/tex]
[tex]\delta x=0.13227\ m=132.27\ mm[/tex]
A pulse waveform with a frequency of 10 kHz is applied to the input of a counter. During 100 ms, how many pulses are counted?
Answer:
n = 1000 pulses
Explanation:
Given that,
The frequency of a pulse waveform, [tex]f=10\ kHz=10^4\ Hz[/tex]
To find,
The number of pulses counted during 100 ms.
Solution,
The frequency of a pulse waveform is equal to the number of pulses per unit time. It is given by :
[tex]f=\dfrac{n}{t}[/tex]
[tex]n=f\times t[/tex]
[tex]n=10^4\ Hz\times 100\times 10^{-3}\ s[/tex]
n = 1000 pulses
So, there are 1000 pulses counted in a pulse waveform.
A rigid container equipped with a stirring device contains 1.5 kg of motor oil. Determine the rate of specific energy increase when heat is transferred to the oil at a rate of 1 W and 1.5 W of power is applied to the stirring device.
To solve this problem we will apply the first law of thermodynamics which details the relationship of energy conservation and the states that the system's energy has. Energy can be transformed but cannot be created or destroyed.
Accordingly, the rate of work done in one cycle and the heat transferred can be expressed under the function,
[tex]\dot{U} = \dot{Q}-\dot{W}[/tex]
Substitute 1W for [tex]\dot{Q}[/tex] and 1.5 W for [tex]\dot{W}[/tex]
[tex]\dot{U} = 1-1(1.5)[/tex]
[tex]\dot{U} = 2.5W[/tex]
Now calculcate the rate of specific internal energy increase,
[tex]\dot{u} = \frac{\dot{U}}{m}[/tex]
[tex]\dot{u} = \frac{2.5}{1.5}[/tex]
[tex]\do{u} = 1.6667W/kg[/tex]
The rate of specific internal energy increase is 1.6667W/kg
To answer the student's question, combine the total power input from heat and stirring, then calculate the temperature increase of the oil using its specific heat capacity with the energy conservation principle.
Explanation:The student is asking about the rate of specific energy increase when heat is transferred to motor oil while work is being done on it by stirring. In thermodynamics, this relates to the conversion of work into thermal energy. The first step is to calculate the total energy input per second (power), which is the sum of heat transfer and the work done by stirring. The next step is to use the specific heat capacity to determine the rate of temperature increase for the oil.
Given the specific heat capacity of water and a weight lifting scenario, the student needs to apply the energy conservation principle, converting the work done into the thermal energy, which will result in raising the temperature of the water. To find this temperature increase, one should use the formula Q = mcΔT, where Q is the energy transferred as heat, m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature.