Answer:
a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. Explanation:
The coefficient of volume expansion is [tex]0.000810 \, \text{per} \, ^\circ\text{C}.[/tex]
The coefficient of volume expansion (\(\beta\)) can be calculated using the formula:
[tex]\[\Delta V = V_0 \beta \Delta T\][/tex]
where:
- [tex]\(\Delta V\)[/tex] is the change in volume,
- [tex]\(V_0\)[/tex] is the initial volume,
- [tex]\(\beta\)[/tex] is the coefficient of volume expansion,
- [tex]\(\Delta T\)[/tex] is the change in temperature.
We need to solve for [tex]\(\beta\):[/tex]
[tex]\[\beta = \frac{\Delta V}{V_0 \Delta T}\][/tex]
Substituting the given values:
[tex]\[\beta = \frac{0.0920 \, \text{m}^3}{2.35 \, \text{m}^3 \times 48.5 \, ^\circ\text{C}}\][/tex]
First, calculate the denominator:
[tex]\[2.35 \, \text{m}^3 \times 48.5 \, ^\circ\text{C} = 113.575 \, \text{m}^3 \cdot ^\circ\text{C}\][/tex]
Now, calculate [tex]\(\beta\)[/tex]:
[tex]\[\beta = \frac{0.0920 \, \text{m}^3}{113.575 \, \text{m}^3 \cdot ^\circ\text{C}}\][/tex]
[tex]\[\beta \approx 0.000810 \, \text{per} \, ^\circ\text{C}\][/tex]
Therefore, the coefficient of volume expansion is approximately [tex]\[\beta \approx 8.10 \times 10^{-4} \, ^\circ\text{C}^{-1}\][/tex].
A 14-kg block rests on a level frictionless surface and is attached by a light string to a 5.0-kg hanging mass where the string passes over a massless frictionless pulley. If g=9.8 m/s2 what is the tension in the connecting string?
Answer:
Explanation:
Let T be the tension in the string .
For hanging mass
net force = m₁g - T
m₁g - T = m₁a
For mass placed on horizontal plane
T = m₂a
m₁g - m₂a = m₁a
m₁g = a ( m₂ +m₁)
a = m₁g / ( m₂ +m₁)
= 5 x 9.8 / 19
= 2.579 m /s²
m₁g - T = m₁a
T = m₁g - m₁a
= m₁ ( g - a )
= 5 ( 9.8 - 2.579)
= 36.10 N
A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motion, after which a horizontal force of 56.0 N is required to keep the block moving with constant speed.Find the coefficients of static and kinetic friction from this information.
Answer: The coefficient of static friction is 3.85 and The coefficient of kinetic friction is 2.8
Explanation:
in the attachment
g 4. A student with a mass of m rides a roller coaster with a loop with a radius of curvature of r. What is the minimum speed the rollercoaster can maintain and still make it all the way around the loop?
Answer:
minimum speed v=[tex]\sqrt({Fr}/m)}[/tex]
Explanation:
Recall the formula for centripetal force;
Centripetal force is the force that is required to keep an object moving in circular part
[tex]F=mv^{2} /r[/tex]
where;
F=centripetal force
m=mass of object
r=radius of curvature
v= minimum speed
To find minimum speed make v the subject of formula;
v=[tex]\sqrt({Fr}/m)}[/tex]
The question is incomplete.
The complete question is:
A student with a mass of m rides a roller coaster with a loop with a radius of curvature of r. What is the minimum speed the rollercoaster can maintain and still make it all the way around the loop? g= 9.8m/s
r=5.5 m= 55kg
Answer: 7.3m/s
Explanation:
Centripetal force is the force acting on a body causing circular motion towards the center which allows to keep the body on track or right path. Any combination of forces in nature can cause centripetal acceleration in various systems.
Where:
V is the tangential velocity
W is the angular velocity
M is the Mass in kg
g is the gravitational force
r is the circular radius
Apply Newton's second law of motion.
The minimum speed is also known as the critical speed is the point where the tension, frictional or normal force is zero and the only thing keeping the object in circular motion is the force of gravity.
This is explained as follows:
N + mg = mw^2r
mg = mv^2/r
Therefore,the minimum speed is:
v^2 = gr
v = square root(gr)
= sqrt(9.8N/kg)(5.5m)
=(9.8N/kg)(5.5m) ^1/2
v = 7.3m/s
During a walk on the Moon, an astronaut accidentally drops his camera over a 20.0-m cliff. It leaves his hands with zero speed, and after 2.00 s it attains a velocity of 3.40 m/s downward. How far has the camera fallen after 9.10 s
Answer:
Δy=70.3885 downwards
Explanation:
First we get the acceleration on moon by substituting the values we have(v=0m/s at t=0s and v=-3.40m/s at t=2.0 s) into equation of simple motion:
[tex]v_{f}-v_{i}=at\\-3.40m/s-0=(2s)a\\a=-1.7m/s^{2}[/tex]
As we know that
Δy=vit+(1/2)at²
Substitute the values of acceleration and time t=9.10s
So
Δy=vit+(1/2)at²
[tex]=(0)(9.10s)+(1/2)(-1.7m/s^{2})(9.10s)^{2}\\ =-70.3885m[/tex]
Δy=70.3885 downwards
The camera has fallen approximately 402.5 meters after 9.10 seconds.
Explanation:To find the distance the camera has fallen after 9.10 s, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance fallen
u = initial velocity (0 m/s)
t = time (9.10 s)
a = acceleration (due to gravity, -9.8 m/s^2)
Plugging the values into the equation, we get:
s = 0 + (1/2)(-9.8)(9.10)^2
Simplifying, we find that the camera has fallen approximately 402.5 meters after 9.10 seconds.
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A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like protons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create x-rays by bombarding a hard metal target with the beam. Consider a beam of protons at 1.85 keV and a current of 5.15 mA produced by the generator.(a) What is the speed of the protons (in m/s)?(b) How many protons are produced each second?
Answer:
595391.482946 m/s
[tex]3.21875\times 10^{6}[/tex]
Explanation:
E = Energy = 1.85 keV
I = Current = 5.15 mA
e = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
t = Time taken = 1 second
m = Mass of proton = [tex]1.67\times 10^{-27}\ kg[/tex]
Velocity of proton is given by
[tex]v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s[/tex]
The speed of the proton is 595391.482946 m/s
Current is given by
[tex]I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C[/tex]
Number of protons is
[tex]n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons[/tex]
The number of protons is [tex]3.21875\times 10^{6}[/tex]
The speed of the protons accelerated by the Van de Graaff generator is approximately 3.19 million meters per second. Additionally, the generator produces roughly 3.21 x 10^16 protons every second.
Explanation:The Van de Graaff generator is a particle accelerator that can be used to accelerate charged particles. In the given scenario, a beam of protons with an energy of 1.85 keV and a beam current of 5.15 mA is being generated.
(a) The energy of a proton (kinetic energy = 1/2 mv²) is given by the equation E = mv²/2, where m is the mass of the proton (1.67262192369 × 10⁻²⁷ kg) and v is the speed of the proton. Solving the equation v = sqrt((2*E)/m), where E is the energy in joules (1.85 keV = 1.85 * 10^-16 joules), we find that v ≈ 3.19 x 10^6 m/s. This implies that the speed of the protons is approximately 3.19 x 10^6 meters per second.
(b) The number of protons produced each second, i.e., the beam current, can be calculated using the formula I = qN/t, where I is the current, q is the charge of a proton (1.602 x 10^-19 Coulombs), N is the number of protons, and t is time. Rearranging the formula, we find that N = It/q. Substituting the values I=5.15 mA = 5.15 x 10^-3 A and t=1s, we get N ≈ 3.21 x 10^16 protons. Therefore, approximately 3.21 x 10^16 protons are produced every second under the mentioned conditions.
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how is Newton’s three laws of motion are used in testing the safety of our automobiles.
Answer
Newton's third law is used for the impact test on the vehicle.
In the frontal collision of the vehicle, the impact of the vehicle is made against the concrete wall then the impact on the passengers and driver is calculated.
In the impact test time of the opening of the airbag is also calculated.
Similar tests also take place when the impact is sideways.
Ballistic pendulum conservation of momentum/mechanical energy?
"Mechanical energy and momentum are conserved only when certain conditions rae met. Explain why the collision between the ball and the pendulum conserves momentum but not mechanical energy. Similarly, explain why the motion of the pendulum during its swing conserves mechanical energy but (apparently) not momentum."
I understand why during the collision kinetic energy is lost and momentum is conserved, but why would mechanical energy be conserving during the swing and not momentum?
Answer:
In the explanation the answers for the two questions are analyzed.
Explanation:
1. When a collision occurs, it can be said that the total energy is not conserved, but the momentum is conserved. Kinetic energy is converted to heat in the case of an inelastic collision, therefore the sum of all energies is the same before and after the collision. Regarding the net moment before the collision is equal to the net moment after said collision.
2. Regarding the impulse, this is not conserved in the case of an oscillating pendulum because the resulting force acting on the pendulum is not equal to zero. While the total energy would be equal to the sum of the potential energy plus the kinetic energy.
Explain whether a solenoid will attract paper clips if there is no current through its coils. Predict what will happen to the force between the solenoid and the paper clips if you increase the current in the coils or increased the number of turns of coil per un
Answer:
magnetic field strength increases.
Explanation:
No, the solenoid will not attract any magnetic substance if it does not carries a direct electric current.
The formula for the magnetic field strength of the solenoid is given as:
[tex]B=\mu.n.I[/tex]
where:
[tex]\mu=[/tex] permeability of free space
[tex]n=[/tex] no. of coil turns in the solenoid
[tex]I=[/tex] current in the coil
So, when the current & no. of coil turns in the coil are increased then the magnetic field strength also increases.
A light plane is headed due south with a speed of 200 km/h relative to still air. After 1.00 hour, the pilot notices that they have covered only 137 km and their direction is not south but 15.0∘ east of south. What is the wind velocity?
Answer: the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.
Explanation: The initial velocity of the plane is 200km/h south (supose that south is our positive x-axis here and east is the positive y-axis)
In one hour, the plane is located 137km away from the initial position, and the position in X is equal to 137km*cos(15°) = 132.33, this means that the velocity in the x axis is equal to 132.33 km/h, knowing that the initial velocity of the plane was 200km in the x-axis, this means that the velocity of the air must be:
132.33km/h - 200km/h = -67.69km/h
km and the position in "y" is equal to 137km*sin(15°) = 35.4km
This means that the velocity of the air in the y-axis is 35.4km/h
So the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.
A block with a mass of 4 kg is sliding across a horizontal surface with an initial speed of 5 m/s. Because of kinetic friction, the energy of this system will decrease linearly with the distance traveled by the block.
What is the average power (in W) supplied by the force of kinetic friction acting on the block if the block moves 4 m before coming to a stop?
Explanation:
Below is an attachment containing the solution.
The average power supplied by the force of kinetic friction acting on the block if the block moves 4 m before coming to stop is -31.25 W.
What is Power?In physics, power is the amount of energy that is transferred or transformed in a given amount of time. The International System of Units uses the watt, or one joule per second, as the unit of power. Ancient literature frequently used the term "activity" to describe power. Powers are scalar variables.
According to the question, the given values are :
Block mass, m = 4 kg,
The initial speed of the block at the horizontal surface, v₁ = 5 m/s,
The final speed of the block will be, v₂ = 0 m/s.
Distance covered, d = 4 m
Let t be the time when the box is going to stop.
d = [(v₁+v₂)/2] × t
4 = (5/2) t
t = 1.6 sec.
The change in kinetic energy of the block will be ΔE.
ΔE = mv₂²/2 - mv₁²/2
ΔE = 0 - (4)(5)²/2
ΔE = -50 J
Average power supplied by the force of kinetic friction acting on the block = P(avg)
P(avg) = ΔE / t
P(avg) = -50 / 1.6
P(avg) = -31.26 W.
So, the average power supplied by the force of kinetic friction on the block will be -31.26 W.
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An airplane is flying at 150 ft/s at an altitude of 2000 ft in a direction that will take it directly over an observer at ground level. Find the rate of change of the angle of elevation between the observer and the plane when the plane is directly over a point on the ground that is 2000 ft from the observer.
Answer:
Explanation:
Let x be the horizontal distance of airplane . angle of elevation of airplane from observer = θ , altitude of airplane = 2000 ft ,
from the construction θ = 45 degree. , x = 2000 ft .
Tanθ = 2000 / x
sec²θ dθ / dt = (2000 / x²) dx / dt
dθ / dt = 2000 /(sec²θ x²) x dx / dt
dx / dt = 150 ft /s
dθ / dt = 2000x 150 /(sec²θ x²)
= 300000 / sec²45 x 2000²
= .15 degree/ s
The rate of change of the angle of elevation when an airplane (traveling at 150 ft/s) is over a point 2000 ft from the observer is approximately -2.15° per second.
Explanation:This problem falls under the department of mathematics known as trigonometry specifically, its applications in real-world scenarios. In this case, we're looking to find the rate of change of the angle of elevation when the airplane is at a certain point.
When the airplane is directly over a point 2000 ft from the observer, it forms a right triangle with the observers' location and the point over which it is flying. The hypotenuse of this triangle is the line between observer and airplane.
The angle of elevation, θ, from the observer to the plane changes over time as the airplane moves overhead so its rate of change(dθ/dt) is what we need to find. We use the trigonometric relation tangent (tan), which in this case equals to the altitude (opposite side, 2000 ft) over the horizontal distance between the observer and the plane (adjacent side, 2000 ft).
The relation is tan θ = opposite/adjacent = 2000/2000 = 1, thus θ = 45 degrees. With the plane's speed (150 ft/s), this changes the horizontal distance over time, so we differentiate tan θ giving us sec² θ*dθ/dt = -150/2000² = -0.0375 radians/sec, by applying the chain rule and remembering sec² θ is 1/cos² θ.
In degrees per second, this is approximately -2.15°/s, so that is the rate of change of the angle of elevation when the airplane is directly over a point 2000 ft from the observer.
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Coherent light with wavelength = 600 nm falls on two very narrow closely spaced slits and the interference pattern is observed on a screen that is 4 m from the slits. Near the center of the secreen the separation between adjacent maxima is 2 mm. What is the distance between the two slits?
Answer:
The distance between the two slits is 1.2mm.
Explanation:
The physicist Thomas Young establishes, through its double slit experiment, a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.
[tex]\Lambda x = L\frac{\lambda}{d} [/tex] (1)
Where [tex]\Lambda x[/tex] is the distance between two adjacent maxima, L is the distance of the screen from the slits, [tex]\lambda[/tex] is the wavelength and d is the separation between the slits.
If light pass through two slits a diffraction pattern in a screen will be gotten, at which each bright region corresponds to a crest, a dark region to a trough, as consequence of constructive interference and destructive interference in different points of its propagation to the screen.
Therefore, d can be isolated from equation 1.
[tex]d = L\frac{\lambda}{\Lambda x} [/tex] (2)
Notice that it is necessary to express L and [tex]\lambda[/tex] in units of millimeters.
[tex]L = 4m \cdot \frac{1000mm}{1m}[/tex] ⇒ [tex]4000mm[/tex]
[tex]\lambda = 600nm \cdot \frac{1mm}{1x10^{6}nm}[/tex] ⇒ [tex]0.0006mm[/tex]
[tex]d = (4000mm)\frac{0.0006mm}{2mm} [/tex]
[tex]d = 1.2mm[/tex]
Hence, the distance between the two slits is 1.2mm.
Johnson is dragging a bag on a ice surface. He pulls on the strap with a force of 112 N at an angle of 45° to the horizontal to displace it 84 m in 3.33 minutes. Determine the work done by Johnson on the bag and the power generated by Johnson
Answer with Explanation:
We are given that
Force=F=112 N
[tex]\theta=45^{\circ}[/tex]
Distance,[tex]s=84 m[/tex]
Time, t=3.33 minutes
We have to find the work done by Johnson on the bag and the power generated by Johnson.
Work done, W=[tex]Fscos\theta[/tex]
Using the formula
Work done, W=[tex]112\times 84cos45=6652.46 J[/tex]
Power, P=[tex]\frac{W}{t}=\frac{6652.46}{3.33\times 60}=33.3 watt[/tex]
Using 1 minute=60 s
Hence, the power generated by Johnson=33.3 watt
In an arcade game a 0.117 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releasing it. If the spring has a spring constant of 194 N/m and is compressed from its equilibrium position by 7 cm, find th
Find the speed with which the disk slides across the surface
Answer:
[tex]\boxed{2.85 m/s}[/tex]
Explanation:
The Potential energy of spring is transformed to kinetic energy hence
[tex]0.5kx^{2}=0.5mv^{2}\\kx^{2}=mv^{2}\\v=\sqrt{\frac {kx^{2}}{m}}[/tex]
Here k is the spring constant, x is the extension of spring, v is the velocity of disk and m is the mass of the disk.
Substituting 0.117 Kg for m, 194 N/m for k and 0.07 m for x then
[tex]v=\sqrt{\frac {194\times 0.07^{2}}{0.117}}=2.850401081 m/s\approx \boxed{2.85 m/s}[/tex]
Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 19 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 29 cm.
What is the wavelength of the sound?
Express your answer using two significant figures.
\lambda ={\rm cm}
Part B
If the distance between the speakers continues to increase, at what separation will the sound intensity again be a maximum?
Express your answer using two significant figures.
Answer:
Explanation:
The the observer must not be stationed between speakers . Let its position be d cm from one of the speaker. When sound has maximum intensity , the the two sounds reaching his ear must have constructive interference and during zero intensity , they must have destructive interference. Distance between sources will be equal to path difference.
For constructive interference
n λ = path diff
= 19 cm
For destructive interference
(2n+1) λ /2= path diff
= 29 cm
n λ + λ /2 = 29
19 + λ /2 = 29
λ = 20 cm
B )
The path difference of 19 cm corresponds to constructive interference or maximum sound intensity . Wave length is 20 cm . Therefore , next path difference that can create condition of constructive interference will be 19 + 20 = 39 cm . Present path difference is 29 cm. So the distance between the speaker must be increased to 39 cm , so that path difference becomes 39 cm and constructive interference takes place.
ANS 39 cm .
ou are unloading a refrigerator from a delivery van. The ramp on the van is 5.0 m long, and its top end is 1.4 m above the ground. As the refrigerator moves down the ramp, you are on the down side of the ramp trying to slow the motion by pushing horizontally against the refrigerator with a force of 370 N . Part A How much work do you do on the refrigerator during its trip down the ramp
Answer:
Explanation:
component of force along the ramp
= 370 cos θ where θ is the slope of the ramp with respect to ground.
sinθ = 1.4 / 5
θ = 16 degree
370 cos16
= 355.67 N
Work done
= component of force along the ramp x length of the ramp
= 355.67 x 5
= 1778.35 J
A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.
A mass weighing 32 pounds stretches a spring 2 feet.
(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.
(b) How many complete cycles will the mass have completed at the end of 4 seconds?
Answer:
[tex]A = 1.803 ft[/tex]
Period = [tex]\frac{\pi}{2}[/tex] seconds
8 cycles
Explanation:
A mass weighing 32 pounds stretches a spring 2 feet;
it implies that the mass (m) = [tex]\frac{w}{g}[/tex]
m= [tex]\frac{32}{32}[/tex]
= 1 slug
Also from Hooke's Law
2 k = 32
k = [tex]\frac{32}{2}[/tex]
k = 16 lb/ft
Using the function:
[tex]\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0[/tex]
[tex]x(0) = -1[/tex] (because of the initial position being above the equilibrium position)
[tex]x(0) = -6[/tex] ( as a result of upward velocity)
NOW, we have:
[tex]x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)[/tex]
However;
[tex]x(0) = -1[/tex] means
[tex]-1 =c_1\\c_1 = -1[/tex]
[tex]x(0) =-6[/tex] also implies that:
[tex]-6 =4(c_2)\\c_2 = - \frac{6}{4}[/tex]
[tex]c_2 = -\frac{3}{2}[/tex]
Hence, [tex]x(t) =-cos4t-\frac{3}{2} sin 4t[/tex]
[tex]A = \sqrt{C_1^2+C_2^2}[/tex]
[tex]A = \sqrt{(-1)^2+(\frac{3}{2})^2 }[/tex]
[tex]A=\sqrt{\frac{13}{4} }[/tex]
[tex]A= \frac{1}{2}\sqrt{13}[/tex]
[tex]A = 1.803 ft[/tex]
Period can be calculated as follows:
= [tex]\frac{2 \pi}{4}[/tex]
= [tex]\frac{\pi}{2}[/tex] seconds
How many complete cycles will the mass have completed at the end of 4 seconds?
At the end of 4 seconds, we have:
[tex]x* \frac{\pi}{2} = 4 \pi[/tex]
[tex]x \pi = 8 \pi[/tex]
[tex]x=8[/tex] cycles
A gas is compressed at constant temperature from a volume of 5.68 L to a volume of 2.35 L by an external pressure of 732 torr. Calculate the work done in J
Answer: The work done in J is 324
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:
[tex]W=-P\Delta V=-P(V_2-V_1)[/tex]
W = amount of work done = ?
P = pressure = 732 torr = 0.96 atm (760torr =1atm)
[tex]V_1[/tex] = initial volume = 5.68 L
[tex]V_2[/tex] = final volume = 2.35 L
Putting values in above equation, we get:
[tex]W=-0.96atm\times (2.35-5.68)L=3.20L.atm[/tex]
To convert this into joules, we use the conversion factor:
[tex]1L.atm=101.33J[/tex]
So, [tex]3.20L.atm=3.20\times 101.3=324J[/tex]
The positive sign indicates the work is done on the system
Hence, the work done for the given process is 324 J
Given values:
Pressure, [tex]P = 732 \ torr \ or \ 0.96 \ atm[/tex]Initial volume, [tex]V_1 = 5.68 \ L[/tex]Final volume, [tex]V_2 = 2.35 \ L[/tex]We know the equation,
→ [tex]W = - P\Delta V[/tex]
[tex]= -P(V_2-V_1)[/tex]
By substituting the values,
[tex]= -0.96\times (2.35-5.68)[/tex]
[tex]= 3.20 \ L.atm[/tex]
By converting it in "J",
[tex]= 3.20\times 101.3[/tex]
[tex]= 324 \ J[/tex]
Thus the above answer is right.
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Two identical capacitors are connected parallel. Initially they are charged to a potential V0 and each acquired a charge Q0. The battery is then disconnected, and the gap between the plates of one capacitor is filled with a dielectric. (a) What is the new potential difference V across the capacitors. possible asnwers: V=(Vo)^2/[kQo+Vo), V=Vo/2k, V=Vo/2, V=kQo/Vo, V=2Vo/[k+1]
(b) If the dielectric constant is 7.8, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.
Answer:
Explanation:
capacitance of each capacitor
C₀= Q₀ / V₀
V₀ = Q₀ / C₀
New total capacitance = C₀ ( 1 + K )
Common potential
= total charge / total capacitance
= 2 Q₀ / [ C₀ ( 1 + K ) ]
2 V₀ / ( 1 + K )
b )
Common potential = 2 x V₀ / ( 1 + 7.8 )
= .227 V₀
charge on capacitor with dielectric
= .227 V₀ x 7.8 C₀
= 1.77 V₀C₀
= 1.77 Q₀
Ratio required = 1.77
The new potential difference V across the capacitors and the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge can be calculated using specific equations.
Explanation:(a) What is the new potential difference V across the capacitors.
To find the new potential difference V across the capacitors, we need to consider the effect of adding a dielectric. The potential difference V is given by the equation:
V = Vo/2k
where Vo is the initial potential difference and k is the dielectric constant. In this case, since the dielectric constant is 7.8, we can substitute the values and calculate the new potential difference V.
(b) If the dielectric constant is 7.8, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.
The ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge can be calculated using the equation:
Q_with_dielectric / Q_initial = k
where Q_with_dielectric is the charge on the capacitor with the dielectric and Q_initial is the initial charge. Given that k is 7.8, we can substitute the values and calculate the ratio.
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A bat strikes a 0.145 kgkg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/sm/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘30∘ above horizontal with a speed of 38.0 m/sm/s. The ball and bat are in contact for 1.75 msms. find the horizontal and vertical components of the average force on the ball.
Answer with Explanation:
We are given that
Mass of ball=m=0.145 kg
Initially horizontal velocity of ball,ux=50 m/s
[tex]\theta=30^{\circ}[/tex]
Final velocity of ball,v=38m/s
Time ,t=1.75 ms=[tex]1.75\times 10^{-3} s[/tex]
[tex]1 ms=10^{-3} s[/tex]
Horizontal component of average force, [tex]F_x=\frac{m(vcos\theta-u_x)}{t}[/tex]
Using the formula
Horizontal component of average force, [tex]F_x=\frac{0.145(-38cos30-50)}{1.75\times 10^{-3}}=-6.9\times 10^3[/tex]N
Vertical component of average force, [tex]F_y=\frac{m(vsin\theta-u_y)}{t}[/tex]
Vertical component of average force,[tex]F_y=\frac{0.145(38sin30-0}{1.75\times 10^{-3}}=1.6\times 10^3 N[/tex]
fter driving a portion of the route, the taptap is fully loaded with a total of 25 people including the driver, with an average mass of 69 kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed
Answer:
The spring compressed is 0.44 m.
Explanation:
Given that,
Number of person = 25
Average Mass of each person [tex]m_{p}= 25\times69 = 1725\ kg[/tex]
Mass of goat [tex]m_{g}= 3\times15 = 45\ kg[/tex]
Mass of chicken [tex]m_{c}= 5\times3 = 15\ kg[/tex]
Mass of bananas = 25 kg
If the spring constant is [tex]4\times10^{4}\ N/m[/tex]
We need to calculate the total mass
[tex]M=m_{p}+m_{g}+m_{c}+m_{b}[/tex]
Put the value in the formula
[tex]M=1725+45+15+25[/tex]
[tex]M=1810\ kg[/tex]
The weight of all these things is
[tex]W=Mg[/tex]
[tex]W=1810\times9.8[/tex]
[tex]W=17738\ N[/tex]
We need to calculate the distance
Using formula of restoring force
[tex]F=kx[/tex]
[tex]x=\dfrac{F}{k}[/tex]
Put the value into the formula
[tex]x=\dfrac{17738}{4\times10^{4}}[/tex]
[tex]x=0.44\ m[/tex]
Hence, The spring compressed is 0.44 m.
A 0.1-kilogram block is attached to an initially unstretched spring of force constant k = 40 newtons per meter as shown above. The block is released from rest at time t=0.
What is the amplitude, in meters, of the resulting simple harmonic motion of the block?
A) 1/40m
B) 1/20m
C) 1/4m
D) 1/2m
Answer:
The amplitude, in meters, of the resulting simple harmonic motion of the block 1/40 m
Explanation:
given information:
the block mass, m = 0.1 kg
spring of force constant, k = 40 N/m
release at t = 0
the formula for simple harmonic motion is
F = kx
where
F = force (N)
k = spring of force constant (N/m)
x = amplitude (m)
thus,
F = kx
x = F/k
= m g/k
= 0.1 x 10/40
= 1/40 m
A 20 centimeter long copper wire sensor is designed to alert if the temperature elevates beyond a safe region causing the wire to make contact with the wall, 1.5 mm away at 25 C. What temperature (in K) will the copper wire make contact with the wall, setting the sensor off? The thermal coefficient of expansion, α, for copper (Cu) is found to be 16.6 ppm/K (16.6x10-6 /K).
Answer:
The temperature is 749.8 K
Explanation:
Final temperature (T2) = (distance apart/thermal coefficient of expansion×length) + initial temperature
distance apart = 1.5 mm = 1.5/1000 = 0.0015 m
thermal coefficient of expansion for copper = 16.6×10^-6/K
Length of copper = 20 cm = 20/100 = 0.2 m
Initial temperature = 25 °C = 25 + 273 = 298 K
T2 = (0.0015/16.6×10^-6×0.2) + 298 = 451.8 + 298 = 749.8 K
A water molecule perpendicular to an electric field has 1.40×10−21 J more potential energy than a water molecule aligned with the field. The dipole moment of a water molecule is 6.2×10−30Cm.
What is the strength of the electric field?
To solve this problem we will apply the concepts related to the potential energy in the molecules and obtain its electric field through the relationship given by the dipole moment. The change in potential energy from one state to another is given by,
[tex]U_2-U_1 = 1.4*10^{-21} J[/tex]
From this difference we can identify that [tex]U_1[/tex] is equivalent to the potential energy when it is perpendicular to the electric field. At the same time, the potential energy [tex]U_2[/tex] would be equivalent when it is aligned with the electric field.
From there the relationship between energy, the dipole moment and the electric field would be subject to
[tex]U_2-U_1 = pE[/tex]
Here,
[tex]p = \text{Dipole moment} = 6.2*10^{-30} C \cdot m[/tex]
Rearranging to find the electric field,
[tex]E = \frac{(U_2-U_1)}{p}[/tex]
[tex]E = \frac{(1.4*10^{-21})}{(6.2*10^{-30})}[/tex]
[tex]E =2.25*10^8 N/C[/tex]
Therefore the electric field is [tex]2.25*10^8N/C[/tex]
(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the battery? V (b) If the same capacitor is connected to another battery and 192.0 µC of charge is stored on the capacitor, what is the voltage of the battery?
Answer:
a.6 V
b.24 V
Explanation:
We are given that
a.[tex]C=8\mu F=8\times 10^{-6} F[/tex]
[tex]1\mu =10^{-6} [/tex]
Q=[tex]48\mu C=48\times 10^{-6} C[/tex]
We know that
[tex]V=\frac{Q}{C}[/tex]
Using the formula
[tex]V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V[/tex]
b.[tex]Q=192\mu C=192\times 10^{-6} C[/tex]
[tex]V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V[/tex]
A tennis ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s. A) How long does it take the ball to reach its maximum height?B) How high does the ball rise?C) What is the magnitude of the acceleration due to gravity on the surface of this planet?D) Determine the velocity of the ball when it returns to its original position. Note: assume the upward direction is positive.E) How long has the ball been in the air when it returns to its original position?
Answer:
(A) t = 4s
(B) H = 40m
(C) g = 5m/s²
(D) V = -20m/s
(E) t = 8s
The detailed solution to this problem requires the knowledge of costant linear acceleration motion.xplanation:
The detailed solution to this problem requires the knowledge of costant linear acceleration motion.
Explanation:
The full solution can be found in the attachment below.
To answer part A, we have been given some values of velocities bounding the time jnterval of 1s from which we can calculate the acceleration due to gravity and then the time.
Part B
Requires just using the acceleration due to gravity and the time taken in the equation
H = ut - 1/2gt²
Part C
Has already been calculated in part A
Part D
V = -20m/s because the tennis ball is coming down as the upward direction was assumed positive.
Part E
When the ball retirns to its original position it is the same as it never left and so H = 0m
The calculation can be found in the attachment below.
Final answer:
Ball reaches peak at 4s, rises 80m, experiences -5m/s² gravity. Returns at -20m/s and spends 8s in the air (4s up + 4s down).
Explanation:
Here are the answers to your questions about the tennis ball on the atmosphere-free planet:
A) Time to reach maximum height: Between the beginning and one second later, the ball's velocity decreases by 5.0 m/s. Since acceleration is constant (no atmosphere), this decrease is due to gravity acting against the initial velocity. We can use the equation:
v = u + at
where:
v = final velocity (15.0 m/s)
u = initial velocity (20.0 m/s)
a = acceleration due to gravity (unknown, negative direction)
t = time (1 second)
Solving for a:
a = (v - u) / t = (15.0 m/s - 20.0 m/s) / 1 s = -5.0 m/s²
Now, to find the time to reach maximum height, we know that at the peak, the velocity is 0 m/s. We can again use the same equation:
0 = u + at
Solving for t:
t = -u / a = -20.0 m/s / (-5.0 m/s²) = 4.0 seconds (positive result since time cannot be negative)
Therefore, it takes 4.0 seconds for the ball to reach its maximum height.
B) Maximum height:
We can use the kinematic equation to find the maximum height (h):
h = ut + 1/2 × at²
Substituting the values:
h = 20.0 m/s × 4.0 s + 1/2 × (-5.0 m/s²) × (4.0 s)² = 40.0 m + 40.0 m = 80.0 meters
C) Acceleration due to gravity:
We already calculated the acceleration due to gravity in part A: -5.0 m/s² (the negative sign indicates downward direction).
D) Velocity upon returning to the original position:
Since the motion is symmetrical (same initial and final positions), the velocity when the ball returns will be the negative of its initial velocity: -20.0 m/s.
E) Time spent in the air:
The total time in the air is the sum of the time to reach the maximum height and the time to fall back to the original position. Since the motion is symmetrical, the time to fall back is also 4.0 seconds. Therefore, the total time in the air is:
4.0 seconds (upward) + 4.0 seconds (downward) = 8.0 seconds
Water from a fire hose is directed horizontally against a wall at a rate of 53.9 kg/s and a speed of 38.9 m/s. Calculate the magnitude of the force exerted on the wall (in N), assuming the water's horizontal momentum is reduced to zero.
Answer:
[tex]F_{avg}=2096.71N[/tex]
Explanation:
As from the given data that:
m/Δt=53.9kg/s
The numerator is the mass and denominator is the change in time
Solve for change in velocity we have:
Δv=v₂-v₁
[tex]=0-38.9m/s\\=-38.9m/s[/tex]
Δv= -38.9m/s
From Newtons second law we know that:
F=ma
We can write this equation as:
Favg=(m/Δt)Δv
Substitute the given values
So
[tex]F_{avg}=(53.9kg/s)(-38.9m/s)\\F_{avg}=-2096.71N[/tex]
Using the average velocity formula.The answer will be positive, the negative only implies that force is coming away from the wall
[tex]F_{avg}=2096.71N[/tex]
The resistivity of a metallic, single-walled carbon nanotube is 2.30 ✕ 10−8 Ω · m. The electron number density is 6.60 ✕ 1028 m−3. What is the mean free time for the electrons flowing in a current along the carbon nanotube?
Answer:
The mean free time is given as z [tex]= 2.3*10^{-14}sec[/tex]
Explanation:
Generally the formula for resistivity
[tex]\rho = \frac{m_e}{e^2 n_ez}[/tex]
Where
[tex]\rho[/tex] is he resistivity of metal
[tex]m_e[/tex] is the mass of the electron
[tex]e[/tex] is the charge of the electron
[tex]n_e[/tex] is electron density
[tex]z[/tex] is the mean free time
Now making z the subject of the formula
=> [tex]z = \frac{m_e}{e^2n_e\rho}[/tex]
Substituting the given values
[tex]z = \frac{9*10^{-31}}{(1.6*10^{-19}(16.1*10^{28}(2.5*10^{-8})))}[/tex]
[tex]= 2.3*10^{-14}sec[/tex]
A 2.0 m ordinary lamp extension cord carries a 3.0 A current. Such a cord typically consists of two parallel wires carrying equal currents in opposite directions.
a. Find the magnitude of the force that the two segments of this cord exert on each other. (You will need to inspect an actual lamp cord at home and measure or reasonably estimate the quantities needed to do this calculation.)
b. Find the direction (attractive or repulsive) of the force that the two segments of this cord exert on each other. (Opposite or same directional currents, and do the wires attract or repel?)
Our values are given are,
[tex]l = 2m[/tex]
[tex]I = 3A[/tex]
[tex]r = 3.5mm = 3.5*10^{-3} m[/tex]
Electromagnetic force can then be defined as,
[tex]F = \frac{\mu_0 I^2 l}{2\pi r}[/tex]
Here,
[tex]\mu_0[/tex] = Permeability free space
I = Current
l = Length
r = Distance between them
Replacing,
[tex]F = \frac{(4\pi *10^{-7})(3)^2(2)}{(2\pi)(3.5*10^{-3})}[/tex]
[tex]F = 0.001028N[/tex]
Therefore the magnitude of the force that two segments of this cord exert on each other is 0.001028N.
Since the magnitude of the force is positive, the direction of both is attractive.
a) The magnitude of the force that the two segments of the cord exert on each other is [tex]\(1.2 \times 10^{-4} \, \text{N}\)[/tex]
b) The force is repulsive.
The force per unit length between two parallel wires carrying currents [tex]\(I_1\)[/tex] and [tex]\(I_2\)[/tex] separated by a distance d is given by Ampère's law:
[tex]\[F/L = \frac{\mu_0 I_1 I_2}{2 \pi d}\][/tex]
Where:
F is the force,
L is the length of the wires,
[tex]\(\mu_0\)[/tex] is the permeability of free space [tex](\(4 \pi \times 10^{-7} \, \text{N/A}^2\))[/tex],
[tex]\(I_1\)[/tex] and [tex]\(I_2\)[/tex] are the currents through the wires (in this case, both are 3.0 A but in opposite directions),
d is the distance between the wires.
a. First, assume the distance d between the wires in the extension cord. A reasonable estimate for the distance between the wires in a standard extension cord might be around 3 mm (0.003 m).
It is given that:
[tex]\(I_1 = I_2 = 3.0 \, \text{A}\),[/tex]
[tex]\(L = 2.0 \, \text{m}\),[/tex]
[tex]\(d \approx 0.003 \, \text{m}\).[/tex]
Using the formula for the force per unit length:
[tex]\[\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}\][/tex]
Substitute the known values:
[tex]\[\frac{F}{L} = \frac{(4 \pi \times 10^{-7} \, \text{N/A}^2) \times (3.0 \, \text{A})^2}{2 \pi \times 0.003 \, \text{m}}\][/tex]
[tex]\[\frac{F}{L} = \frac{4 \pi \times 10^{-7} \times 9}{2 \pi \times 0.003}\][/tex]
[tex]\[\frac{F}{L} = \frac{36 \times 10^{-7}}{0.006}\][/tex]
[tex]\[\frac{F}{L} = 6 \times 10^{-5} \, \text{N/m}\][/tex]
Now, multiply by the length L to find the total force:
[tex]\[F = \left(6 \times 10^{-5} \, \text{N/m}\right) \times 2.0 \, \text{m}\][/tex]
[tex]\[F = 1.2 \times 10^{-4} \, \text{N}\][/tex]
b. Since the currents in the two parallel wires are equal but in opposite directions, the magnetic force between them will be repulsive. This is because currents flowing in opposite directions repel each other according to the right-hand rule for magnetic fields and forces between current-carrying wires.
The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s).
Find the speed of the passengers when the Ferris wheel is rotating at this rate.
Answer:
V = 5.24m/s
Explanation:
See attachment below.