The question is incomplete, the complete question is;
A spherical steel ball bearing has a diameter of 2.540cm at 25.00∘C.
(a) What is its diameter when it's temperature is raised to
100∘C?
(b) What temperature change is required to increase its volume by
1.000% ?
Answer:
a) 2.542 cm
b) 303.03°C
Explanation:
Given;
Diameter of the ball= 2.540cm
Initial temperature= 25.0°C
Final temperature= 100.0°C
Percentage increase in volume = 1.000%
Temperature coefficient of expansion for steel =11.0×10^−6/∘C
d2= d1[1 + α(T2-T1)]
d2= 2.540[1 + 11.0×10^−6(100-25)]
d2= 2.540[1 + 8.25×10^-4]
d2= 2.542 cm
From;
%V ×1/100 = V ×3α ×∆T/ V
Substituting values;
1.000 ×1/100= 3× 11.0×10^−6 × ∆T
∆T= 0.01/3× 11.0×10^−6
∆T= 303.03°C
A railroad boxcar rolls on a track at 2.90 m/s toward two identical coupled boxcars, which are rolling in the same direction as the first, but at a speed of 1.20 m/s. The first reaches the second two and all couple together. The mass of each is 3.05 ✕ 104 kg.(a)What is the speed (in m/s) of the three coupled cars after the first couples with the other two? (Round your answer to at least two decimal places.)Incorrect: Your answer is incorrect.What is the momentum of the two coupled cars? What is the momentum of the first car in terms of its mass and initial speed? Note all cars are initially traveling in the same direction. Apply conservation of momentum to find the final speed. m/s(b)Find the (absolute value of the) amount of kinetic energy (in J) converted to other forms during the collision.J
Answer:
momentum of the coupled cars V = 1.77 m/s
kinetic energy coverted to other forms during the collision ΔK.E = -2.892×10⁴J
Explanation:
given
m₁ =3.05 × 10⁴kg
u₁ =2.90m/s
m₂=6.10× 10⁴kg
u₂=1.20m/s
using law of conservation of momentum
m₁u₁ + m₂u₂ = (m₁ + m₂) V
3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)V
V = 1.617×10⁵/9.15×10⁴
V = 1.77m/s
K.E =1/2mV²
ΔK.E = K.E(final) - K.E(initial)
ΔK.E = ¹/₂ × 9.15×10⁴ ×(1.77)² - ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²
ΔK.E = ¹/₂ × (28.67-25.65-8.784) ×10⁴
ΔK.E = -2.892×10⁴J
The final speed is 1.77 m/s
The initial momentum is 8.84 × 10⁴ kgm/s [first car] and 7.3 × 10⁴ kgm/s [coupled car]
2.892×10⁴J of energy is converted.
Inelastic collision:Since the first boxcar collides and couples with the two coupled boxcars, the collision is inelastic. In an inelastic collision, the momentum of the system is conserved but there is a loss in the total kinetic energy of the system.
Let the mass of the railroad boxcar be m₁ =3.05 × 10⁴kg
The initial speed of the railroad boxcar is u₁ = 2.90m/s
Mass of the two coupled boxcars m₂ = 2 × 3.05 × 10⁴kg = 6.10× 10⁴kg
And the initial speed be u₂ = 1.20m/s
The initial momentum of the first car is:
m₁u₁ = 3.05 × 10⁴ × 2.90 = 8.84 × 10⁴ kgm/s
The initial momentum of the coupled car is:
m₁u₁ = 6.10 × 10⁴ × 1.20 = 7.3 × 10⁴ kgm/s
Let the final speed after all the boxcars are coupled be v
From the law of conservation of momentum, we get:
m₁u₁ + m₂u₂ = (m₁ + m₂)v
3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)Vv
v = 1.617×10⁵/9.15×10⁴
v = 1.77m/s
The difference between initial and final kinetic energies is the amount of energy converted into other forms, which is given as follows:
ΔKE = K.E(final) - K.E(initial)
ΔKE = ¹/₂ × 9.15×10⁴ ×(1.77)² - ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²
ΔKE = ¹/₂ × (28.67-25.65-8.784) ×10⁴
ΔKE = -2.892×10⁴J
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A pulley system is used at a dock to lift shipments of fish off a boat. If you apply a force of 100 N to the pulley, it pulls the shipment with a force of 830 N. a. What is the mechanical advantage of the pulley? b. The pulley has an efficiency of 80%. If you perform 600 J of work, how much useful work does the pulley do?
Explanation:
a. Mechanical advantage = force out / force in
MA = 830 / 100
MA = 8.3
b. Efficiency = work out / work in
0.80 = W / 600 J
W = 480 J
Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and total work transfer
Answer:
Total heat transfer is positive
Total work transfer is positive
Explanation:
The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.
Heat transfer to a system is positive and that transferred from the system is negative.
Also, work done by a system is positive while the work done on the system is negative.
Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)
Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).
A particle with kinetic energy equal to 282 J has a momentum of magnitude 26.4 kg · m/s. Calculate the speed (in m/s) and the mass (in kg) of the particle.
Answer:
[tex]v=21.36\,\,\frac{m}{s}\\[/tex]
[tex]m=1.2357\,\,kg[/tex]
Explanation:
Recall the formula for linear momentum (p):
[tex]p = m\,v[/tex] which in our case equals 26.4 kg m/s
and notice that the kinetic energy can be written in terms of the linear momentum (p) as shown below:
[tex]K=\frac{1}{2} m\,v^2=\frac{1}{2} \frac{m^2\,v^2}{m} =\frac{1}{2}\frac{(m\,v)^2}{m} =\frac{p^2}{2\,m}[/tex]
Then, we can solve for the mass (m) given the information we have on the kinetic energy and momentum of the particle:
[tex]K=\frac{p^2}{2\,m}\\282=\frac{26.4^2}{2\,m}\\m=\frac{26.4^2}{2\,(282)}\,kg\\m=1.2357\,\,kg[/tex]
Now by knowing the particle's mass, we use the momentum formula to find its speed:
[tex]p=m\,v\\26.4=1.2357\,v\\v=\frac{26.4}{1.2357} \,\frac{m}{s} \\v=21.36\,\,\frac{m}{s}[/tex]
A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t squared plus 2 t, where v (t )is in meters per second. a) How far does the particle travel during the first 2 sec? b) How far does it travel during the second 2 sec?
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
which statement did Ernest Rutherford make about atoms?
Answer:
Option A
Explanation:
Ernest Rutherford concluded that the atom has a small, dense center which constitutes the mass of the whole atom. He called it a "Nucleus". He also said that most of the space in the atom is empty.
A radiograph of the forearm is produced using 4 mA at 75 kVp. What kvp would be required to double the exposure?
A) 86 kVp
B) 66 kVp
76 kVp
D) 8,6 kVp
Answer:
Required KVP = 86 KVP (Approx)
Explanation:
Given:
Current KVP = 75 KVP
Find:
KVP required to double exposure
Computation:
According to 15% rule of KVP,
15% change increse in KVP required to get double exposure.
So,
Required KVP = Current KVP + Current KVP(15%)
Required KVP = 75 KVP + 75 (15%)
Required KVP = 75 KVP + 11.25 KVP
Required KVP = 86.25 KVP
Required KVP = 86 KVP (Approx)
Why can a magnetic monopole not exist, assuming Maxwell's Equations are currently correct and complete?
Answer:
Because closed magnetic field loops have to be formed between both ends of the magnet, a magnet will always have two poles.
Explanation:
Magnetic Monopoles do not exist in nature because a magnetic field always forms a loop that runs from one end of the magnet to the other.
Since this loop of the magnetic field has an origination and termination point which are at the two ends of the magnet (North and South poles). A magnet will always be bipolar which is in this case, North and South; even at an atomic level.
A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate v=0.5⋅10−9 m3 per unit time. Use differentials to estimate the maximum error in the viscosity η given by
Answer:
The maximum error is [tex]\Delta \eta = 2032.9[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 1\ m[/tex]
The radius is [tex]r = 0.002 \pm 0.0002 \ m[/tex]
The pressure is [tex]P = 4 *10^{5} \ \pm 1750[/tex]
The rate is [tex]v = 0.5*10^{-9} \ m^3 /t[/tex]
The viscosity is [tex]\eta = \frac{\pi}{8} * \frac{P * r^4}{v}[/tex]
The error in the viscosity is mathematically represented as
[tex]\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v[/tex]
Where [tex]\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}[/tex]
and [tex]\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}[/tex]
and [tex]\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}[/tex]
So
[tex]\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v][/tex]
substituting values
[tex]\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ][/tex]
[tex]\Delta \eta = \frac{\pi}{8} [56 + 5120 ][/tex]
[tex]\Delta \eta = 647 \pi[/tex]
[tex]\Delta \eta = 2032.9[/tex]
A parallel-plate air capacitor is connected to a constant-voltage battery. If the separation between the capacitor plates is doubled while the capacitor remains connected to the battery, the energy stored in the capacitor
1) drops to one-fourth its previous value.
2) quadruples.
3) becomes six times its previous value.
4) doubles.
5) drops to one-third its previous value.
6) Not enough information is provided.
7) triples.
8) drops to half its previous value.
9) drops to one-sixth its previous value.
10) remains unchanged.
Answer:
Drop to half of the previous value
Explanation:
Energy stored in capacitor is inversly propotional to the distance between the plates.
If the separation between the capacitor plates is doubled while the capacitor remains connected to the battery, the energy stored in the capacitor drops to half its previous value.
What is parallel plate capacitor?The two parallel plates placed at a distance apart used to store charge when electric supply is on.
The capacitance of a capacitor is given by
C = ε₀ A/d
where, ε₀ is the permittivity of free space, A = area of cross section of plates and d is the distance between them.
Capacitance is inversely proportional to the distance between them. So, if distance is doubled, the capacitance decreases to half its original value.
Thus, the correct option is 8.
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An amusement park ride has a vertical cylinder with an inner radius of 3.4 m, which rotates about its vertical axis. Riders stand inside against the carpeted surface and rotate with the cylinder while it accelerates to its full angular velocity. At that point the floor drops away and friction between the riders and the cylinder prevents them from sliding downward. The coefficient of static friction between the riders and the cylinder is 0.87. What minimum angular velocity in radians/second is necessary to assure that the riders will not slide down the wall?
Answer:
The minimum angular velocity necessary to assure that the riders will not slide down the wall is 1.58 rad/second.
Explanation:
The riders will experience a centripetal force from the cylinder
[tex]F_{C}[/tex] = mrω^2 .... equ 1
where
m is the mass of the rider
r is the inner radius of the cylinder = 3.4 m
ω is the angular speed of of the rider
For the riders not to slide downwards, this centripetal force is balanced by the friction between the riders and the cylinder. The frictional force is given as
[tex]F_{f}[/tex] = μR ....equ 2
where
μ = coefficient of friction = 0.87
R is the normal force from the rider = mg
where
m is the rider's mass
g is the acceleration due to gravity = 9.81 m/s
substitute mg for R in equ 2, we'll have
[tex]F_{f}[/tex] = μmg ....equ 3
Equating centripetal force of equ 1 and frictional force of equ 3, we'll get
mrω^2 = μmg
the mass of the rider cancels out, and we are left with
rω^2 = μg
ω^2 = μg/r
ω = [tex]\sqrt{\frac{ug}{r} }[/tex]
ω = [tex]\sqrt{\frac{0.87*9.81}{3.4} }[/tex]
ω = 1.58 rad/second
The minimum angular velocity necessary so that the riders will not slide down the wall is 1.58 rad/s
The riders will experience a centripetal force from the cylinder
[tex]F = mrw^2[/tex]
where m is the mass of the rider
r is the inner radius of the cylinder = 3.4 m
ω is the angular speed of the rider
For the riders not to slide downwards, this centripetal force must be balanced by friction. The frictional force is given as
f = μN
where
μ = coefficient of friction = 0.87
N is the normal force = mg
f = μmg
Equating centripetal force of and frictional force of we'll get
[tex]mrw^2 = umg[/tex]
[tex]rw^2 = ug[/tex]
[tex]w^2 = ug/r[/tex]
[tex]w= \sqrt{ug/r}[/tex]
[tex]w= \sqrt{0.87*9.8/3.4}[/tex]
ω = 1.58 rad/s is the minimum angular velocity needed to prevent the rider from sliding.
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Which observation have scientists used to support Einstein's general theory of relativity?
The orbital path of Mercury around the Sun has changed.
O GPS clocks function at the same rate on both Earth and in space.
O The Sun has gotten more massive over time.
Objects act differently in a gravity field than in an accelerating reference frame.
Answer:
Objects act differently in a gravity field than in an accelerating reference frame.
Explanation:
The main thrust of the theory general relativity as proposed by Albert Einstein boarders on space and time as the two fundamental aspects of spacetime. Spacetime is curved in the presence of gravity, matter, energy, and momentum. The theory of general relativity explains gravity based on the way space can 'curve', that is, it seeks to relate gravitational force to the changing geometry of space-time.
The Einstein general theory of relativity has replaced Newton's ideas proposed in earlier centuries as a means of predicting gravitational interactions. This concept is quite helpful but cannot be fitted into the context of quantum mechanics due to obvious incompatibilities.
Answer:
A - The orbital path of mercury around the sun has changed.
Explanation:
got right on edg.
Two plane mirrors are stood vertically making a right angle between them. How many images of an object close to and in front of the mirrors can be seen
Answer:
3
Explanation:
When two plane mirrors are placed side by side such that they make some angle, θ, with each other, the number of images, n, of an object placed close to and in front of these mirrors is given by;
n = (360 / θ) - 1 ------------(i)
From the question;
θ = 90° [since they stood making a right angle with each other]
Substitute this value into equation (i) as follows;
n = (360 / 90) - 1
n = 4 - 1
n = 3
Therefore, the number of images formed is 3
A building is located on earth's equator. As the earth rotates about its axis, which floor of the building has the greatest angular speed?
Answer:
The angular speed of the earth rotation is equal. Therefore
Our angular speed due to Earth’s rotation is same at every point on the earth irrespective of the elevation. So your angular speed due to earth’s rotation on the top floor of the building will be same as it is on the ground floor.
Explanation:
What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T magnetic field?
Answer:
E = 1.50 × [tex]10^{8}[/tex] V/m
Explanation:
given data
B = 0.50 T
solution
we know that energy density by the magnetic field is express as
[tex]\mu _b = \frac{B}{2\mu _o}[/tex] ...............1
and
energy density due to electric filed is
[tex]\mu _e = \frac{\epsilon _o E^2}{2}[/tex] ...............2
and here [tex]\mu _b = \mu _ e[/tex]
so that
E = [tex]\frac{B}{\sqrt{\mu _o \times \epsilon _o}}[/tex] ...................3
put here value and we get
[tex]E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}[/tex]
E = 3 × [tex]10^{8}[/tex] × 0.50
E = 1.50 × [tex]10^{8}[/tex] V/m
In an experiment different wavelengths of light, all able to eject photoelectrons, shine on a freshly prepared (oxide-free) zinc surface. Which statement is true
Answer:
the energy of the photons is greater than the work function of the zinc oxide.
h f> = Ф
Explanation:
In this experiment on the photoelectric effect, it is explained by the Einstein relation that considers the light beam formed by discrete energy packages.
K_max = h f - Ф
in the exercise phase, they indicate that different wavelengths can inject electrons, so the energy of the photons is greater than the work function of the zinc oxide.
h f > = Ф
How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
Question:
A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes
How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
Answer:
S = 5.508 × 10¹¹m
V = 2.62 × 10⁸ m/s
Explanation:
The radius of the orbit of Jupiter, Rj is 43.2 light-minutes
radius of the orbit of Mars, Rm is 12.6 light-minutes
Distance travelled S = (Rj - Rm)
= 43.2 - 12.6 = 30.6 light- minutes
= 30.6 × (3 ×10⁸m/s) × 60 s
= 5.508 × 10¹¹m
time = 35mins = (35 × 60 secs)
= 2100 secs
speed = distance/time
V = 5.508 × 10¹¹m / 2100 s
V = 2.62 × 10⁸ m/s
You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current in the wire is 1.2A. What is the radius of the coils and how many loops, N, are there
Answer:
radius of the loop = 7.9 mm
number of turns N ≅ 399 turns
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ = [tex]4\pi *10^{-7} T-m/A[/tex] = 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ 399 turns
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop = [tex]2\pi r[/tex]
0.005 = 2 x 3.142 x r
r = 0.005/6.284 = [tex]7.9*10^{-4} m[/tex] = 0.0079 m = 7.9 mm
A tightly wound toroid of inner radius 1.2 cm and outer radius 2.4 cm has 960 turns of wire and carries a current of 2.5 A.
Requried:
a. What is the magnetic field at a distance of 0.9 cm from the center?
b. What is the field 1.2 cm from the center?
Answer:
a
[tex]B = 0.0533 \ T[/tex]
b
[tex]B = 0.04 \ T[/tex]
Explanation:
From the question we are told that
The inner radius is [tex]r = 1.2 \ cm = 0.012 \ m[/tex]
The outer radius is [tex]r_o = 2.4 \ cm = \frac{2.4}{100} = 0.024 \ m[/tex]
The nu umber of turns is [tex]N = 960[/tex]
The current it is carrying is [tex]I = 2. 5 A[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * N* I }{2 * \pi * r }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a constant value
[tex]\mu = 4\pi * 10^{-7} N/A^2[/tex]
And the given distance where the magnetic field is felt is r = 0.9 cm = 0.009 m
Now substituting values
[tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.009 }[/tex]
[tex]B = 0.0533 \ T[/tex]
Fro the second question the distance of the position considered from the center is r = 1.2 cm = 0.012 m
So the magnetic field is
[tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.012 }[/tex]
[tex]B = 0.04 \ T[/tex]
The magnetic field at a distance of 0.9 cm from the center of the toroid is 0.053 T.
The magnetic field at a distance of 1.2 cm from the center of the toroid is 0.04 T.
The given parameters;
radius of the toroid, r = 1.2 cm = 0.012 mouter radius of the toroid, R = 2.4 cm = 0.024 mnumber of turns, N = 960 turnscurrent in wire, I = 2.5 AThe magnetic field at a distance of 0.9 cm from the center of the toroid is calculated as follows;
[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.009} \\\\B = 0.053 \ T[/tex]
The magnetic field at a distance of 1.2 cm from the center of the toroid is calculated as follows;
[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.012} \\\\B = 0.04 \ T[/tex]
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A ball is dropped from the top of an eleven-story building to a balcony on the ninth floor. In which case is the change in the potential energy associated with the motion of the ball the greatest
Answer:
at the top of the 9 story building i think
Explanation:
When the ball starts to move, its kinetic energy increases and potential energy decreases. Thus the ball will experience its maximum potential energy at the top height before falling.
What is potential energy?Potential energy of a massive body is the energy formed by virtue of its position and displacement. Potential energy is related to the mass, height and gravity as P = Mgh.
Where, g is gravity m is mass of the body and h is the height from the surface. Potential energy is directly proportional to mass, gravity and height.
Thus, as the height from the surface increases, the body acquires its maximum potential energy. When the body starts moving its kinetic energy progresses and reaches to zero potential energy.
Therefore, at the sate where the ball is at the top of the building it have maximum potential energy and then changes to zero.
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a certain plane parallel capacitor stores energy E when the plates have a charge Q on each plate. Then distance between the plates is double. In order to store triply as much energy, how much charge should it have in its plates
Answer:
[tex]Q'=\sqrt{6}Q[/tex]
Explanation:
You have that a parallel plate capacitor has a total energy of E when the distance between the plates is d and the charge on each plate is Q.
You take into account the following formula for the stored energy in the capacitor:
[tex]E=\frac{1}{2}\frac{Q^2}{C}[/tex] (1)
The capacitance C of the parallel plate capacitor is given by the following formula is:
[tex]C=\epsilon_o\frac{A}{d}[/tex] (2)
A: area of the plates
ε0: dielectric permittivity of vacuum
You replace the expression (2) into the equation (1):
[tex]E=\frac{1}{2}\frac{Q^2A}{\epsilon_o d}[/tex] (3)
the previous formula is the expression for the total energy stored for the given parameters A, d and Q.
If the distance between the plates is twice and it is required that the energy is three times the initial energy, to find the value of the charge you use the equation (3):
[tex]E'=\frac{1}{2}\frac{Q'^2A}{\epsilon_o d'}[/tex] (4)
d' = 2d
E' = 3E
Q': required charge
You replace the values of d' and E' in the equation (4) and then divide the result with the equation (3):
[tex]3E=\frac{1}{2}\frac{Q'^2A}{\epsilon_o(2d)}=\frac{1}{4}\frac{Q'^2A}{\epsilon_od}\\\\\frac{3E}{E}=\frac{1/4\frac{Q'^2A}{\epsilon_od}}{1/2\frac{Q^2A}{\epsilon_o d}}\\\\3=\frac{1}{2}\frac{Q'^2}{Q^2}[/tex]
Finally, you solve for Q':
[tex]3=\frac{1}{2}\frac{Q'^2}{Q^2}\\\\Q'=\sqrt{6}Q[/tex]
Then, the required charge is √6Q , to obtain three times the initial energy E, when the distance between plates is doubled.
A 5000 kg railcar hits a bumper (a spring) at 1 m/s, and the spring compresses 0.1 meters. Assume no damping. a) Find the spring constant k.
Answer:
k = 0.5 MN/m
Explanation:
Mass of the railcar, m = 5000 kg
Speed of the rail car, v = 1 m/s
The Kinetic energy(KE) of the railcar is given by the equation:
KE = 0.5 mv²
KE = 0.5 * 5000 * 1²
KE = 2500 J
The spring's compression, x = 0.1 m
The potential energy(PE) stored in the spring is given by the equation:
PE = 0.5kx²
PE = 0.5 * k * 0.1²
PE = 0.005k
According to the principle of energy conservation, Kinetic energy of the railcar equals the potential energy stored in the spring
KE = PE
2500 = 0.005k
k = 2500/0.005
k = 500000 N/m
k = 0.5 MN/m
Raven throw a baseball directly downward from a terrace froma speed of 5.0 m/s. How fast it will be moving when it hits the path way 3.0 m below
Answer:
The speed of the ball at this distance is 9.15 m/s
Explanation:
Given;
initial speed of the baseball, U = 5.0 m/s.
distance traveled along the path way, h = 3 m
final speed of the baseball at this distance, V = ?
The baseball is falling under the influence of gravity.
Acceleration due to gravity, g is positive, since the baseball is falling towards its direction.
g = 9.8 m/s²
Apply the third kinematic equation;
V² = U² + 2gh
V² = 5² + 2 x 9.8 x 3
V² = 25 + 58.8
V² = 83.8
V = √83.8
V = 9.15 m/s
Therefore, the speed of the ball at this distance is 9.15 m/s
If a pickup is placed 16.25 cm from one of the fixed ends of a 65.00-cm-long string, which of the harmonics from n=1 to n=12 will not be "picked up" by this pickup?
Answer:
The answer to this question can be defined as follows:
Explanation:
Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.
Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.
Answer:
b
Explanation:
because:/
A length of organ pipe is closed at one end. If the speed of sound is 344 m/s, what length of pipe (in cm) is needed to obtain a fundamental frequency of 50 Hz
Answer:
The length = 27.52m
Explanation:
v=f x wavelength
What accurately depicts the change in average kinetic energy of the particles undergoes in matter as the temperature of the sample is decreased?
Answer:
As a sample of matter is continually cooled, the average kinetic energy of its particles decreases. Eventually, one would expect the particles to stop moving completely. Absolute zero is the temperature at which the motion of particles theoretically ceases.
Explanation:
Water flowing through a garden hose of diameter 2.76 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water leaving the end of the hose
Answer:
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
Explanation:
Given;
Diameter of hose d = 2.76 cm
Volume filled V = 20.0 L = 20,000 cm^3
Time t = 1.45 min = 105 seconds
The volumetric flow rate of water is;
F = V/t = 20,000cm^3 ÷ 105 seconds
F = 190.48 cm^3/s
The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.
F = Av
v = F/A
Area A = πd^2/4
Speed v = F/(πd^2/4)
v = 4F/πd^2 ......1
Substituting the given values;
v = (4×190.48)/(π×2.76^2)
v = 31.83767439628 cm/s
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
The radius of he Earth orbit around the sun (assumed circular) is 1.50 X 10^8km, with T=365d. What is the radial acceleration of Earth towards the sun?
Answer:
ar = 5.86*10^-3 m/s^2
Explanation:
In order to calculate the radial acceleration of the Earth, you first take into account the linear speed of the Earth in its orbit.
You use the following formula:
[tex]v=\sqrt{\frac{GM_s}{r}}[/tex] (1)
G: Cavendish's constant = 6.67*10^-11 m^3 kg^-1 s^-2
Ms: Sun's mass = 1.98*10^30 kg
r: distance between Sun ad Earth = 1.50*10^8 km = 1.50*10^11 m
Furthermore, you take into account that the radial acceleration is given by:
[tex]a_r=\frac{v^2}{r}[/tex] (2)
You replace the equation (1) into the equation (2) and replace the values of all parameters:
[tex]a_r=\frac{1}{r}\frac{GM_s}{r}=\frac{GM_s}{r^2}\\\\a_r=\frac{(6.67*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}{(1.50*10^{11}m)^2}\\\\a_r=5.86*10^{-3}\frac{m}{s^2}[/tex]
The radial acceleration of the Earth, towards the sun is 5.86*10^-3 m/s^2
HELP AGAIN A car going .80m/s accelerates uniformly at .20m/s^2 . What is the distance covered in 1.3 minutes? brainliest goes to person who shows formula/work
Answer:
initial speed (u) = 0.8 m/s
acceleration (a) = 0.2 m/s/s
time (t) = 1.3 min OR 1.3*60 seconds
= 78 seconds
we will use the second equation of motion to find the distance
distance (s) = ut + 1/2 a(t^2)
s = 0.8 * 78 + 1/2 * 0.2 * (6084)
s = 62.4 + 608.4
s = 670.8 m
what is drift speed ? {electricity}
Answer: In physics a drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
Explanation: