The total work W done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
W = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
W = ∆K = K
where K is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
W = 1/2 mv ²
where m is the mass of the object and v is the speed you want to find. Solving for v, you get
v = √(2W/m) ≈ 0.46 m/s
Integrated Concepts Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another.
(a) Calculate the potential energy of two singly charged nuclei separated by 1.00×10–12 m by finding the voltage of one at that distance and multiplying by the charge of the other.
(b) At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?
Answer:
(a) 2.3 x 10^-16 J
(b) 1.1 x 10^7 K
Explanation:
charge, q = 1.6 x 10^-19 C
distance, r = 10^-12 m
(a) Let the potential energy is U.
[tex]U = \frac{k q q}{r^2}\\\\U = \frac{9\times 10^{9}\times 1.6\times 1.6\times 10^{-38}}{10^{-12}}\\\\U = 2.3\times 10^{-16} J[/tex]
(b) Let the temperature is T.
[tex]U = K = \frac{3}{2} kT\\\\2.3 \times 10^{-16} = 1.5\times 1.38\times 10^{-23} T\\\\T = 1.1\times 10^7 K[/tex]
PLEASE HELP ME WITH THIS PHYSICS QUESTION PLSSS!!!
Vf^2 = v0^2 + 2a (xf -x0)
Solve for v0
b. solve for Vo
[tex] vf ^{2} = vo^{2} + 2a(xf - xo) \\ vf ^{2} = vo ^{2} + 2axf - 2axo\\ vo ^{2} = vf ^{2} - 2axf + 2axo \\ vo = \sqrt{vf ^{2} - 2axf + 2axo } \\ vo = - \sqrt{vf^{2} - 2axf + 2axo } [/tex]
I hope I helped you ^_^
A mass of 240 grams oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm.
a. What is the effective spring constant for this motion?
b. How much energy is involved in this motion?
Answer:
(a) The spring constant is 59.23 N/m
(b) The total energy involved in the motion is 0.06 J
Explanation:
Given;
mass, m = 240 g = 0.24 kg
frequency, f = 2.5 Hz
amplitude of the oscillation, A = 4.5 cm = 0.045 m
The angular speed is calculated as;
ω = 2πf
ω = 2 x π x 2.5
ω = 15.71 rad/s
(a) The spring constant is calculated as;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m[/tex]
(b) The total energy involved in the motion;
E = ¹/₂kA²
E = (0.5) x (59.23) x (0.045)²
E = 0.06 J
If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
Hz 50
Hz 70.7
Hz 100
Hz 25
Answer:
100Hz
Explanation:
In a full wave rectifier, the fundamental frequency of the ripple is twice that of input frequency. Given the input frequency of 50 Hz, the fundamental frequency will be 2 × 50 = 100Hz
Answer:
HZ 100 is the right answer hope you like it
1. A flywheel begins rotating from rest, with an angular acceleration of 0.40 rad/s. a) What will its angular velocity be 3 seconds later? b) What angle will it have turned through in that time?
Answer:
(a) 1.2 rad/s
(b) 1.8 rad
Explanation:
Applying,
(a) α = (ω-ω')/t................ Equation 1
Where α = angular acceleration, ω = final angular velocity, ω' = initial angular velocity, t = time.
From the question,
Given: α = 0.40 rad/s², t = 3 seconds, ω' = 0 rad/s (from rest)
Substitute these values into equation 1
0.40 = (ω-0)/3
ω = 0.4×3
ω = 1.2 rad/s
(b) Using,
∅ = ω't+αt²/2.................. Equation 2
Where ∅ = angle turned.
Substitutting the values above into equation 2
∅ = (0×3)+(0.4×3²)/2
∅ = 1.8 rad.
Outline five ways of varying the force on a current-carrying conductor in a magnetic field. (7 marks)
A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction between the tires and the horizontal road is 0.735, how long will the skid marks be
Answer:
126.56 m
Explanation:
Applying,
-F = ma............. Equation 1
Where F = frictional force, m = mass of the car, a = acceleration.
Note: Frictional force is negative because it act in opposite direction to motion
But,
F = mgμ.......... Equation 2
Where g = acceleration due to gravity, μ = coefficient of friction
Substitute equation 2 in equation 1
-mgμ = ma
a = -gμ.............. Equation 3
From the question,
Given: μ = 0.735
Constant: 9.8 m/s²
Substitute these values in equation 3
a = -9.8×0.735
a = -7.203 m/s²
Finally,
Applying
v² = u²+2as.............. Equation 4
Where v = final velocity, u = initial velocity, s = distance
From the question,
Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²
Substitute these values into equation 4
0² = 42.7²+2(-7.203)s
-1823.29 = -14.406s
s = -1823.29/-14.406
s = 126.56 m
With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how many grid marks is the wavelength of the wave (use the pause button and step button as you need to in order to get a good measure, and round to the nearest whole grid mark)?
Answer:
The wavelength stays the same.
Explanation:
When the amplitude is increased, the wavelength stays the same.
Here the wavelength doesn't depend upon the amplitude.
Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are doubled and their separation is halved, what happens to the electrical force that each charge exerts on the other one
Answer:
F' = 16 F
Hence, the electric force between charges becomes sixteen times its initial value.
Explanation:
The electric force between the two charges is given by the Colomb's Law:
[tex]F = \frac{KQ_1Q_2}{R^2}[/tex] ------------------- eq(1)
where
F = electric force
K = Colomb's Constant
Q₁ = magnitude of the first charge
Q₂ = magnitude of the second charge
R = Distance between charges
Now the magnitudes of the charges are doubled and the distance between them is halved. Therefore:
[tex]F' = \frac{K(2Q_!)(2Q_2)}{(\frac{R}{2})^2}\\\\F' = 16 \frac{KQ_1Q_2}{R^2}[/tex]
using equation (1):
F' = 16 F
Hence, the electric force between charges becomes sixteen times of its initial value.
A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a magnetic force of 0.16 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 72.0° with respect to the wire.
Answer:
7.28×10⁻⁵ T
Explanation:
Applying,
F = BILsin∅............. Equation 1
Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°
Substitute these values into equation 2
B = 0.16/(68×34×sin72°)
B = 0.16/(68×34×0.95)
B = 0.16/2196.4
B = 7.28×10⁻⁵ T
A typical radio frequency is 99.5 MHz, while a HD TV broadcast frequency is 600 MHz. Which one is your body position more likely to interfere with and why
I NEED YOUR HELP!!
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.
52 cm4
72 cm4
32 cm4
24 cm4
2 cm4
Answer:
Minimum area of rectangle = 24 centimeter²
Explanation:
Given:
Length of rectangle = 6 centimeter
Width of rectangle = 4 centimeter
Find:
Minimum area of rectangle
Computation :
Minimum area of rectangle = Length of rectangle x Width of rectangle
Minimum area of rectangle = 6 x 4
Minimum area of rectangle = 24 centimeter²
The speed of a car decreases uniformly as it passes a curve point where normal component of acceleration is 4 ft/sec2. If the car total acceleration of 5ft/sec2 is the same as it passes a hump, the tangential component of acceleration is _______________ ft/sec2.
Answer:
45
Explanation:
ft/sec2
write state of matter with 5 example of each
There are broadly 3 states of matter (there are 5, but they don't teach 2 of them at school).
1. Solid
Examples: Iron, wood, steel, ice, paper
2. Liquid
Examples: Water, mercury, milk, soup, juice
3. Gas
Examples: Oxygen, Chlorine, Carbon dioxide, Sulphur dioxide, Nitrogen
A ball is thrown upward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it moving 0.5 s later? In what direction? (b) How fast is it moving 2 s later? In what direction?
Answer:
Explanation:
Kinematic equation
v = u + at
If UP is assumed to be the positive direction and we let gravity be 10 m/s² which will be in the downward direction so will be negative.
a) v = 6 + (-10)(0.5) = 1 m/s the result is positive, so upward
b) v = 6 + (-10)(2) = -14 m/s the result is negative, so downward
What is the feature known as the "Great Dark Spot" of Neptune? It is an apparently permanent feature about five times the size of Earth, similar to the Great Red Spot of Jupiter, near Neptune's south pole. It was a dark hole in the upper atmosphere left by the collision of the comet Shoemaker-Levy 9. It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years. It is a dark surface feature on the surface snow layers caused by radiation discoloration of the older layers. It is a permanent discoloration of the north polar region of Neptune caused by locally prevailing lower surface temperatures there.
Answer:
It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years.
Explanation:
The Great Dark Spot of Neptune was an immense spinning storm in the southern atmosphere of Neptune. The size of the entire Earth, it had the strongest winds ever recorded on any planet in the solar system. It was discovered by the Voyager 2 spacecraft in 1989, but by 1994 the Hubble Space Telescope saw it was gone.
The Great Red Spot is a storm found in Jupiter's southern hemisphere, with similar characteristics to the Great Dark Spot.
What would be the consequences if the animated structures suddenly become non-polar
Answer:
the lipid bilayer would not be able to hold its shape in water and the cell membrane would disassemble. Only a lipid monolayer would be possible. The fatty acid "tails' would fall off The phospholipids would flip upside down.
Explanation:
hope this helps
distance of distinct vision.
is placed at a distance less than the distance of near point, its image o
will be blurred. Hence human eye can not see such object clearly.
ADDITIONAL INFORMATION
distance of distinct vision for a normal eye of different age groups
Babies = 7 cm
Adults = 25 cm
erson of age 55 years and above = 100 cm
ever, in our discussion we are concerned with a normal eye of an adult so least
The foulart position of an ahiect from a human eve so that the sh
The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.
Answer:
25 you said ? thats incorecct
Explanation:
Calculate area moment of inertia for a circular cross-section with 3 mm diameter:
Answer:
circles
A=7.07multiple 10-6 m2
I hope you understand and help
The area of the circular cross-section will be 7068 × [tex]10^{-6}m^{2}[/tex].
What is the area?The measurement that represents the size of a region on a plane or curved surface is called an area.
What is cross-section?A cross-section would be the non-empty point where a solid body intersects a plane in three dimensions or its equivalent in higher dimensions.
Given data:
Diameter = 3 × [tex]10^{-3} m[/tex].
It is known that. Diameter = 2 radius.
The area can be calculated by using the formula:
A = 1/4 [tex]\pi[/tex][tex]d^{2}[/tex] = 1/4 (3.14) [tex](3 * 10^{-3})^{2}[/tex]= 7068 × [tex]10^{-6}m^{2}[/tex].
To know more about area and cross-section
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Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3, is stirred until its temperature is 500 K. Assuming the ideal gas model, for the air, and ignoring kinetic and potential energy, determine
Answer:
The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".
Explanation:
As we know,
PV = nRT
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]
then,
⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]
⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]
[tex]=\frac{10}{3} \ Bar[/tex]
Thus the above is the correct answer.
which option is correct n why?
6. The projectile motion is a good example of
A. one dimensional motion.
B. two dimensional motion.
C. three dimensional motion.
D. four dimensional motion.
2. two dimensional motion
Because it has just 2 dimensions x and y
A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of the magnetic field if there is a 45o angle between it and the proton's velocity
Answer:
the strength of the magnetic field is 3 x 10⁻⁵ T
Explanation:
Given;
velocity of the cosmic ray, v = 5 x 10⁷ m/s
force experienced by the ray, f = 1.7 x 10⁻¹⁶ N
angle between the ray's velocity and the magnetic field, θ = 45⁰
The strength of the magnetic field is calculated as;
[tex]F = qvB \ sin(\theta)\\\\B = \frac{F}{qv\times sin(\theta)} \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 \times 10^{-19} \ C\\\\B = \frac{1.7\times 10^{-16}}{(1.602 \times 10^{-19})\times (5\times 10^7) \times sin \ (45)} \\\\B = 3 \times 10^{-5} \ T[/tex]
Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T
I REALLY NEED HELP WITH PHYSICS ASAP!!!
Vf^2 = v0^2 + 2a (xf - x0)
Solve for a
Answer:
a. solve for a
[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]
I hope I helped you ^_^
a model car moves round a circular path of radius 0.3m at 2 revolutions per secs what is its angular speed, the period of the car and the speed of the car
Answer:
a) T = 0.5 s
b) v = 1.2π m/s ≈ 3.77 m/s
Explanation:
It makes two revolutions in one second so makes one revolution in ½ second
circumference of the circle is
C = 2πr = 0.6π m
which it traverses in one time period
0.6π m / 0.5 s = 1.2π m/s
To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
What is speed?Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.
Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.
Mathematically, the formula for speed can be given as
speed= distance/time
Values that are given
Time period= 0.5 s
Circumference = 2πr = 0.6π m
substituting all the given values in the above equation, we get
speed =0.6π m / 0.5 s
On calculations, we get
= 1.2π m/s
=3.77 m/s
Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
To learn more about speed, here:
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One ball is aprojected in the uapward directon with a certain velocity ‘v’ and other
is thrown downwards with the same velocity
Complete question is;
One ball is projected in the upward direction with a certain velocity ‘v’ and other is thrown downwards with the same velocity at an angle θ.
The ratio of their potential energies at highest points of their journey, will be:
Answer:
u² : (u cos θ)²
Explanation:
Maximum potential energy for the first ball will be at a maximum height of;
H = u²/2g
Thus;
PE = mg(u²/2g)
For second ball at an angle θ, maximum PE will occur at a max height of (u cos θ)²/2g
PE = mg((u cos θ)²/2g)
The ratios of the potential energies are;
mg(u²/2g) : mg((u cos θ)²/2g)
mg will will cancel out since they are of same mass.
Thus;
(u²/2g) : (u cos θ)²/2g
Again 2g will cancel out to give;
u² : (u cos θ)²
Use the general formulas for gravitational force and centripetal force to derive the relationship between speed (v) and orbital radius (r) for circular orbits. Show the relationship you derived to your instructor before going on.
Solution :
We know that :
Formula for Gravitational force is given by :
[tex]$F_g=\frac{Gmn}{r^2}$[/tex]
where, G is the gravitational constant
M is the mass of the bigger body
m is the mass of the smaller body
r is the distance between the two bodies.
And the formula for the centripetal force is given by :
[tex]$F_c=\frac{mv^2}{r}$[/tex]
where, m is the mass of the rotating body
v is the velocity
r is the radius of rotation of the body.
We know that mathematically, the gravitational force is equal to the centripetal force of the body.
Therefore,
[tex]$F_g=F_c$[/tex]
[tex]$\frac{GMm}{r^2}=\frac{mv^2}{r}$[/tex]
[tex]$\sqrt{\frac{GM}{r}}=v$[/tex]
Hence derived.
write a use of magnetic force and frictional force each
If "FRICTION" means breaking up of relationships, then how can you reduce friction among friends?
A. Respect the opinion of your friends
B. Learn how to be understanding
C. Avoid bullying your friends
D. Force your friends to do things they don't like
Answer:
A
Explanation:
respect the opinion of your friends
Newton's law of gravity describes the gravitational force between __________. Earth and the moon. All of the listed answers are correct. Earth and the sun. the sun and the planets. a person and the earth.
Answer:
the force is known as gravity or gravitational force
The paper dielectric in a paper-and-foil capacitor is 0.0785 mm thick. Its dielectric constant is 2.35, and its dielectric strength is 49.5 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.
Required:
a. What area of each plate is required for for a 0.300 uF capacitor?
b. If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor?
Answer:
a) required area is 1.1318 m²
b) the maximum potential difference that can be applied across the compactor is 1931.1 V
Explanation:
Given the data in the question;
dielectric constant εr = 2.35
distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m
dielectric strength = 49.5 MV/m
a)
given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F
To find the Area, we use the following the expression.
C = ε₀εrA / d
we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹² (F/m)
we substitute
0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A ] / 7.85 × 10⁻⁵
A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]
A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹
A = 1.1318 m²
Therefore, required area is 1.1318 m²
b)
the maximum potential difference that can be applied across the compactor.
We use the following expression;
⇒ 1/2 × dielectric strength × thickness d
we substitute
⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )
⇒ 1931.1 V
Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V
