Answer:
Explanation:
11 oscillations in 18.2 s
Time period is defined as the time taken to complete one oscillation.
T = 18.2 / 11 = 1.655 s
mass, m = 290 g = 0.29 kg
Δx = 21.1 cm = 0.211 m
ω = 2π / T = (2 x 3.14) / 1.655 = 3.796 rad/s
[tex]\omega =\sqrt{\frac{K}{m}}[/tex]
Where, K is the spring constant
K = ω² m = 3.796 x 3.796 x 0.29 = 4.18 N/m
Now, mg = K Δx
0.29 x g = 4.18 x 0.211
g = 3.04 m/s²
A rock is dropped in a pond, causing circular ripples. The radius increased with a rate 1 foot per second. When the radius is 4 feet, the area is increasing by 8π square feet per second.1. True2. False
Answer:
true
Explanation:
A = πr²
Differentiating in respect to r
we get
dA/dr = 2πr
since r = 4feet
so ,
2π(4) = 8π
The loss of an electron from a neutral helium atom results in
A mass M is attached to spring, which exerts no force at position C. The spring is compressed until the mass is in position A. Then the mass is released. At what position is the velocity at a maximum and positive? At what position is the acceleration at a maximum and positive?
Answer:
Maximum velocity: Position C, Maximum positive acceleration: Position A.
Explanation:
Let consider that spring is compressed in the negative direction. Then, the maximum velocity occurs at position C, when spring is not compressed nor stretched. Since force in spring is of reactive nature, that is, the direction of force is opposed to the direction of movement, a maximum positive acceleration occurs at position A.
When you jump from a height to the ground, you let your legs bend at the knees as your feet hit the floor.Why we do this in terms in physics in momentum?
Final answer:
Bending the knees when landing from a jump increases the impact time and decreases the force experienced by the body, as per the impulse-momentum theorem. This technique prevents injuries by distributing the force over a longer period, thus reducing the force's intensity on the bones.
Explanation:
When we jump from a height to the ground and bend our knees upon landing, we are effectively reducing the impact force experienced by our body. This is explained by the impulse-momentum theorem, which states that the change in momentum (or impulse) of an object is equal to the force applied multiplied by the time the force is applied. The momentum of a person jumping from a height is quite large due to their mass and the speed at which they are moving toward the ground due to gravity. If we land with straight legs, that momentum is transferred very quickly to our legs and the ground, which results in a very high force over a short time. This can lead to injuries as bones can fracture if the force on them is too large.
By bending our knees, we increase the time over which our momentum is brought to zero, which means the force exerted on our legs and subsequently on the bones is lessened, as force is inversely proportional to the time over which the momentum changes. This principle applies to various real-world situations like rolling on the ground after a jump, using a parachute, or even the crumple zones in cars, all designed to extend the impact time and reduce the force felt by the occupants.
The gravitational attraction between Earth and the person causes the person to accelerate towards the Earth. However, due to Earth's significantly greater mass, Earth's movement is negligible compared to that of the person. This results in the person experiencing a much greater acceleration and subsequently a larger change in momentum when they hit the ground after a jump.
Two boxes connected by a light horizontal rope are on a horizontal surface. The coefficient of kinetic fric-tion between each box and the surface is ilk = 0.30. Box B has mass 5.00 kg, and box A has mass m. A force F with magnitude 40.0 N and direction 53.1° above the horizontal is applied to the 5.00-kg box, and both boxes move to the right with a = 1.50 m/s2. A) What is the tension T in the rope that connects the boxes? B) What is m?
Given, Two boxes(Let's say A and B) are connected horizontally by a light rope. Coefficient of kinetic friction between each box is 0.30. Mass of Box B is 5.00 kg and Box A is 'm'. The force is 40.0 N in the direction [tex]53.1^{o}[/tex].
To find the tension in the rope and the mass of one of the boxes, Newton's Second Law is applied separately to each box, accounting for tension, friction, and the applied force. By resolving the applied force and calculating the friction, we can solve for tension and then for the unknown mass.
To solve for these, we must apply Newton's Second Law of Motion (F = ma) to each box separately.
Part A: Tension T in the Rope
For Box B (5.00 kg), which has the force applied:
Resolve the applied force F into horizontal [tex](F_{x})[/tex] and vertical [tex](F_{y})[/tex] components.Calculate the friction force [tex](f_{k})[/tex] acting against Box B using the coefficient of kinetic friction (μ) and the normal force (N).Write Newton's Second Law for horizontal motion, accounting for tension (T), friction [tex](f_{k})[/tex], and the applied force component [tex](f_{x})[/tex].Solve for tension (T) based on the known acceleration (a) and the mass of Box B.Part B: Mass m of Box A
For Box A (with unknown mass m):
Calculate the friction force [tex](f_{k})[/tex] acting against Box A using μ_k and the weight (mg).Apply Newton's Second Law for Box A, considering only tension (T) and friction [tex](f_{k})[/tex].Solve for the unknown mass (m) based on the calculated tension (T) and the known acceleration (a).Appropriate use of Newton's laws and kinetic friction calculations will yield answers to both parts of the question.
Two different wave groups with the same height travel together in the same direction. The wavelength of one group is twice as long as the wavelength of the other group. A combined wave of extra height will be produced _______.
Answer:
A combined wave of extra height will produce every other wave.
Explanation:
One wave Crest can catch up to another wave Crest if the group of waves are from different places.
Here the two group of waves have different wavelength but travelling together in the same direction so when they combine,they will produce every other wave.
(1 point) A spotlight on the ground is shining on a wall 20m20m away. If a woman 2m2m tall walks from the spotlight toward the building at a speed of 0.8m/s,0.8m/s, how fast is the length of her shadow on the building decreasing when she is 8m8m from the building
Answer:
The answer to the question is;
When she is 8 m from the building fast the length of her shadow on the building is decreasing at [tex]\frac{2}{9} m/s[/tex] or 0.22 m/s.
Explanation:
We have
Distance of the spotlight from the building = 20 m
Distance of woman from the building when her speed is measured = 8 m
Height of the woman = 2 m
Actual speed of the woman = 0.8 m/s
Comparing the distance of the woman from the spotlight and the wall from the spotlight, we have when the woman is 8 m from the building she is 12 m from the spotlight
Therefore we have
[tex]\frac{12}{20} = \frac{2}{y}[/tex] where y is the shadow cast by the woman on the building = 10/3
When the woman is x distance from the building, she is 20 - x meters from the spotlight
Therefore the above equation can be written as
[tex]\frac{20-x}{20} = \frac{2}{y}[/tex] which gives [tex]1 - \frac{1}{20}*x = 2* \frac{1}{y}[/tex] finding the derivative of both sides gives
[tex]-\frac{1}{20}dx =-2*\frac{1}{y^2}dy[/tex] hence we have by dividing by dt gives [tex]-\frac{1}{20}\frac{dx}{dt} =-2*\frac{1}{y^2}\frac{dy}{dt}[/tex]
However we know that [tex]\frac{dx}{dt} = 0.8 m/s[/tex]
Therefore [tex]-\frac{0.8}{20} = -0.18\frac{dy}{dt}[/tex]
The rate of decrease of her shadow [tex]\frac{dy}{dt}[/tex] is given by
[tex]\frac{dy}{dt} = \frac{0.8}{3.6} =\frac{2}{9} m/s[/tex] or 0.222 m/s.
A 1.44-mole sample of an ideal gas is allowed to expand at a constant temperature of 258 K. The initial volume is 14.5 L and the final volume is 27.0 L. How much work does the gas perform on its container? Let the ideal-gas constant R = 8.314 J/(mol • K).
1920 J
2340 J
1550 J
1040 J
Answer:
1920 J
Explanation:
Answer:
19.6 L
Explanation:
In this process we have an isothermal process, a transformation in which the temperature of the gas remains constant.
In an isothermal process, the work done by the ideal gas on the surrounding is:
[tex]W=nRT ln\frac{V_f}{V_i}[/tex]
where:
n is the number of moles of the gas
R is the gas constant
T is the absolute temperature of the gas
[tex]V_i[/tex] is the initial volume of the gas
[tex]V_f[/tex] is the final volume
In this problem, we have:
n = 1.44 mol
T = 258 K is the gas temperature
[tex]V_i=14.5 L[/tex] is the initial volume of the gas
[tex]V_f=27.0 L[/tex] is the final volume
Solving the equation for W, we find the work done by the gas:
[tex]W=(1.44)(8.314)(258)ln\frac{27.0}{14.5}=1920 J[/tex]
Which circle passes through more continents antarctic or arctic
Answer:Arctic circle
Explanation:
Arctic circle is one of the circles of latitude. It is located at the northern region if the equator and passes through more than one continent. These continent includes: North Asia,
Northern America, and
Europe.
In these continent, it passes through eight known counties. These countries includes: Canada, Russia, Norway, Sweden, Finland, Denmark, Alaska and Iceland. It covers 4% of the Earth surface and it's climatic conditions are extreme.
A particle of mass 4.5 × 10-8 kg and charge +5.4 μC is traveling due east. It enters perpendicularly a magnetic field whose magnitude is 2.7 T. After entering the field, the particle completes one-half of a circle and exits the field traveling due west. How much time does the particle spend in the magnetic field?
Answer:
0.00970 s
Explanation:
The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field
Force due to magnetic field = qvB sin θ
q = charge on the particle = 5.4 μC
v = velocity of the charge
B = magnetic field strength = 2.7 T
θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1
F = qvB
Centripetal force responsible for circular motion = mv²/r = mvw
where w = angular velocity.
The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field
mvw = qvB
mw = qB
w = (qB/m) = (5.4 × 10⁻⁶ × 2.7)/(4.5 × 10⁻⁸)
w = 3.24 × 10² rad/s
w = 324 rad/s
w = (angular displacement)/time
Time = (angular displacement)/w
Angular displacement = π rads (half of a circle; 2π/2)
Time = (π/324) = 0.00970 s
Hope this Helps!!!
The electrode comes into direct contact with the workpiece, and some degree of force is applied Group of answer choices Electron Beam Welding (EBW) Gas Tungsten Arc Welding (GTAW) Resistance Spot Welding (RSW) Laser Welding (LBW) Plasma Arc Welding (PAW) Oxyacetylene Welding (OAW)
Answer:
Gas Tungsten Arc Welding (GTAW) Resistance Spot Welding (RSW)
Explanation:
Only the Gas tungsten arc welding and the spot resistance welding have a physical electrode and both of them make a contact with the work-piece. Both of the techniques use a non-filling electrode that generates heat energy into the welding area of the work-piece. In the GTAW the tungsten electrode is brought directly into the contact of two closely spaced interfaces of the work-piece and the electric current generated by the arc produced and the inert gas (usually argon) is used to shield the welding zone to prevent the oxidation of the material. A force of compressive nature is applied on the joint to keep the fused material at the joint to complete the welding. This welding process is also called Tungsten inert gas welding.In the spot resistance welding the electrode comes into contact with the pieces of the workpiece (usually thin sheets upto 3mm) from both the sides of the workpiece and generates the heat of electric power without any spark intended. The filler material may be placed between the two joining pieces. This generates only the weld spots and not a continuous wekd seam.
In the other welding techniques the heat is generated via the non-solid energy beams such as electron beam in EBM, plasma in PAW, gas flame in Oxyacetylene welding.
You throw a ball straight upward. As it leaves your hand, its speed is 15 m/s. (a) How much time does it take for the ball to reach the top of its trajectory? Start from a fundamental principle and show all your work.
Answer:
the ball takes 1.53 s to reach the top of its trajectory.
Explanation:
given information:
the speed, v = - 15 m/s (moving upward)
(a) How much time does it take for the ball to reach the top of its trajectory?
we know that the speed for the vertical motion is
v = v₀ - gt, v₀ = 0
where
v = speed (m/s)
g = gravitational force (9.8 m/s²)
t = time (s)
thus
v = - gt
-15 = - 9.8 t
t = 15/9.8
= 1.53 s
so, the time that is needed by the ball to reach the top its trajectory is 1.53 s
Answer:
The ball takes 1.53 seconds to reach its top trajectory
Explanation:
The velocity of the ball will keep pushing it upwards until the velocity becomes zero. Therefore, the ball will reach the top of its trajectory when velocity i.e. V=0,
Fundamental principal of velocity is V = Vo + g*t
where, V=0
Vo = 15 m/s
g = -9.8 m/s^2 (since ball is going upwards against the gravity)
t = ?
0 = 15 + (-9.8 * t)
-15 = -9.8t
-15 / -9.8 = t
t = 1.53 Seconds
The ball takes 1.53 seconds to reach its top trajectory
LC oscillators have. been used to circuits. connected to loud speakers to create some of the sounds of electronic musc. What indductance must be usedd with a 6.7 uF capacito r to prodice a frequncy of 10
Answer:
[tex]3.8\times 10^{-8}\ H[/tex]
Explanation:
Given the [tex]6.7\mu F[/tex] capacitor is used in an LC circuit to produce [tex]10\ kHz[/tex] frequency.
We need to find the value of inductance required.
As we know the relation between angular frequency in [tex]rad/sec[/tex] and frequency in [tex]Hz[/tex] is.
[tex]\omega =2\times \pi\times f[/tex]
Where [tex]\omega[/tex] is angular frequency and [tex]f[/tex] is frequency.
[tex]\omega=2\times \pi\times 10\times 1000=20000\pi\ rad/sec\\\omega=62832\ rad/sec[/tex]
Also, the relation between the angular frequency, capacitance and inductance is given by.
[tex]\omega^2=\frac{1}{LC}\\\\L=\frac{1}{\omega^2C} \\\\L=\frac{1}{62832^2\times6.7\times 10^{-6} } \\\\L=\frac{1}{26451}.\\ \\L=3.8\times 10^{-8}\ H[/tex]
So, [tex]3.8\times 10^{-8}\ H[/tex] inductance will be required to produce [tex]10\ kHz[/tex].
How much work is required to turn an electric dipole 180° in a uniform electric field of magnitude E = 46.0 N/C if the dipole moment has a magnitude of p = 3.02 × 10−25 C·m and the initial angle is 64°?
Answer:
[tex]W=1.22*10^{-23}J[/tex]
Explanation:
Torque and energy of an electric dipole in an electric field we find:
[tex]W=U(\alpha_{o}+\pi )-U(\alpha_{o} )=-pE(cos(\alpha_{o}+\pi )-cos(\alpha_{o} ))\\W=2pECos\alpha_{o}\\ W=2(3.02*10^{-25}C.m )(46.oN/C)Cos64\\W=1.22*10^{-23}J[/tex]
The work required to turn an electric dipole 180° in a uniform electric field is -4.89 × 10^-24 J.
Explanation:To calculate the work required to turn an electric dipole 180° in a uniform electric field, we can use the formula:
W = -pE(1 - cosθ)
where W is the work done, p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field.
Plugging in the given values:
W = - (3.02 × 10-25 C·m)(46.0 N/C)(1 - cos(180° - 64°))
Simplifying the equation gives the work done to be -4.89 × 10-24 J.
Learn more about Work done on an electric dipole here:https://brainly.com/question/32478301
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Two boxes connected by a light horizontal rope are on a horizontal surface. The coefficient of kinetic fric-tion between each box and the surface is ilk = 0.30. Box B has mass 5.00 kg, and box A has mass m. A force F with magnitude 40.0 N and direction 53.1° above the horizontal is applied to the 5.00-kg box, and both boxes move to the right with a = 1.50 m/s2. A) What is the tension T in the rope that connects the boxes? B) What is m?
Answer:
(A). The tension in the rope that connects the boxes is 10.50 N.
(B). The value of m is 7 kg.
Explanation:
Given that,
Mass of box B = 5.00 kg
Mass of box A = m
Force = 40.0 N
Direction= 53.1°
Acceleration = 1.50 m/s²
Coefficient of kinetic friction = 0.30
(A). We need to calculate the tension in the rope that connects the boxes
Using balance equation
[tex]T=ma+m\cos\theta[/tex]
Put the value into the formula
[tex]T=5\times1.50+5.00\cos53.1[/tex]
[tex]T=10.50\ N[/tex]
(B). We need to calculate the value of m
Using formula of tension
[tex]T=ma[/tex]
[tex]m=\dfrac{T}{a}[/tex]
Put the value into the formula
[tex]m=\dfrac{10.50}{1.50}[/tex]
[tex]m=7\ kg[/tex]
Hence, (A). The tension in the rope that connects the boxes is 10.50 N.
(B). The value of m is 7 kg.
A speedboat accelerates at 2m/s^2. If the boat starts from rest, how fast will the boat be traveling, in m/s, if it accelerates for 5s? Make sure you use the correct unit for your answer and do not put a space between the number and the unit.
The boat travels with a speed of 10m/s
Explanation:
Given-
Acceleration, a = 2m/s²
Time, t = 5s
Speed, s = ?
We know,
[tex]Acceleration = \frac{speed}{time} \\\\s = at[/tex]
[tex]s = 5 * 2m/s\\s = 10m/s[/tex]
Therefore, Boat travels with a speed of 10m/s
Which of the following best defines a nation's labor force? the total number of persons who are willing and able to work but cannot find a job the total number of persons between the ages of 16 and 65 the total number of employed and unemployed persons the total number of persons working full time and part time
Answer:
the total number of persons between the ages of 16 and 65
Explanation:
The labour force consists of all the people who are able to work in a country or area, or all the people who work for a particular company.
The workforce of a country includes both the employed and the unemployed (labour force)
The labor force of a nation includes all employed and unemployed persons, excluding those not actively seeking work. Employed individuals have a job, while unemployed individuals are jobless but actively looking for work. The unemployment rate measures the percentage of the labor force that is unemployed.
Explanation:A nation's labor force is best defined as the total number of employed and unemployed persons. Specifically, it includes all individuals who are either currently working (employed) or actively seeking employment (unemployed). The labor force does not include those who are not seeking work, such as students, stay-at-home parents, individuals with disabilities preventing them from working, or those who have taken early retirement.
The unemployment rate is a key economic indicator that represents the percentage of the labor force that is unemployed. It's important to note that to be considered unemployed, a person must be actively looking for work and available to work. This excludes people who are not actively seeking employment, who are then categorized as out of the labor force.
Lastly, the definition of 'employed' in the United States is quite broad, including those working part-time or temporarily, as well as individuals on leave but who have a job.
When an ultraviolet photon is absorbed by a molecule of DNA, the photon's energy can be converted into vibrational energy of the molecular bonds. Excessive vibration damages the molecule by causing the bonds to break. Ultraviolet light of wavelength less than 290 nmnm causes significant damage to DNA; ultraviolet light of longer wavelength causes minimal damage.What is the threshold photon energy, in eV, for DNA damage?
Answer:
4.28405172412 eV
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
[tex]\lambda[/tex] = Wavelength = 290 nm
Energy is given by
[tex]E=h\dfrac{c}{\lambda}\\\Rightarrow E=6.626\times 10^{-34}\times \dfrac{3\times 10^{8}}{290\times 10^{-9}}\\\Rightarrow E=6.8544827586\times 10^{-19}\ J[/tex]
Converting to eV
[tex]\dfrac{6.8544827586\times 10^{-19}}{1.6\times 10^{-19}}=4.28405172412\ eV[/tex]
The threshold energy is 4.28405172412 eV
Final answer:
The threshold photon energy for DNA damage is approximately 4.29 eV, calculated using the relationship E = h * c /
, where h is Planck's constant, c is the speed of light, and
is the given threshold wavelength for UV damage to DNA (290 nm).
Explanation:
To calculate the threshold photon energy in electron volts (eV) necessary to cause damage to DNA, we can use the relationship between a photon's energy (E), its wavelength (
), and Planck's constant (h). The equation is E = h * c /
, where c is the speed of light. Using the given wavelength of 290 nm (which is the threshold for significant DNA damage), we can convert this value into meters (290 nm = 290 x 10-9 m). Planck's constant (h) is approximately 6.626 x 10-34 J*s, and the speed of light (c) is about 3.00 x 108 m/s.
First, we find the energy in joules:
E = (6.626 x 10-34 J*s) * (3.00 x 108 m/s) / (290 x 10-9 m) = 6.863 x 10-19 J
Next, we convert joules to electron volts using the conversion factor 1 eV = 1.602 x 10-19 J:
E = (6.863 x 10-19 J) / (1.602 x 10-19 J/eV)
4.29 eV
Therefore, the threshold photon energy for DNA damage is approximately 4.29 eV.
A 1500 kg car skids to a halt on a wet road where μk = 0.47. You may want to review (Pages 141 - 145) . Part A How fast was the car traveling if it leaves 68-m-long skid marks?
Final answer:
To find the initial velocity of a 1500 kg car that left 68-m-long skid marks, the kinetic energy (½mv²) equal to the work done by friction is used, resulting in the equation v = √(2μkgd). Plugging in the values gives the car's speed.
Explanation:
To determine how fast a 1500 kg car was traveling before it skidded to a halt on a wet road with a coefficient of kinetic friction (μk) of 0.47, leaving 68-meter-long skid marks, we can use the work-energy principle. The kinetic energy the car had before skidding must equal the work done by friction to halt the car.
Work done by friction (W) = μk × normal force (N) × distance (d)
Since the normal force (N) is equal to the weight of the car (mg) where ‘m’ is the mass of the car and ‘g’ is the acceleration due to gravity, normal force can be replaced in this equation:
W = μk × mg × d
The kinetic energy (KE) the car initially had is given by:
KE = ½ m × v2
Setting work done by friction equal to kinetic energy:
½ m × v2 = μk × mg × d
We can then solve for velocity (‘v’), which ends up being:
v = √(2 × μk × g × d)
By plugging in the values (m = 1500 kg, μk = 0.47, g = 9.8 m/s2, d = 68 m) we get the initial velocity.
Calculation:
v = √(2 × 0.47 × 9.8 m/s2 × 68 m)
After completing the calculation, we can determine the car's initial velocity that resulted in the 68-m-long skid marks on the wet road.
Harbor seals, like many animals, determine the direction from which a sound is coming by sensing the difference in arrival times at their two ears. A small difference in arrival times means that the object is in front of the seal; a larger difference means it is to the left or right. There is a minimum time difference that a seal can sense, and this leads to a limitation on a seal's direction sense. Seals can distinguish between two sounds that come from directions 3∘ apart in air, but this increases to 9∘ in water.Explain why you would expect a seal's directional discrimination to be worse in water than in air.
Answer:
that the angle must be increased to maintain the minimum time of discrimination due to the increase in the speed of sound in material
Explanation:
The direction of sound is detected by the difference in time of reception of each ear, the speed of the wave is
v = d / t
t = d / v
In air the velocity is v = 330 m / s, let's use trigonometry
Cos 3 = d / L
L = d / cos 3
The difference in distance is
Δd = d - d / cos 3 = d (1- 1 / cos3)
t = Δd / 330
When the animal is in the water the speed of sound is
v = 1540 m / s
So time is
t' = Δd ’/ 1540
t ’= Dd’ / 4.67 330
So if t = t’ is the minimum response time, the distance must be increased
Δd ’= 4.6 Δd
1-1 / cos θ = 4.6 (1- 1 / cos 3) = -4.6 0.00137 = -0.00631
1 + 0.0063 = 1 / cos θ
1.00631 = 1 / cos θ
Cos θ = 1 / 1.00631
Tea = 6.5
We see that the angle must be increased to maintain the minimum time of discrimination due to the increase in the speed of sound in material
An archer puts a 0.285 kg arrow in a bow and uses an average force of 182 N to draw the string back 1.32 m. Assume the energy stored in the bow is transferred to the arrow when it is shot. (a) What is the speed of the arrow as it leaves the bow
Answer:
speed of the arrow as it leaves is 41.05 m/sec
Explanation:
We have given mass m = 0.285 kg
Average force F = 182 N
Distance traveled d = 1.32 m
We know that work done = force [tex]\times[/tex]distance
Sp work done = [tex]182\times 1.32=240.24J[/tex]
Now according to work energy theorem work done will be equal to kinetic energy
So [tex]\frac{1}{2}mv^2=240.24[/tex]
[tex]\frac{1}{2}\times 0.285\times v^2=240.24[/tex]
[tex]v^2=1685.89[/tex]
v = 41.05 m/sec
So speed of the arrow as it leaves is 41.05 m/sec
Tony drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took hours. When Tony drove home, there was no traffic and the trip only took hours. If his average rate was miles per hour faster on the trip home, how far away does Tony live from the mountains? Do not do any rounding.
Answer:
Question not completed, so I analysed the question first
Tony drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 6 hours. when tony drove home, there was no traffic and the trip only took 4 hours. if his average rate was 22 miles per hour faster on the trip home, how far away does tony live from the mountains?
Explanation:
Let use variables to solve the problems
Let the first trip to be mountain take x hours
Let the trip back home take y hours
Let the speed to while going to the mountain be a miles/hour
Then, while going home it was b miles/hour faster than while going to the mountain.
Then, speed going home is (a+b)miles / hour
The formula for speed is given as
Speed=distance/time
The constant through out the journey is distance, the two journey has the same distance.
Then,
Distance =speed×time
For first journey going to the mountain
Distance = a×x=ax miles
For the second journey going home
Distance =y×(a+b)
Distance Mountain= distance home
ax=y(a+b)
Make a subject of the formula
ax=ya+yb
ax-ya=yb
a(x-y)=yb
a=yb/(x-y)
Therefore, distance from mountain is
Distance=speed ×time
Distance= a×x=ax
Now, applying the questions
So from the questions
x=6hours, y=4hours
Also, b=22miles/hour
Then,
a=yb/(x-y)
a=4×22/(6-4)
a=88/2
a=44miles/hour
Then, the house distance from the mountain is
Distance=ax
Distance =44×6
Distance =264miles
Answer:
480 miles.
Explanation:
Let, S = rate on his way to the mountains.
Assume, Sgoing x time going = Sreturning x time returning
= S × 12 hours = (S + 20mph) ×8 hours
= 12 × S = 8 × S + 160.
4 × S = 160
S = 40 miles/hour
The trip took 12 hours at 40 miles per hour, so distance was:
= 12 hours × 40 mph
= 480 miles.
If the equipotential surfaces due to some charge distribution are vertical planes, what can you say about the electric field direction in this region: (a) it is vertically upward, (b) it is vertically downward, (c) it is horizontally to the left, (d) it is horizontally to the right, or (e)either (c) or (d) could be correct?
Answer:
The correct option is
(e)either (c) or (d) could be correct.
Explanation:
The electric field of a charge radiates out in all directions and the intensity of the electric field strength given by E = F/q₀, diminishes as the lines of force moves further away from the source. The direction of F and E is in the line of potential motion of the source charge in the field.
Equipotential surfaces are locations in the radiated electric that have the same field strength or electric potential. The work done in moving within an equipotential surface is zero and as such since
Work = Force × distance = 0 where distance ≠ 0.
The force acting between two points on an equipotential surface is also zero or the component of the force within an equipotential surface is zero and since there is a force in the electric field, it is acting at right angles to the equipotential surface which could be horizontally to the left or right directions where the equipotential surfaces due to the charge distribution are in the vertical plane.
Therefore it is either horizontally to the left, or horizontally to the right.
Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets. Which of the following best explains why finding 1 planet with such a moon is consistent with the nebular theory, while finding 6 planets with such moons is not consistent?
Answer:
Explanation:
Solution:
- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.
- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible.
A student sits on a rotating stool holding two 2.6 kg masses. When his arms are extended horizontally, the masses are 0.71 m from the axis of rotation, and he rotates with an angular velocity of 1.8 rad/sec. The student then pulls the weights horizontally to a shorter distance 0.23 m from the rotation axis and his angular velocity increases to ω2. For simplicity, assume the student himself plus the stool he sits on have constant combined moment of inertia Is = 3.8 kg m2 . Find the new angular velocity ω2 of the student after he has pulled in the weights. Answer in units of rad/s.
Answer:
2.8 rad/s
Explanation:
In absence of external forces, the total angular momentum of the system must be conserved.
The angular momentum when the arms of the student are extended horizontally is given by:
[tex]L_1 = (I_0 + 2I)\omega_1[/tex]
where:
[tex]I_0=3.8 kg m^2[/tex] is the moment of inertia of the student+stool
[tex]I=mr^2[/tex] is the moment of inertia of each mass, where:
m = 2.6 kg is one mass
r = 0.71 m is the distance of each mass from the rotation axis
[tex]\omega_1=1.8 rad/s[/tex] is the initial angular velocity
So we have
[tex]L_1=(I_0+2mr^2)\omega_1[/tex]
When the student pulls the weights to a distance of r' = 0.23 m, the angular momentum is:
[tex]L_2=(I_0+2I')\omega_2[/tex]
where:
[tex]I'=mr'^2[/tex] is the new moment of inertia of each mass, with
r' = 0.23 m
Since the angular momentum must be constant, we have:
[tex]L_1=L_2\\(I_0+2mr^2)\omega_1 = (I_0+2mr'^2)\omega_2[/tex]
and solving for [tex]\omega_2[/tex], we find the final angular velocity:
[tex]\omega_2 = \frac{I_o+2mr^2}{I_0+2mr'^2}\omega_1=\frac{3.8+2(2.6)(0.71)^2}{3.8+2(2.6)(0.23)^2}(1.8)=2.8 rad/s[/tex]
A supersonic jet is at an altitude of 14 kilometers flying at 1,500 kilometers per hour toward the east. At this velocity, how far will the jet fly in 1.6 hours?
Answer:
The jet will fly 2400 km.
Explanation:
Given the velocity of the jet flying toward the east is 1,500 kmph toward the east.
We need to find the distance covered in 1.6 hours.
In our problem we are given speed and time, we can easily determine the distance using the following formula.
[tex]Distance=Speed\times Time[/tex]
[tex]Distance=1500\times 1.6=2400\ km[/tex]
So, the supersonic jet will travel 2400 km in 1.6 hours toward the east from its starting point.
You put a 3 kg block in the box, so the total mass is now 8 kg, and you launch this heavier box with an initial speed of 4 m/s. How long does it take to stop
Answer:
0.7 secs
Explanation:
In this question, the speed does not change as the mass changes. So we can use
Δt=Δ∨x/χgμ............................equ 1
To stop, the final speed will be 0
Therefore,
Δvx=vf-vt
Δvx=0-4m/s
= -4m/s
Now substitute the various values in equ 1
Δt=Δ∨x/χgμ
Δt= -4m/s/(9.8m/s∧2) (0.6)
Δt=0.7 secs
Answer:
Explanation:
The box stops at zero speed.
Final Velocity = 0 ,Initial speed (s)= -4 m/s
Therefore= change in velocity = Vf - Vi. ( 0 m/s- 4 m/s) = -4 m/s
Change in velocity = -0.4 m/s
Gravity g = 9.8 m/s^2
Mass= 0.8 g
-4 ms divided by 9.8 ms^2 * (0.8) = 0.51 s
It takes 0.51 seconds to stop
The 38-mm-diameter shaft ab is made of a grade of steel for which the yield strength is 250 mpa. v y using the maximum-shearing-stress criterion, determine the magnitude of the torque t for which yield occurs when p 240 kn.
Answer:
Answer is 717 N . m
Refer below for the explanation.
Explanation:
As per the question,
38 mm diameter shaft,
Yield strength 250 mpa,
P 240kn.
Refer to the picture for complete explanation.
The maximum torque that the steel shaft can withstand before yielding, according to the maximum-shearing-stress criterion, is 8224 N.m.
Explanation:The problem can be solved by first calculating the maximum shear stress that the steel shaft can withstand before yielding. The shear strength is given by the yield strength divided by the square root of 3, as the maximum-shearing-stress criterion indicates that failure occurs when the maximum shear stress equals half the yield stress. Therefore, the shear strength τ of the steel shaft is τ = 250 MPa/√3 = 144.34 MPa.
Next, we can find the maximum torque T that the shaft can handle before yielding by using the formula T=τ*J/r, where J is the polar moment of inertia and r is the radius. Because the shaft is a circular cross section, its polar moment of inertia J = π*r⁴/2. Substituting the given diameter of the shaft d=38mm, we find r = d/2 = 19mm, so J = π*(19mm)⁴/2 = 1075.21 mm⁴.
Thus, the critical torque T causing yield by the maximum-shearing-stress criterion can be calculated: T = τ*J/r = (144.34 MPa)*(1075.21 mm⁴)/(19 mm)=8224 N.m
Learn more about Yield Strength & Torque here:https://brainly.com/question/36889008
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What are the correct methods of heat transfer
Explanation:
The transfer of heat generally depends on the difference in the temperature in the surroundings.The best methods for the transfer of heat are conduction convection, radiation and sometimes evaporation also.
Conduction: This is the transfer of heat through the system that is solid. Convection: Convection is the process in which warm surfaces of a liquid or gas rises to cooler surfaces in the liquid or gas that is the transfer of heat from the surface. Natural convection occurs as air is heated: it expands, rises, and is replaced by cooler air. Radiation: This is a process where energy is radiated among the surroundings in the form of electromagnetic radiation. Evaporation: The latent heat of a liquid is used to transfer heat by absorbing the energy needed to evaporate that liquid. The heat absorbed is released by condensing the liquid outside the enclosure.Answer:Condution radiation convection
Explanation:
You hold a metal ring horizontally above a bar magnet standing on its end. You drop the ring and catch it before it reaches the magnet. When, if ever, is a current induced in the ring?
Only when the ring stops falling
only when the ring starts falling
while the ring is falling
never, current is not induced
Answer:
never, current is not induced
Explanation:
The induced emf in the ring equals the rate of change of magnetic flux in the ring.
E = -dФ/dt = -dAB/dt = -BdA/dt since B the magnetic field is constant.
E = -BdA/dt
Now dA/dt = dA/dy × dy/dt where dA/dy = rate of change of area with vertical distance as the ring is falling. dy/dt = speed of ring.
Since the ring is falling freely before being caught, its speed v is gotten from v = u + at where u = 0 and a = -g
v = 0 - gt = -gt
v = -gt
So, dA/dt = dA/dy × dy/dt = vdA/dy = -gtdA/dy
So E = -BdA/dt = -B × -gtdA/dy = BgtdA/dy
Since dA/dy = 0 since the area of the ring does not change with vertical distance. So,
E = BgtdA/dy = Bgt × 0 = 0
E = 0
So, emf is never induced because the flux through the ring remains constant