Answer:
speed of the bicycle = 2.81 m/s
Explanation:
given data
tuning fork rated f = 488 Hz
beat frequency fb = 8 Hz
speed of sound vs = 343 m/s
solution
we get here frequency that rider is listening that is
rider hearing frequency = 488 Hz + 8 Hz
rider hearing frequency = 496 Hz
and when it bouncing from sheer wall at frequency of related to stationary object
frequency = 488 Hz + 4 Hz
frequency = 492 Hz
so now we get speed of the bicycle that is
speed of the bicycle = ( [tex]\frac{492}{488}[/tex] × 343 ) - 343 Hz
speed of the bicycle = 2.81 m/s
The speed of the bicycle is 2.81 m/s.
Based on the given information,
• The tuning fork rated at frequency (f) 488 Hz.
• The beat frequency (fb) given is 8Hz.
• The speed of the sound is 343 m/s.
Now with the help of the given information, the rider is hearing a frequency of,
[tex]=(f+fb)\\=(488+8)\\=496 Hz[/tex]
However, it is bounding from sheer wall at the frequency of,
[tex]=(488+\frac{8}{2})\\= 492 Hz[/tex]
Now the speed of the bicycle will be,
[tex]=(\frac{492}{488} * 343)-343\\=2.811 m/s\\= 2.81 m/s[/tex]
Thus, the speed of the bicycle is 2.81 m/s.
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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) while returning to base camp. He was supposed to travel due north for 5.3 km, but when the snow clears, he discovers that he actually traveled 8.4 km at 47o north of due east. (a) How far and (b) in what direction (south of due west) must he now travel to reach base camp
Answer:
[tex]s=5.79\ km[/tex]
[tex]\theta=47^{\circ}[/tex] east of south
Explanation:
Given:
distance of the person form the initial position, [tex]d'=8.4\ km[/tex]direction of the person from the initial position, [tex]47^{\circ}[/tex] north of eastdistance supposed to travel form the initial position, [tex]d=5.3\ km[/tex]direction supposed to travel from the initial position, due NorthNow refer the schematic for visualization of situation:
[tex]y=d'.\sin47^{\circ}-d[/tex]
[tex]y=8.4\times \sin47-5.3[/tex] ...............(1)
[tex]x=d'.\cos47^{\circ}[/tex]
[tex]x=8.4\times \cos47^{\circ}[/tex] .................(2)
Now the direction of the desired position with respect to south:
[tex]\tan\theta=\frac{y}{x}[/tex]
[tex]\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}[/tex]
[tex]\theta=47^{\circ}[/tex] east of south
Now the distance from the current position to the desired position:
[tex]s=\sqrt{x^2+y^2}[/tex]
[tex]s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}[/tex]
[tex]s=5.79\ km[/tex]
A ball is dropped and bounces off the floor. Its speed is the same immediately before and immediately after the collision. Which of the following is true about the collision between the ball and the floor?
a) The ball's momentum is conserved.
b) The ball's momentum changes direction but not magnitude.
c) The ball's momentum changes magnitude but not direction.
d) The impulse delivered to the ball by the floor is zero.
e) The impulse delivered to the floor by the ball is zero.
Answer:
option B.
Explanation:
The correct answer is option B.
when the ball drops, the velocity of the ball before the collision is v
After the collision, the velocity of the ball is the same but in the opposite direction.
Impulse delivered to the ball and the floor, in this case, is not zero.
The magnitude of the momentum remains the same but the direction of the ball changes.
The true statement about the collision between the ball and the floor is that the ball's momentum is conserved.
What is momentum?The term momentum is physics is the product of mass and velocity. Momentum is a vector quantity hence when discussing momentum, we have to take the magnitude and direction into account.
The true statement about the collision between the ball and the floor is that the ball's momentum is conserved.
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Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block 2. A force is applied to R2 and the blocks accelerate forward. Is the magnitude of the force exerted by the rope R1 on block 1 larger, smaller, or equal to the magnitude of the force exerted by R2 on block 2?
Final answer:
The force exerted by rope R1 on block 1 is equal to the tension in that rope, and if both blocks have the same acceleration and there are no other external forces, this tension is also equal to the force R1 exerts on block 2. However, the force applied through rope R2 on block 2 may be larger because it must account for the total mass and acceleration of both blocks.
Explanation:
When two blocks are connected by a rope and a force is applied to one of the blocks (via a second rope), the tensions in the ropes are subjects of interest in Physics, specifically classical mechanics. In this scenario, the force applied to rope R2 is used to accelerate both blocks 1 and 2. The tension in rope R1, which connects blocks 1 and 2, should be analyzed using a free-body diagram for each block.
Rope R1 exerts a force on block 1, and rope R2 exerts a force on block 2. The force exerted by rope R2 must be sufficient not only to accelerate block 2 but also to account for the force needed to accelerate block 1. Therefore, the force exerted by R2 on block 2 is equal to the sum of the forces necessary to accelerate both blocks. This total force must be greater than or equal to the force that R1 exerts on block 1, as rope R1 only has to provide the force needed to accelerate block 1. However, if we assume the acceleration of the two blocks is the same and the system is not experiencing any external forces other than the applied force, the tension in rope R1 will be equal to the force R1 exerts on block 1 because the massless ropes can only transmit the force equally along their lengths.
Therefore, it can be concluded that the magnitude of the force exerted by rope R1 on block 1 is equal to the magnitude of the tension on block 2 by rope R1, but the magnitude of the applied force through R2 on block 2 could be larger as it has to account for the combined mass of block 1 and block 2.
As magnification increases, the area of the field of view _______, the depth of the field of view _________, the working distance ______, and the amount of light required ____________.
Answer:
As magnification increases, the area of the field of view increases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.
As magnification increases, the area of the field of view decreases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.
Area of the field of vision: As magnification is increased, the field of view narrows, and you can see less of the object or scene that you are watching. The area you can see reduces as a result of the field of view is smaller.
Depth of field: The range of distances inside an object or scene that simultaneously seem in focus is referred to as depth of field. The depth of field tends to get shallower as magnification gets bigger. This makes it more difficult to simultaneously examine objects at various depths because only a limited range of distances will ever be in focus.
The working distance is the separation between the thing being observed and the objective lens (or magnifying device). The working distance often reduces as magnification rises. To keep the object in focus, you must move it closer to the lens or microscope.
Higher magnification frequently necessitates additional light in order to preserve image quality and visibility. The features become more noticeable when you zoom in on a smaller region, although they might need more light to be clearly visible. At greater magnifications, insufficient light can produce a hazy or dim image, therefore more light would be required.
Hence, As magnification increases, the area of the field of view decreases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.
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An unbalanced force is defined as a force that __________. allows an object to remain at rest cause an object to accelerate in a direction opposite to the unbalanced force is not canceled out by one or more other forces acting in other directions allows an object to move at constant speed
Answer:
An unbalanced force is defined as a force that is not canceled out by one or more other forces acting in other directions.
Explanation:
According to Newton's first law of motion, “An object in a constant state of motion remains in that constant state of motion unless acted upon by an unbalanced force.” An unbalanced force is a force that changes the speed, position, or direction of an object. It is not canceled out by other forces acting in other directions.
An unbalanced force is defined as a force that is not canceled out by other forces acting in different directions. This results in a net force that causes an object to accelerate, changing its state of motion.
An unbalanced force is defined as a force that is not canceled out by other forces acting in different directions. allows an object to remain at rest cause an object to accelerate in a direction opposite to the unbalanced force is not canceled out by one or more other forces acting in other directions allows an object to move at constant speed.
When forces on an object are unbalanced, it means that the net force acting on the object is not zero, leading to a change in the object's motion. This unbalanced force results in acceleration according to Newton's second law of motion, which states that [tex]\( F = ma \)[/tex] force equals mass times acceleration.
What is the angular velocity of the second hand on a clock? (Hint: It takes the second hand 60 seconds to rotate 2π radians. Divide the number of radians by the number of seconds so that your answer has units of radians/second.)
Explanation:
Time taken to rotate 2π radians = 60 seconds
Angular displacement = 2π radians
Time taken = 60 seconds
Angular displacement = Angular velocity x Time
2π = Angular velocity x 60
Angular velocity = 0.105 rad/s
Angular velocity of the second hand on a clock is 0.105 rad/s
Kellie jogs 6.0 km in 54 minutes, then 1.0 km in 16 minutes. What is her average speed in km/min please show your work
Answer:
Average speed: 0.1 km/min
Explanation:
The average speed of a body is the ratio between the total distance covered and the time taken:
[tex]v=\frac{d}{t}[/tex]
where
d = distance covered
t = time taken
In this problem, we have:
- Distance covered is the sum of the two distances:
d = 6.0 + 1.0 = 7.0 km
- Time taken is the sume of the two times:
t = 54 + 16 = 70 min
Therefore, the average speed is
[tex]v=\frac{7.0}{70}=0.1 km/min[/tex]
The displacement of thoughts, feelings, fears, wishes, and conflicts from past relationships onto new relationships is called _____.
The displacement of thoughts, feelings, fears, wishes, and conflicts from past relationships onto new relationships is called transference.
A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one piece is 7.0 times the resistance of the other
Determine the length of the short piece.(% of length of the wire)?
Determine the resistance of the short piece?
Determine the resistance of the long piece?
Answer with Explanation:
Let r be the resistance of short piece of copper wire.
Resistance of copper wire=R=[tex]13\Omega[/tex]
Resistance is directly proportional to length.
If a wire has greater resistance then,the wire will be greater in length.
Therefore,resistance of long piece of wire=7r
Total resistance of copper wire=Sum of resistance of two piece of wires
[tex]r+7r=13[/tex]
[tex]8r=13[/tex]
[tex]r=\frac{13}{8}[/tex]ohm
Resistance of long piece of wire=[tex]7\times\frac{13}{8}=\frac{91}{8}\Omega[/tex]
Resistance of short piece of wire =[tex]\frac{13}{8}\Omega[/tex]
Resistivity of wire and cross section area of wire remains same .
Let L be the total length of wire and L' be the length of short piece of wire.
We know that
[tex]R=\frac{\rho L}{A}[/tex]=[tex]\frac{\rho}{A}L=KL[/tex]
[tex]\frac{R}{L}=K[/tex]
Where K=[tex]\frac{\rho}{A}[/tex]=Constant
Using the formula
[tex]\frac{13}{L}=\frac{\frac{13}{8}}{L'}[/tex]
[tex]\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}[/tex]
[tex]L'=\frac{L}{8}[/tex]
Length of short piece of wire=L'=[tex]\frac{L}{8}[/tex]
Length of long piece of wire=[tex]L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L[/tex]
% of length of short piece of wire=[tex]\frac{\frac{L}{8}}{L}\times 100=12.5%[/tex]%
The resistance of the short piece=[tex]\frac{13}{8}\Omega[/tex]
The resistance of the long piece=[tex]\frac{91}{8}\Omega[/tex]
The length of short piece wire is [tex]12.5[/tex] % of total wire length.
The resistance of the short piece wire is [tex]\frac{13}{8}ohms[/tex]
The resistance of the long piece wire is, [tex]7*\frac{13}{8}=\frac{91}{8}ohms[/tex]
Resistance :Let us consider that resistance of short piece wire is r.
So that, resistance of long piece wire is [tex]7r[/tex].
Total resistance of wire is [tex]13[/tex] ohms.
[tex]7r+r=13\\\\8r=13\\\\r=\frac{13}{8}Ohm[/tex]
The resistance of the short piece wire is [tex]\frac{13}{8}ohms[/tex] The resistance of the long piece wire is, [tex]7*\frac{13}{8}=\frac{91}{8}ohms[/tex]The resistance of wire is directly proportional to length of wire.Let us consider that total length of wire is 13L.
So that, length of short piece wire [tex]=\frac{13}{8} L[/tex]
length of short piece wire is,
[tex]=\frac{\frac{13}{8}L }{13L}*100 =\frac{1}{8} *100=12.5[/tex]
The length of short piece wire is 12.5 % of total wire length.
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You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the "oars") toward you, it moves a distance of 1.5 m in a time of 1.0 s. The readout on the display indicates that the average power you are producing is 82 W. What is the magnitude of the force that you exert on the handle?
Answer:
54.67 N
Explanation:
The total energy produced is the product of power and time duration:
E = Pt = 82 * 1 = 82 J
Which is converted from work, product of forced extended over a displacement
W = E = Fs = F*1.5 = 82
F = 82 / 1.5 = 54.67 N
So the magnitude of the force exerting on the handle is 54.67 N
Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At which points in time does a zero net force act on the ball? ignore air resistance
A. When you hold the ball still in your hands after catching it
B. Just after the ball first leaves your hands.
C. At the instant the ball reaches its highest point.
D. At the instant the falling ball hits your hands.
E. When you hold the ball still in your hands before it is thrown.
Answer:
C. At the instant the ball reaches its highest point.
Explanation:
When a body is thrown up, it tends to come down due to the influence of gravitational force acting on the body. The body will be momentarily at rest at its maximum point before falling. At this maximum point, the velocity of the body is zero and since force acting on a body is product of the mass and its acceleration, the force acting on the body at that point will be "zero"
Remember, F = ma = m(v/t)
Since v = 0 at maximum height
F = m(0/t)
F = 0N
This shows that the force acting on the body is zero at the maximum height.
Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 1.9 cm. You observe that it takes 3.0 s to make one round trip. You construct another vertical oscillator with a mass 6 times as heavy and a spring 10 times as stiff. You take it to a planet where gplanet = 8.8 N/kg. You start it oscillating with amplitude 4.8 cm. How long does it take for the mass to make one round trip?
Answer:
T = 3.23 s
Explanation:
In the simple harmonic movement of a spring with a mass the angular velocity is given by
w = √ K / m
With the initial data let's look for the ratio k / m
The angular velocity is related to the frequency and period
w = 2π f = 2π / T
2π / T = √ k / m
k₀ / m₀ = (2π / T)²
k₀ / m₀ = (2π / 3.0)²
k₀ / m₀ = 4.3865
The period on the new planet is
2π / T = √ k / m
T = 2π √ m / k
In this case the amounts are
m = 6 m₀
k = 10 k₀
We replace
T = 2π√6m₀ / 10k₀
T = 2π √6/10 √m₀ / k₀
T = 2π √ 0.6 √1 / 4.3865
T = 3.23 s
A toy roller coaster cart has a mass of 2kg. The track of the roller coaster has a semicircular valley with a radius of 1.2 meters. If the cart has a speed of 8 m/s at the very bottom of the valley, find the amount of centripetal force necessary to complete the motion
Answer:106.6N
Explanation:
Centripetal force is a force that cause a body to move along a circular path in a field. The body possesses a mass and velocity(since it's moving) and also radius alone the circular path on which it moves.
Given mass (m) = 2kg, velocity (v) = 8m/s and radius (r) = 1.2m
Centripetal acceleration(a)= v²/r
Centripetal force = ma = mv²/r
Substituting this values in the formula above we have;
Centripetal force = 2×8²/1.2m
Centripetal acceleration = 128/1.2
= 106.6N
Heat is a) the amount of thermal energy in an object. b) the energy that moves from a hotter object to a colder object c) a fluid-like substance that flows from a hotter object to a colder object. d) both A and B. e) both B and C.
Answer: Option (b) is the correct answer.
Explanation:
Heat is defined as the energy obtained by the molecules of an object that travels from a hotter body to a colder body.
For example, a metal spoon placed in a cup full of hot tea. Then transfer of heat will take place from the tea to the spoon.
And, this transfer of heat will continue till the temperature of both the objects become equal.
Thus, we can conclude that heat is the energy that moves from a hotter object to a colder object.
A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 244 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume that distilled water is an insulator.
(a) What is the charge on the plates before immersion?
What is the charge on the plates after immersion?
(b) What is the capacitance after immersion?
What is the potential difference after immersion?
(c) What is the change in energy of the capacitor?
?U = Uwater - Uair = 5 nJ
Answer:
(a) Since net charge remains same,after immersion Q is same
(b) I. 14.56pF ii. 3.05V
(c) ΔU = 5.204nJ
Explanation:
a)
C = kεA/d
k=1 for air
ε is 8.85x10-12F/m
A = .0025m2
d = .125m
C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF
Q = CV = .177pF * 244V = 43.188pC
Since net charge remains same,after immersion Q is same
b)
C = kεA/d, for distilled water k is approx. 80
Cwater = Cair x k
=0.177pF x 80 = 14.16pF
Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V
c) Change in energy: ΔU = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
ΔU = 5.204nJ
A student who weighs 550 N is wearing a backpack that weighs 80 N. The student is standing still on level ground. Give your answers to the following questions in newtons. Remember, enter only the number; don't try to enter the units.a) What is the magnitude of the net force on the student?
b) What is the magnitude of the contact force on the student by the backpack?
c) What is the magnitude of the contact force on the student by the ground?
Answer:
Explanation:
a) The magnitude of the net force on the student = 0 N since the student is standing still on level ground and upward reaction force = downward force.
b) the magnitude of the contact force on the student by the backpack = 80 N since the student was backing the backpack
c) the magnitude of the contact force o the student by the ground = 550 N + 80 N = 630 N reactional force on the student
How many conditions does the NEC list whereby conductors shall be considered to be outside of a building or other structure?
Answer:
4 conditions
Explanation:
The National Electrical Code (NEC) provides safety guidelines for the installation of electrical wiring and electrical equipment in the United States. The purpose behind NEC is to standardize the safe electrical installation practices.
There is always a confusion among people to distinguish between inside or outside of a building or structure. Therefore, NEC has listed 4 conditions in the Article 230.6 where it has mentioned conductors to be considered outside a building or structure.
Conductors are considered outside a building when they are installed:
1. Under not less than 2 inches of concrete beneath a building or structure.
2. Within a building or structure in a raceway that is encased in no less than 2 inches thick of concrete or brick.
3. Installed in a vault that meets the construction requirements of Article 450, Part III.
4. In conduit under not less than 18 inches of earth beneath a building or structure.
Suppose you want to move a big rock in your yard. It is about waist high and 4 feet long, has a volume of 1.2 m3 and a density of 2.5 g/cm3. How much does it weigh, in units of lbs?
Answer:
6613.87 lbs
Explanation:
1.2 m³ = 1200000 cm³
Mass = Density * Volume
M = (2.5 g/cm³) * 1200000 cm³ = 3000000 g
1 lb = 453.592 g
3000000 g * (1 lbs / 453.592 g) = 6613.87 lbs
The mass of the rock is 6613.87 lbs.
How do you calculate the Weigh?Mass is the quantity of matter in a physical body and the weigh is used to measure ascertain the heaviness of by or as if by a balance.
Given that volume of the rock is 1.2 meter cube and density is 2.5 gram/cm cube. The height of the rock is 4 feet. The mass can be calculated as given below.
Mass m = Density w [tex]\times[/tex] Volume v
Substituting the values in the above equation.
[tex]m = 2.5 \;\rm g/cm^3 \times 1.2\;\rm m^3[/tex]
[tex]m = 2.5 \times 1.2 \times 1000000[/tex]
[tex]m = 3000000\;\rm g[/tex]
The mass of the rock is 3000000 g. We know that
[tex]1\;\rm lb = 453.592\;\rm g[/tex]
[tex]1\;\rm g = \dfrac {1}{453.592} \;\rm lb[/tex]
Now convert the mass (grams) into mass (lbs).
[tex]m = \dfrac {1}{453.592}\times 3000000\;\rm lbs[/tex]
[tex]m = 6613.87 \;\rm lbs[/tex]
Hence we can conclude that the mass of the rock is 6613.87 lbs.
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The greenhouse effect plays a major role in the climate of a planet. Visit the Greenhouse Effect AstroTour (Links to an external site.), and use what you learn there to choose all the statements below that are correct. Note that 273 K = 0°C = 32°F.
Answer:
If there were no greenhouse effect, liquid water would not exist on the surface of the Earth.
The Earth has reached thermal equilibrium, emitting the same amount of energy into space as it absorbs from the Sun.
The more carbon dioxide there is in an atmosphere, the stronger the greenhouse effect will be.
Explanation:
Without greenhouse gases like carbon dioxide, the entire Earth would be well below the freezing temperature of water. These gases slow the radiation of infrared light into space, warming the planet until it reaches thermal equilibrium.
The greenhouse effect is a natural phenomenon that plays a major role in the Earth's climate. It involves the trapping of heat by greenhouse gases in the atmosphere, which raises the Earth's surface temperature. Without the greenhouse effect, the Earth would be much colder.
The greenhouse effect is a natural phenomenon that plays a major role in the Earth's climate. It is similar to the heating of a greenhouse or a car left in the sun with the windows rolled up. The greenhouse gases in the atmosphere, such as carbon dioxide and methane, trap heat from the sun, preventing it from escaping back into space. Without the greenhouse effect, the Earth's average surface temperature would be well below freezing and Earth would be locked in a global ice age.
Greenhouse gases like carbon dioxide and water vapor absorb and emit radiation, which is an important factor in the greenhouse effect. Approximately half of the radiation from the sun passes through these gases in the atmosphere and strikes the Earth. This radiation is converted into thermal radiation on the Earth's surface, which is then partially reflected back to the Earth's surface by greenhouse gases. The more greenhouse gases there are in the atmosphere, the more thermal energy is reflected back to the Earth's surface, causing it to heat up.
The greenhouse effect is responsible for raising the Earth's surface temperature by about 23 °C. Changes in global climate due to the greenhouse effect can lead to more intense storms, changes in precipitation patterns, reduction in biodiversity, and rising sea levels.
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A brass lid screws tightly onto a glass jar at 20 degrees C. To help open the jar, it can be placed into a bath of hot water. After this treatment, the temperature of the lid and the jar are both 60 degrees C. The inside diameter of the lid is 8.0 cm at 20 degrees C. Find the size of the gap (difference in radius) that develops by this procedure.
Answer:
0.0016 cm
Explanation:
[tex]\alpha_b[/tex] = Thermal coefficient of expansion of brass = [tex]19\times 10^{-6}\ /^{\circ}C[/tex]
[tex]\alpha_g[/tex] = Thermal coefficient of expansion of glass = [tex]9\times 10^{-6}\ /^{\circ}C[/tex]
[tex]\Delta T[/tex] = Change in temperature = [tex](60-20)^{\circ}C[/tex]
[tex]R_0[/tex] = Initial radius = 4 cm
Change in radius of material is given by
[tex]R=R_0(1+\alpha\Delta T)[/tex]
Difference in radii of the lid and jar
[tex]\Delta R=R_b-R_g\\\Rightarrow \Delta R=R_0(1+\alpha_b\Delta T)-R_0(1+\alpha_g\Delta T)\\\Rightarrow \Delta R=R_0(\alpha_b-\alpha_g)\Delta T\\\Rightarrow \Delta R=4\times (19\times 10^{-6}-9\times 10^{-6})\times (60-20)\\\Rightarrow \Delta R=0.0016\ cm[/tex]
The size of the gap is 0.0016 cm or 0.000016 m
Final answer:
Heating the brass lid and glass jar causes the lid to expand more than the jar, creating a gap between them. To find the size of the gap, we can calculate the change in radius of the lid. Using the given information, we can find that the size of the gap is approximately 19.2 x [tex]10^-^3[/tex] cm.
Explanation:
When the brass lid and glass jar are heated in hot water, their temperatures increase. The lid, being made of metal, expands more than the glass jar due to the higher coefficient of expansion for metals. This causes the lid to loosen its grip on the jar, creating a gap between them.
To find the size of the gap, we need to calculate the difference in radius of the lid before and after heating. The inside diameter of the lid is given as 8.0 cm at 20 degrees C. We can calculate the radius using the formula r = ½ d, where d is the diameter. The radius at 20 degrees C is therefore 4.0 cm.
Now, let's calculate the radius at 60 degrees C. To do this, we need to know the coefficient of linear expansion for brass. Assuming it is 19x10^-6 /°C, we can use the formula: ∇r = αr0∇T, where ∇r is the change in radius, α is the coefficient of linear expansion, r0 is the initial radius at a reference temperature, and ∇T is the change in temperature. Plugging in the values, we get: ∇r = (19x[tex]10^-^6[/tex] /°C) x (4.0 cm) x (60 °C - 20 °C) = 19.2 x [tex]10^-^3[/tex] cm.
Therefore, the size of the gap that develops by heating the lid and jar is approximately 19.2 x [tex]10^-^3[/tex] cm.
While surveying a cave, a spelunker follows a passage 190 m straight west, then 220 m in a direction 45.0 ∘ east of south, and then 270 m at 30.0 ∘ east of north. After a fourth unmeasured displacement, she finds herself back where she started. Use vector components to find the magnitude and direction of the fourth displacement.
Answer:
The magnitude of the displacement is 127.43m and the direction is 38°North of East
Explanation:
Final answer:
To find the magnitude and direction of the fourth displacement in a cave survey, vector components can be used. By breaking down the given displacements into their north and east components, calculating the net displacements, and using the Pythagorean theorem and trigonometry, the magnitude and direction of the fourth displacement can be determined.
Explanation:
To find the magnitude and direction of the fourth displacement, we can use vector components. First, let's break down the given displacements into their north and east components:
Displacement 1: 190 m west (0 north, -190 east)
Displacement 2: 220 m at 45.0° east of south (-220 sin(45°) north, 220 cos(45°) east)
Displacement 3: 270 m at 30.0° east of north (270 sin(30°) north, 270 cos(30°) east)
Next, we can add up the north and east components separately to find the net north and east displacements. Lastly, we can use these net displacements to calculate the magnitude and direction of the fourth displacement using the Pythagorean theorem and trigonometry.
A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N and points due west. She then turns around and rides due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 3.3 N and points due east.?
Final answer:
To find the displacement, subtract the distance traveled in one direction from the distance traveled in the opposite direction.
Explanation:
To solve this problem, we need to consider the concept of displacement. Displacement is a vector quantity that represents the change in position of an object. To find the displacement, we can subtract the distance traveled in one direction from the distance traveled in the opposite direction. In this case, the girl rides 5.4 km due east and then turns around and rides back to her starting point. So her displacement is 5.4 km - 5.4 km = 0 km.
Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some program and computer B has a clock cycle time of 250 ps and an effective CPI of 1.7 for the same program. Which computer is faster and by how much?
Answer:
Computer A is 1.41 times faster than the Computer B
Explanation:
Assume that number of instruction in the program is 1
Clock time of computer A is [tex]CT_{A} =200 ps[/tex]
Clock time of computer B is [tex]CT_{B} =250 ps[/tex]
Effective CPI of computer A is [tex]CPI_{A} =1.5[/tex]
Effective CPI of computer B is[tex]CPI_{B} =1.7[/tex]
CPU time of A is
[tex]CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec[/tex]
CPU time of B is
[tex]CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec[/tex]
Hence Computer A is Faster by [tex]\frac{425}{300} =1.41[/tex]
Computer A is 1.41 times faster than the Computer B
A post is wrapped two full turns around with a belt. The tension in the belt is 7500 N by exerting a force of 150 N on its free end. Determine the coefficient of static friction between the belt and the post.
Answer:
Coefficient of friction is
Ū = 0.31
Explanation:
T2 = T1* e^(ūơ)
Where T2 = 7500n = tension in the belt, T1 = 150n = reaction force,
ū = coefficient of friction
Ơ = 2pai * N
Where N = number of turns = 2
Ơ = 4pai
7500 = 150e^(ū*4pai)
50 = e^(ū *4pai)
lin 50 = 4pai * ū
Ū = 3.91/4pai
Ū = 0.31
Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at a wavelength of 350 nm. Each photon of the ultraviolet light has:________.
a) half the energy of each photon of the red light.
b) twice the energy of each photon of the red light.
c) four times the energy of each photon of the red light.
d) either more or less energy than each photon of the red light, depending on the intensities of the two light sources.
Answer:
b) twice the energy of each photon of the red light.
Explanation:
[tex]\lambda[/tex] = Wavelength
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
Energy of a photon is given by
[tex]E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}[/tex]
Let [tex]\lambda_1[/tex] = 700 nm
[tex]\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}[/tex]
For red light
[tex]E_1=\dfrac{hc}{\lambda_1}[/tex]
For UV light
[tex]E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}[/tex]
Dividing the equations
[tex]\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1[/tex]
Hence, the answer is b) twice the energy of each photon of the red light.
Answer:
b) twice the energy of each photon of the red light.
Explanation:
Given:
wavelength of red light, [tex]\lambda_r=7\times 10^{-7}\ m[/tex]
wavelength of ultraviolet light, [tex]\lambda_{uv}=3.5\times 10^{-7}\ m[/tex]
We know that the energy of a photon is given as:
[tex]E=h.\nu[/tex] ..............(1)
where:
h = plank's constant [tex]=6.626\times 10^{-34}\ J.s[/tex]
[tex]\nu=[/tex] frequency of the wave
we have the relation:
[tex]\nu=\frac{c}{\lambda}[/tex] ...................(2)
From (1) and (2) we have:
[tex]E=\frac{h.c}{\lambda}[/tex]
Energy of photons for ultraviolet light:
[tex]E_{uv}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{3.5\times 10^{-7}}[/tex]
[tex]E_{uv}=5.6794\times 10^{-19}\ J[/tex]
Since the energy of photons is inversely proportional to the wavelength of the light hence the photon-energy of ultraviolet light is double of the photon-energy of the red light as the wavelength of the red light is twice to that of ultraviolet light.
The kinetic energy of a photoionized electron increases linearly with the wavelength of the ionizing radiation, but is independent of the intensity of the radiation. If False, explain in one sentence what makes the statement False.A. True
B. False
Answer: False
Explanation:
It is dependent on the intensity of the radiation i.e the photo electron should increase with the light amplitude.
From whom should a departing VFR aircraft request radar traffic information during ground operations?
Answer:From whom should a departing VFR aircraft request radar traffic information during ground operations? ... Answer: Sequencing to the primary Class C airport, traffic advisories, conflict resolution, and safety alerts.
Explanation:
Final answer:
During ground operations, a departing VFR aircraft should request radar traffic information from ground control. Once airborne, the responsibility switches to tower control for traffic updates.
Explanation:
A departing VFR (Visual Flight Rules) aircraft should request radar traffic information from the ground control (also known as ground movement control) before taxiing. Ground control is responsible for the safety on the taxiways and runways, except the active runway. Once airborne or when holding short of the runway, the aircraft switches to tower frequency, at which point the aircraft may receive radar traffic updates relevant to their departure.
Technology utilizing a network of radio frequency waveguides is part of managing air traffic, which is crucial for maintaining a safe separation between aircraft. Furthermore, radar technology can provide directional information, and as noted in research such as Hasselmann, 1991, traffic information can be directly calculated from radar data even during the ground operations of VFR aircraft.
Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back and releases it when the seat is 1.09 m above the ground. The acceleration of gravity is 9.8 m/s 2 . How fast is Betty moving when the swing passes through its lowest position? Answer in units of m/s.
Answer:
4.15 m/s
Explanation:
As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:
ΔE = ΔK + ΔU =0
If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:
h₀ = 1.09 m -0.21 m = 0.88 m
⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J
As Uf = 0, ΔU = Uf -U₀ = -352 J
As the swing starts from rest, K₀=0, so we can say:
ΔK = Kf = [tex]\frac{1}{2} *m*vf^{2}[/tex] (1)
As ΔK = -ΔU ⇒ ΔK = 352 J (2)
From (1) and (2) we can solve for vf, as follows:
[tex]vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s[/tex]
So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.
Final answer:
Betty is moving at a speed of approximately 2.86 m/s when the swing passes through its lowest position.
Explanation:
To calculate the speed at the lowest position of Betty on the swing, we can use the principle of conservation of mechanical energy. At the highest point, all of Betty's potential energy is converted into kinetic energy. So we can write:
Initial potential energy = Final kinetic energy
mgh = 0.5mv^2
where m is the mass of Betty, g is the acceleration due to gravity, h is the initial height, and v is the velocity at the lowest point.
Plugging in the values, we get:
(400 N)(0.21 m) = 0.5(400 N)v^2
After simplifying and solving for v, the velocity at the lowest point is approximately 2.86 m/s.
Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ramps. Which statement is TRUE?
Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ramps.Which statement is TRUE?
answer choices
Marble 1 has a faster speed than Marble 2.
Marble 2 has a faster speed than Marble 1.
Both the marbles travel at the same speed.
There is not enough data to compare the speeds of marbles.
Answer:Marble 2 has a faster speed than Marble 1.
Option B.
Explanation:The speed is defined as the distance covered per unit time. Here in the question, 2 balls cover equal distances in same time.
Time taken by the ball = 10 seconds.
Distance covered by 1st ball = 20 cm.
Distance covered by 2nd ball = 3cm.
So speed of the 1st ball = 2cm/sec.
Speed of the 2nd ball = 3 cm /sec.
So,it's very much evident that speed of 2nd Marble is much higher than the speed of the 1st marble.
A box slides down a slope described by the equation y=(0.05 x2)m, where x is in meters. If the box has x components of velocity and acceleration vX= -4m/s and aX= -1.2m/s@ at x = 3m, determine the corresponding y components of velocity and acceleration at this instant
Answer:
[tex]v_y=-1.2m/s,a_y=-0.76m/s^2[/tex]
Explanation:
We are given that
[tex]y=(0.05x^2)[/tex]
Where x (in m)
x component of velocity, [tex]v_x=-4m/s[/tex]
x- component of acceleration,[tex]a_x=-1.2 m/s^2[/tex]
x=3 m
We have to find the y-component of velocity and acceleration at this instant.
Differentiate w.r.t. time
[tex]\frac{dy}{dt}=0.05\times 2x\frac{dx}{dt}[/tex]
[tex]\frac{dy}{dt}=0.1x\frac{dx}{dt}=0.1xv_x[/tex]
By using the formula
[tex]\frac{dx^n}{dx}=nx^{n-1}[/tex]
[tex]v_x=\frac{dx}{dt}[/tex]
Substitute the values
[tex]v_y=\frac{dy}{dt}=0.1(3)\times (-4)=-1.2m/s[/tex]
Again differentiate w.r.t. t
[tex]a_y=\frac{dv_y}{dt}=0.1v_x+0.1x\frac{d(v_x)}{dt}[/tex]
Substitute the values
[tex]a_y=0.1(-4)+0.1(3)(-1.2)=-0.4-0.36=-0.76m/s^2[/tex]
Where [tex]a_x=\frac{dv_x}{dt}[/tex]
Hence, the y- component of velocity and acceleration
[tex]v_y=-1.2m/s,a_y=-0.76m/s^2[/tex]
To find the y components of velocity and acceleration, differentiate the equation y = 0.05x^2 with respect to time. Plug in the given values to find vy and ay at x = 3m.
Explanation:To find the corresponding y components of velocity and acceleration at x = 3m, we need to differentiate the equation y = 0.05x^2 with respect to time. Differentiating y with respect to time gives vy = dy/dt = d/dt(0.05x^2) = 0.1x(dx/dt). Plugging in the given values dx/dt = -4m/s and x = 3m, we can find vy. Similarly, to find the y-component of acceleration, we differentiate vy with respect to time, which gives ay = dvy/dt = d/dt(0.1x(dx/dt)) = 0.1(dx/dt)(dx/dt) + 0.1x(d^2x/dt^2). Plugging in the given values dx/dt = -4m/s and d^2x/dt^2 = -1.2m/s^2, we can find ay.
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