The problem is to calculate the kinetic energy of the puck at the moment it hits the spring. We assume that the puck begins to move at frame 82. From the graph of x versus t, the initial position of the puck is at x = 0. Thus, we use the equation:
[tex]$$K_{f} - K_{i} = W_{NC}$$where $$K_{f}$$[/tex] is the final kinetic energy of the puck, [tex]$$K_{i}$$[/tex] is the initial kinetic energy of the puck, and [tex]$$W_{NC}$$[/tex] is the nonconservative work done by friction. We can use the conservation of mechanical energy as follows:
[tex]$$E_{i} = E_{f}$$$$K_{i} + U_{i} = K_{f} + U_{f}$$[/tex]
where[tex]$$E_{i}$$[/tex] is the initial mechanical energy, [tex]$$E_{f}$$[/tex] is the final mechanical energy, [tex]$$U_{i}$$[/tex] is the initial potential energy, and [tex]$$U_{f}$$[/tex] is the final potential energy.
At the moment the puck hits the spring, the height of the puck is zero, so the potential energy of the puck is zero. Thus, we can write:
[tex]$$K_{i} = K_{f}$$[/tex]
We can use the equation for the kinetic energy of an object:
[tex]$$K = \frac{1}{2}mv^{2}$$where $$m$$[/tex]
is the mass of the object, and [tex]$$v$$[/tex] is its velocity. We need to calculate the velocity of the puck at the moment it hits the spring. we can calculate the kinetic energy of the puck:
[tex]$$K = \frac{1}{2}mv^{2} = \frac{1}{2}(0.05\ \text{kg})(0.088\ \text{m/s})^{2} = 1.94\times10^{-4}\ \text{J}$$[/tex]
Therefore, the kinetic energy of the puck at the moment it hits the spring is [tex]$$1.94\times10^{-4}$$[/tex] J, to three significant figures.
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How is capacitance related to the distance between the plates of a capacitor?
It is directly proportional, so the capacitance increases as the distance increases.
It is inversely proportional, so the capacitance increases as the distance increases.
It is directly proportional, so the capacitance decreases as the distance increases.
It is inversely proportional, so the capacitance decreases as the distance increases.
Answer:C
Explanation: I studied, and C is correct
Answer:
D
Explanation:
what is the velocity of a car that traveled 6 meters in .96 seconds
The velocity of the car is approximately 6.25 meters per second.
To determine the velocity of a car that traveled 6 meters in 0.96 seconds, we can use the formula for velocity: velocity = distance / time. In this case, the distance traveled is 6 meters and the time taken is 0.96 seconds.
Plugging these values into the formula, we have:
velocity = 6 meters / 0.96 seconds
= 6.25 meters per second.
Velocity is a measure of the rate at which an object changes its position. In this context, the car is traveling at a constant speed of 6.25 meters per second. It means that for every second that passes, the car moves 6.25 meters forward.
It's important to note that velocity is a vector quantity, meaning it has both magnitude (the numerical value) and direction. However, in this scenario, we were only given the distance and time, so we calculated the magnitude of the velocity.
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Purpose:You will use the radioactive decay rate and original-daughter element ratios of carbon-14and uranium-238 to determine the ages of different objects.
Procedure:1.Load PhET Radioactive Dating Game
2.Click on the tab for Decay Rates
3.Select Carbon-14.Using the graph, the estimated half-life for C-14 is5,200years.
4.Move the bucket slider all the way to the right.This will place 1000 C-14 atoms onto the screen.a.Click on the Start/Stop to stop the C-14 decay.Click on Reset All Nucleib.Click on theStart/Stop to start the C-14 decay.Stop the decay as you get close to one half-life.c.Use the Step button to stop decay at one half-life.After 1 half-life, how many C-14 atoms of the 1000 original remain?
After one half-life, the number of remaining C-14 atoms can be calculated by multiplying the initial number of atoms (1000) by 0.5 (since half of the atoms decayed).
Based on the given procedure, after one half-life of carbon-14 (C-14), the number of C-14 atoms remaining can be determined. Since the estimated half-life of C-14 is 5,200 years, we can use this information to answer the question. After one half-life, the number of remaining C-14 atoms can be calculated as half of the original number of C-14 atoms. Given that the initial number of C-14 atoms is 1000, after one half-life: Remaining C-14 atoms = (1/2) * 1000. Remaining C-14 atoms = 500
after one half-life of carbon-14 (C-14), the number of C-14 atoms remaining can be determined.
Therefore, after one half-life, 500 C-14 atoms of the 1000 original atoms remain.
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Can someone pls answer
Answer:
im still in elementry so you got to do it yo self
Explanation:
so me confused
A 29 kg child jumps to the ground from the top of a fence that is 1.8 m high. You analyze the problem using upward as the positive x direction.
A. Taking x=0 to be at the bottom of the fence, what are the initial potential energy of the child-Earth system and the chnage in the system kinetic energy during the jump? (Ui , change of U)
B.Repeat the previous part for x=0 at the top of the fence (Ui , change of U)
The initial potential energy of the child-Earth system is 509.4 J, and the change in system kinetic energy during the jump is 509.4 J.
In the first scenario, with x=0 at the bottom of the fence, we can calculate the initial potential energy (Ui) using the formula Ui = mgh, where m is the mass of the child (29 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the fence (1.8 m). Substituting the values, Ui = 29 kg × 9.8 m/s^2 × 1.8 m = 509.4 J.
Since there is no external work done on the child during the jump, the change in system kinetic energy (change of U) is equal to the negative of the initial potential energy. Therefore, the change of U = -509.4 J.
In the second scenario, with x=0 at the top of the fence, the initial potential energy (Ui) is still the same, i.e., 509.4 J. However, since the child is starting from a higher position, the change in system kinetic energy (change of U) will be different. The change of U will still be equal to -509.4 J since it depends on the initial potential energy, regardless of the reference point.
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If it is known that a motor battery has an input voltage of 12V and a capacity of 6 Ah, how much power and resistor value is required to turn on 8 lamps with a parallel circuit, with the specifications of each lamp having a maximum voltage of 3V and an electric current of 140 mA? How long did all the lights go on until they off?
Answer:
Part A
The power to turn on the lamp, ∑P = 3.36 W
Part B
The Resistor required is approximately 8.04 Ohms
Part C
The time for all the lights to go out is approximately 21.43 hours
Explanation:
The input voltage of the motor battery , V = 12 V
The capacity of the battery, Q = 6 Ah
The number of lamps in parallel = 8 lamps
The maximum voltage of each lamp, = 3 V
The electric current in each lamp = 140 mA
The energy available in a battery, E = Q × V
For the battery, we have;
E = 6 Ah × 12 V = 72 Wh
The energy available in a battery, E = 72 Wh
Part A
The power used by the lamps, [tex]P_i[/tex] = [tex]I_i[/tex] × [tex]V_i[/tex]
∴ The total power used by the lamp, ∑P = 8 × 0.14 A × 3 V = 3.36 W
The power to turn on the lamp, ∑P = 3.36 W
Part B
The resistance required, is given as follows;
Resistor required = (Battery voltage - Lamp voltage)/(The sum of bulb current)
∴ Resistor required = (12 V - 3 V)/(8 × 0.14 A)
The Resistor required = 8.03571429 Ohms
The Resistor required ≈ 8.04 Ohms
Part C
The time for all the lights to go out = The time for the lamps to use all the power available in the battery
The time for all the lights to go out, t = E/∑P
∴ t = 72 Wh/(3.36 W) = 21.4285714 h
∴ The time for all the lights to go out, t ≈ 21.43 h
The time for all the lights to go out = The time for the lamps to use all the power available in the battery = t ≈ 21.43 h
∴ The time for all the lights to go out ≈ 21.43 hours.
The electric field strength is 5.50×10⁴ N/C inside a parallel-plate capacitor with a 2.30 mm spacing. A proton is released from rest at the positive plate.What is the proton's speed when it reaches the negative plate?
To determine the proton's speed when it reaches the negative plate, we can utilize the relationship between electric field, force, and acceleration.
The force experienced by a charged particle in an electric field is given by the equation F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength. In this case, the proton has a charge of +e (1.6 × 10⁻¹⁹ C) and experiences a force in the direction of the electric field.
Using Newton's second law, F = ma, we can relate the force to the proton's acceleration (a) and mass (m). Since the proton is released from rest, its initial velocity (v₀) is zero. The distance traveled by the proton (d) is equal to the spacing of the parallel plates, which is 2.30 mm (2.30 × 10⁻³ m).
The force on the proton is F = qE = (1.6 × 10⁻¹⁹ C) × (5.50×10⁴ N/C) = 8.80 × 10⁻¹⁵ N. By equating the force to mass times acceleration, we have ma = 8.80 × 10⁻¹⁵ N.
Rearranging the equation to solve for acceleration, we get a = (8.80 × 10⁻¹⁵ N) / (m). The mass of a proton is approximately 1.67 × 10⁻²⁷ kg.
Substituting the values, we find the acceleration of the proton. Using the kinematic equation v² = v₀² + 2ad, we can find the final velocity (v) of the proton when it reaches the negative plate.
Since the initial velocity (v₀) is zero and the distance (d) is known, we can solve for the final velocity. Calculating the expression gives us the speed of the proton when it reaches the negative plate.
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Two objects collide elastically. The first has a mass of 5.00kg and a
velocity of 8.00 m/s. The second has a mass of 2.50 kg and a velocity
of -4.00m/s. If the velocity of object 1 after the collision is -4.00m/s,
what is the velocity of object 2?
Answer:
Explanation:
General formula
m1 * vi + m2*v2 = m1*v3 + m2*v4
Givens
m1 = 5
m2 = 2.5
v1 = 8 m/s
v2 = - 4 m/s
v3 = -4 m/s
v4 = x
Solution
5 * 8 - 2.5 * 4 = 5 * -4 + 2.5*x
40 - 10 = -20 + 2.5x
30 = - 20 + 2.5x
50 = 2.5x
x = 50/2.5
x = 20 m/s in the positive direction
Remark
Does this answer make sense? It should. You have 5 kg moving 8m/s in the plus direction. That's a lot of momentum. In addition after the collision, it turns around which is more momentum needed.
It has to give up that extra momentum to the 2.5 kg mass.
Jenise is buying a car for $7,020. The TAVT rate is 9.1%.
What is the amount of tax that Jenise will have to pay on her car?
Answer:
$7,658.82
Explanation:
Sales Tax Calculations:
Sales Tax Amount = Net Price x (Sales Tax Percentage / 100)
Total Price = Net Price + Sales Tax Amount
Net Price: $ 7,020.00
+Sales Tax (9.1%): $ 638.82
Total Price: $ 7,658.82
Therefore, the amount of tax that Jenise has to pay on her car is $7,658.82
Please help me with 17 and 18!!!!!! (It's related to 16) It's due today!!!!! NO LINK PLEASE!!!!!!!!
Answer:17: A wave can be defined as follows: It is important to realize that a wave is quite a different object than a particle. A baseball thrown though a window transfers energy from one point to another, but this involves the movement of a material object between two points.
Explanation:
18: In this way, we classify waves into electromagnetic and mechanical waves. The main difference between mechanical and electromagnetic waves is that electromagnetic waves do not require a medium to propagate whereas mechanical waves require a medium in order to propagate.
Please help me if you can.
I can't figure out how the answer for number 30 is C and 31 is C. My question is how did they get that answer?
Answer:
30. $3.85
31. $2.22
Explanation:
30. The time duration of the kitchen clock is left on = All day = 24 hours
The power rating of the kitchen clock, P = 4 watts
The electricity cost in Alberta, R = $0.11 per kilowatt hour
The total number of hours the kitchen clock is on during the year, 't', is given as follows;
t = Number of hours per day × Number of days per year
∴ t = 24 hours/day × 365 days/year = 8,760 hours
The energy consumption of the kitchen clock, per year, E = P × t
∴ E = 4 watts × 8,760 hours = 35,040 watts-hour = 35.040 kw·h
The cost of operating the clock in one year, C = E × R
∴ Operating cost for the kitchen clock
∴ C = 35.040 kw·h × $0.11/(kw·h) = $3.8544 ≈ $3.85
The cost of operating the clock in one year, C ≈ $3.85
31. The power rating of the ghetto blaster, P = 28 watts
The time duration the ghetto blaster was on during an average month = All day = 24 hours
The number of days during an average month, n = 30 days
The cost of electricity in Alberta, R = $0.11 per kilowatt hour
The time in hours the ghetto blaster was on, t = 30 days/month × 24 hours/day
∴ t = 720 hours per month
The cost of operating the ghetto blaster in one month, C = P × t × R
∴ C = 28 W × 720 hours × 1 kw·h/(1000 w·h)× $0.11/(kw·h) = $2.2176
∴ The cost of operating the ghetto blaster in one month, C ≈ $2.22
an iron anchor of density 7810.00 kg/m3 appears 152 n lighter in water than in air. (a) what is the volume of the anchor? (b) how much does it weigh in air
The volume of the anchor is approximately 0.0195 m^3. The weight of the anchor in air is approximately 1492 N.
To find the volume of the anchor, we can use Archimedes' principle, which states that the buoyant force experienced by an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
Given that the anchor appears 152 N lighter in water than in air, we can equate this weight difference to the buoyant force experienced by the anchor in water.
Weight difference = Buoyant force
= Weight in air - Weight in water
Let's assume the weight of the anchor in air is W_air and the weight of the anchor in water is W_water.
W_air - W_water = 152 N
We know that weight = mass × acceleration due to gravity (W = m × g), and density is defined as mass divided by volume (ρ = m/V), where ρ is the density, m is the mass, and V is the volume.
Therefore, W_air = ρ_anchor × g × V × (1), and
W_water = ρ_water × g × V × (2).
Given that the density of water, ρ_water, is 1000 kg/m^3, and the density of the anchor, ρ_anchor, is 7810.00 kg/m^3, we can substitute these values into equations (1) and (2):
7810.00 × g × V - 1000 × g × V = 152
Simplifying the equation:
6810.00 × g × V = 152
V = 152 / (6810.00 × g)
Using the standard acceleration due to gravity, g = 9.8 m/s^2:
V = 152 / (6810.00 × 9.8)
≈ 0.0195 m^3
Therefore, the volume of the anchor is approximately 0.0195 m^3.
To calculate the weight of the anchor in air, we can use the formula:
Weight in air = ρ_anchor × g × V
Substituting the values:
Weight in air = 7810.00 × 9.8 × 0.0195
≈ 1492 N
Therefore, the weight of the anchor in air is approximately 1492 N.
The volume of the anchor is approximately 0.0195 m^3, and its weight in air is approximately 1492 N.
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(a) Does the index of refraction vary as you change the wavelength of light. (B) What is the angle of the ray that leaves the glass square (emergent ray) relative to the ray that enters it? (c) What can you say about the path of emergent ray relative to that of the incident ray?
(a) Yes, the index of refraction varies with the wavelength. (b) The emergent ray is refracted at a different angle. (c) The path of the emergent ray deviates from the incident ray.
(a) Yes, the index of refraction varies as you change the wavelength of light.
The index of refraction (n) of a material is a measure of how much the speed of light is reduced when it passes through that material compared to its speed in a vacuum.
The index of refraction is wavelength-dependent and typically varies slightly with different wavelengths of light. This phenomenon is known as dispersion.
One way to express this variation is through the refractive index as a function of wavelength, often represented by a refractive index versus wavelength graph.
In general, different wavelengths of light are bent or refracted by different amounts when passing through a medium due to their interaction with the material's atoms or molecules.
This bending is a result of the change in the speed of light, which is dictated by the refractive index.
The index of refraction does vary as you change the wavelength of light. This variation is responsible for phenomena like dispersion, where different colors of light are separated when passing through a prism, for example.
(b) The angle of the emergent ray leaving a glass square relative to the incident ray depends on the angle of incidence and the refractive index of the glass.
According to Snell's law, the relationship between the angle of incidence (θ₁), the angle of refraction (θ₂), and the refractive indices of the two media involved can be expressed as:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
In the case of a glass square, let's assume light is incident on one of its faces. If we know the angle of incidence (θ₁) and the refractive index of the glass (n₂), we can calculate the angle of the emergent ray (θ₂) using Snell's law.
The angle of the emergent ray leaving the glass square relative to the incident ray depends on the angle of incidence and the refractive index of the glass, and it can be calculated using Snell's law.
(c) The path of the emergent ray relative to the incident ray can be different due to refraction.
When light passes from one medium to another, it changes direction due to the change in its speed caused by the change in the refractive index. This change in direction is called refraction. Therefore, the emergent ray may have a different direction compared to the incident ray.
The emergent ray will still follow the law of refraction (Snell's law) and will be bent towards or away from the normal depending on the refractive indices of the two media involved and the angle of incidence.
The amount of bending depends on the difference in refractive indices and the angle at which the light strikes the boundary between the two media.
The path of the emergent ray relative to the incident ray can be different due to refraction, as the emergent ray changes direction upon passing from one medium to another.
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basketball center steve tootall is 7 feet 2 inches in height. what is steve’s height in inches?
Steve Tootall's height is 86 inches. To calculate his height in inches, we convert feet to inches and then add the remaining inches.
Steve Tootall's height is given as 7 feet 2 inches. To calculate his height in inches, we convert feet to inches and then add the remaining inches.
1 foot is equal to 12 inches. So, 7 feet would be 7 * 12 = 84 inches.
Adding the remaining 2 inches, Steve's height in inches would be:
84 inches + 2 inches = 86 inches.
Therefore, Steve Tootall's height is 86 inches.
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Is O2 classified as a compound?
Answer: NO.
Explanation: Oxygen is not a compound. It has only one element in it.
during each cycle, a heat engine ejects 75 j of thermal energy for every 115 j of input thermal energy. this engine is used to lift a 375 kg load a vertical distance of 27.0 m at a steady rate of 52.5 mm/s. how many cycles of the engine are needed to accomplish this task?
Hence, 1309 cycles of the engine are needed to accomplish this task.
Given values:
Work done = Force × Distance moved = mgh,
Force = mg = 375 × 9.8 = 3675 N,
Distance moved, s = 27 m
Rate of work = Power = Work done ÷ time
Rate of work = Fs/t
rate of work = 3675 × 0.0525
rate of work = 193.69 W.
Potential energy = Work done = mgh = 375 × 9.8 × 27 = 98175 J.
Heat ejected by the engine per cycle = 75 J, Heat input to the engine per cycle = 115 J.
We can find the number of cycles of the engine needed to accomplish this task by dividing the potential energy by the amount of heat ejected per cycle.
Therefore: Number of cycles = Potential energy ÷ Heat ejected per cycle= 98175 ÷ 75 = 1309 cycles.
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A box m=87 kg is being pulled by a constant force F=124 N at an angle of θ =43 degrees. The initial speed of the box is zero. A 33\% Part (a) Write an expression for the work done by force F as the block moves a horizontal distance d. A3\% Part (b) How much work, in joules, was done in moving the block 4.1 m ? A 33% Part (c) What is the speed of the box at d=4.I m if the surface is frictionless?
Part (a) involves writing an expression for the work done by force F as the box moves a horizontal distance d. Part (b) requires calculating the amount of work done in moving the box 4.1 m. Part (c) asks for the speed of the box at a distance of 4.1 m, assuming a frictionless surface.
(a) The work done by a force is given by the formula W = Fd cos(θ), where W represents work, F is the force applied, d is the distance moved, and θ is the angle between the force and the displacement. In this case, since the force is applied horizontally ([tex]\theta[/tex] = 0), the expression for the work done by force F becomes W = Fd cos(0) = Fd.
(b) To calculate the work done in moving the box 4.1 m, we can substitute the given values into the equation from part (a). Thus, the work done is W = (124 N)(4.1 m)(cos 0) = 124 N * 4.1 m = 508.4 J (joules).
(c) If the surface is frictionless, the work done is converted entirely into kinetic energy. We can use the work-energy principle to find the speed of the box. The work done (508.4 J) is equal to the change in kinetic energy. Assuming the initial speed is zero, the final kinetic energy is 508.4 J. We can calculate the speed using the equation [tex]KE = (1/2)mv^2[/tex], where KE is the kinetic energy, m is the mass, and v is the speed. Rearranging the equation, [tex]v = \sqrt(2KE/m)[/tex]. Given the mass m = 87 kg, the speed at d = 4.1 m can be calculated.
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for the truss bridge shown, a) sketch the influence lines for the force in members be, bc, and da.
a) Sketching the influence lines for the force in members BE, BC, and DA in a truss bridge requires a visual representation. Influence lines show the variation of forces in a structure due to the movement of a load.
Please refer to structural engineering resources or software for accurate sketches and graphical representations of the influence lines for these specific members in a truss bridge. These resources will provide detailed illustrations based on the structural dimensions, member properties, and load positions. Influence lines are valuable tools for structural engineers as they help identify critical load positions, assess the structural response, and determine the maximum forces experienced by different members. Please consult reliable structural engineering references, textbooks, or appropriate software that specifically address truss bridge analysis and design to obtain accurate sketches of the influence lines for members BE, BC, and DA. These resources will provide the necessary diagrams and detailed explanations based on the specific truss bridge configuration and load positions.
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The distance between two cities A and B is 180 km. a car moved from the city A towards the city :
with a velocity of 25 Km/hr , at the same moment another car moved from the citv B towards the city A with a uniform velocity of 65 Km/hr. Then: when and where do the two cars meet?
The two cars will meet at a distance of 50 km from city A, and the meeting will occur 2 hours after they start moving.
To determine when and where the two cars meet, we need to calculate the time it takes for them to meet and then use that time to find the meeting location.
In this case:
Distance between cities A and B = 180 km
Velocity of the car starting from city A = 25 km/hr
Velocity of the car starting from city B = 65 km/hr
Let's assume the meeting point is at a distance of x km from city A. Since the total distance between the two cities is 180 km, the distance traveled by the car starting from city A is x km, and the distance traveled by the car starting from city B is (180 - x) km.
Using the formula:
Time = Distance / Velocity
The time taken by the car starting from city A to reach the meeting point is:
Time for car from A = x km / 25 km/hr = x/25 hr
The time taken by the car starting from city B to reach the meeting point is:
Time for car from B = (180 - x) km / 65 km/hr = (180 - x)/65 hr
Since the two cars meet at the same time, we can set their time equations equal to each other:
x/25 = (180 - x)/65
Now, we can solve this equation to find the value of x:
65x = 25(180 - x)
65x = 4500 - 25x
90x = 4500
x = 50
Therefore, the meeting point is 50 km from city A.
To find the time it takes for the cars to meet, we can substitute this value of x back into either of the time equations:
Time = Distance / Velocity
Time = 50 km / 25 km/hr
Time = 2 hours
So, the two cars will meet after 2 hours.
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True or False: A balanced force acts on different objects, and action-
eaction forces act on the same object. *
A. True
B. False
Answer:
true
Explanation:
according to the Newton's third law
I need help with one question on my homework. This is on the Specific Heat Capacity required practical.
Sample c has no solid residue left when evaporated. Suggest why it has a boiling point 1.7 degrees Celsius lower than distilled water.
Answer:
will mostly accord at the top of the boiling water my kind sir
Explanation:
Evaporation takes place only at the surface of a liquid, whereas boiling may occur throughout the liquid. In boiling, the change of state takes place at any point in the liquid where bubbles form. The bubbles then rise and break at the surface of the liquid.
12. Which of the following statements is accurate?
A. If an object's velocity is changing, it's experiencing either acceleration or deceleration.
B. If an object's velocity decreases, then the object is accelerating
C. If an objects said to be decelerating, its velocity must be increasing,
D. If an object's velocity remains constant, its acceleration must be increasing.
Answer:
Option (a) is correct
Explanation:
The acceleration of an object is defined as the rate of change of velocity. Mathematically, it can be written as :
[tex]a=\dfrac{v-u}{t}[/tex]
Where
v and u are final and initial velocity
It is clear that if there is some change in velocity, it means the object is experiencing either acceleration or deceleration. Hence, the correct option is (a).
Answer:
a
Explanation:
Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature T and 2.0 times the diameter of the hotter star. Part A What is the temperature of the hotter star in terms of T? VO AED h ? TT = Submit Previous Answers Request Answer Part B What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star? VO AED ? = Submit Request Answer Provide Feedback
The temperature of the hotter star ([tex]T_h[/tex]) is equal to the square root of the surface temperature of the cooler star (T), and the ratio of the peak-intensity wavelengths is proportional to the inverse cube of the temperature ratio.
Part A: Let's denote the temperature of the hotter star as [tex]T_h[/tex]. According to the Stefan-Boltzmann law, the total energy radiated by a blackbody is proportional to the fourth power of its temperature. Since both stars radiate the same total energy per second, we can write:
[tex]T_h^4 = T^4[/tex]
Taking the fourth root of both sides, we get:
[tex]T_h = T^{(\frac {1}{4})}[/tex]
Part B: The peak intensity wavelength (λmax) of a blackbody radiation is inversely proportional to its temperature.
According to Wien's displacement law, we can express the ratio of peak-intensity wavelengths ([tex]\lambda_{max, hot}/ \lambda_{max, cool}[/tex]) as the ratio of their temperatures:
[tex]\frac{\lambda_{max, hot}}{ \lambda_{max, cool}} = \frac{T_h}{T}[/tex]
Substituting the relationship we derived in Part A, we have:
[tex]\frac{\lambda_{max, hot}}{ \lambda_{max, cool}} = \frac{T^{\frac{1}{4}} }{T}[/tex]
Simplifying, we get:
[tex]\frac{\lambda_{max, hot}}{ \lambda_{max, cool}} = T^{\frac{-3}{4}} }[/tex]
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what tangential speed v must the bob have so that it moves in a horizontal circle with the string making an angle 21.0 ∘ with the vertical?
To find the tangential speed required for the bob to move in a horizontal circle with the string making an angle of 21.0 degrees with the vertical, we can use the concept of centripetal force.
The centripetal force required to keep the bob moving in a circular path is provided by the tension in the string. The tension can be resolved into two components: the vertical component and the horizontal component. The vertical component of the tension balances the weight of the bob, which is given by: T * cos(21.0°) = mg. where T is the tension in the string, m is the mass of the bob, and g is the acceleration due to gravity. The horizontal component of the tension provides the centripetal force required for circular motion, and it can be expressed as: T * sin(21.0°) = mv^2 / r. where v is the tangential speed of the bob and r is the radius of the circular path. Dividing the two equations: [T * sin(21.0°)] / [T * cos(21.0°)] = (mv^2 / r) / (mg). tan(21.0°) = v^2 / (rg). Solving for v: v = √(rg * tan(21.0°)) Now, we can substitute the values of the gravitational acceleration (g) and the angle (21.0°) to calculate v. Note: It is assumed that the bob is moving in a horizontal circle without any additional external forces affecting the system.
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the focal length of a converging lens is 0.50 meters. an object is placed 1.0 meters from the lens. the distance between the lens and the image is
The distance between the lens and the image is 1.0 meter.
To find the distance between the lens and the image formed by a converging lens, we can use the lens formula:
1/f = 1/v - 1/u
Where:
f is the focal length of the lens
v is the distance of the image from the lens (positive if the image is on the same side as the observer, negative if the image is on the opposite side)
u is the distance of the object from the lens (positive if the object is on the same side as the observer, negative if the object is on the opposite side)
In this case:
Focal length (f) = 0.50 meters
Distance of the object (u) = 1.0 meter
Let's substitute the given values into the lens formula:
1/0.50 = 1/v - 1/1.0
2 = 1/v - 1
2v = v - 1
v = 1
Therefore, the distance between the lens and the image = 1.0 m.
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How much work is done when a car is moved 10 m with a force of 3.4 N?
Answer:
34J I assume
Explanation:
force×distance is work done. 10×3.4 is 34. therefore its 34 joules of work done
Answer:
34J
Explanation:
The formula for work is W=Force x Distance
W=FxD
F=3.4N
D=10m
W=10x3.4
W=34 Joules
assume+that+kish+inc.+hired+you+as+a+consultant+to+help+estimate+its+cost+of+capital.+you+have+obtained+the+following+data:+d0+=+$0.90;+p0+=+$47.50;+and+g+=+7.00%
The cost of capital is the rate of return on a firm's investments that must be earned to meet the cost of financing. The cost of capital refers to the opportunity cost of making a specific investment. This opportunity cost is the rate of return that could have been earned by placing the same capital into a different investment that has equivalent risk.
Consultant is a professional who provides expert advice in a specific area such as management, accounting, human resources, and information technology. They provide guidance to an organization to assist them in improving their performance or solving particular problems.
The components of the cost of capital are the cost of debt and the cost of equity. Cost of DebtCost of debt is the interest rate that a firm pays on its debt. It is calculated as follows: Cost of debt = (Interest rate) x (1 - Tax rate)Here, D0 = $0.90, P0 = $47.50, and G = 7.00%.The current dividend is D0.
The next dividend is calculated as follows:D1 = D0 (1 + G) = $0.90 (1 + 0.07) = $0.963Dividend yield can be calculated as follows:Dividend yield = D1 / P0= $0.963 / $47.50= 0.0203 = 2.03%.
The cost of equity can be calculated using the following formula: Cost of Equity = (Dividend Yield) + (Growth Rate of Dividends).
Cost of Equity = 2.03% + 7.00% = 9.03%.
The cost of capital for Kish Inc. is the weighted average of the cost of debt and the cost of equity.
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a light ray can change direction when going from one material into another. that phenomenon is known as __________.
A light ray can change direction when going from one material to another. That phenomenon is known as refraction.
The phenomenon you are referring to is known as refraction. Refraction occurs when a light ray transitions from one medium to another, causing a change in its direction.
This change in direction is a result of the difference in the speed of light between the two media. When light passes through a medium with a different optical density or refractive index, it experiences a change in speed, causing the light ray to bend or deviate from its original path.
Refraction can be observed in various everyday situations. For example, when light travels from air into water or glass, it undergoes refraction.
The bending of light at the interface between these media is responsible for phenomena like the apparent shift in position of objects submerged in water, the bending of a pencil when placed in a glass of water, or the formation of rainbows.
The amount of bending that occurs during refraction depends on the angle at which the light ray enters the interface and the refractive indices of the two media involved.
Snell's law, which describes the relationship between the incident angle, the refracted angle, and the refractive indices, governs the behavior of light during refraction.
Refraction plays a crucial role in various optical devices, including lenses, prisms, and fiber optics. Understanding and controlling the phenomenon of refraction is essential in fields such as optics, physics, and engineering, enabling the development of technologies and applications that rely on manipulating light for imaging, communication, and scientific research.
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It is observed that 55.00 mLmL of water at 20∘C∘C completely fills a container to the brim. When the container and the water are heated to 60∘C∘C, 0.35 gg of water is lost.
What is the coefficient of volume expansion of the container? Density of water at 60 ∘C∘C is 0.98324 g/mLg/mL.
The coefficient of volume expansion of the container is 1.64 x 10⁻⁴ °C⁻¹.
Initial volume of water, V₁ = 55 mL
Initial temperature of the water, T₁ = 20°C
Final temperature of the water, T₂ = 60°C
Density of water at 60°C, d = 0.983 g/mL
Mass of water lost during heating, m = 0.355 g
The change in volume of water is,
ΔV = m/d
ΔV = 0.355/0.983
ΔV = 0.361 mL
Volume expansion occurs when a solid, whether it be in the form of a cube, cuboid, sphere, or another shape, rises in volume as a result of heating.
The expression for the coefficient of volume expansion of the container is given by,
α = ΔV/VΔT
α = 0.361/[55 x (60 - 20)]
α = 0.361/(55 x 40)
α = 0.361/2200
α = 1.64 x 10⁻⁴ °C⁻¹
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An electron and a proton are both released from rest, midway between the plates of a charged parallel-plate capacitor. The only force on each of the two particles is the force from the uniform electric field due to the capacitor. Each particle accelerates until striking one of the plates of the capacitor. (There is no gravity in this problem and we ignore the small force between the electron and the proton.) How do the final kinetic energies and final speeds (just before striking a plate) compare
Answer:
Explanation:
Let the potential difference between the middle point and one of the plate be ΔV .
electric potential energy will be lost and it will be converted into kinetic energy .
Electrical potential energy lost = Vq , where q is charge on charge particle .
For proton
ΔV× q = 1/2 M V² ( kinetic energy of proton )
where M is mass and V be final velocity of proton .
For electron
ΔV× q = 1/2 m v² ( kinetic energy of electron )
where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .
As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same
1/2 M V² = 1/2 m v²
(V / v )² = ( m / M )
(V / v ) = √ ( m / M )
In other words , their velocities are inversely proportional to square root of their masses .