Answer:
(a) [tex]P=33000W[/tex]
(b) [tex]P=51000W[/tex]
Explanation:
The average power is defined as the amount of work done during a time interval:
[tex]P=\frac{W}{t}(1)[/tex]
According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:
[tex]W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)[/tex]
Recall that the weight is given by:
[tex]w=mg\\m=\frac{w}{g}(3)[/tex]
The car accelerates uniformly from rest ([tex]v_0=0[/tex]). Replacing (3) in (2), we have:
[tex]W=\frac{wv_f^2}{2g}[/tex]
(a) Finally, we replace this in (1):
[tex]P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W[/tex]
(b)
[tex]P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W[/tex]
(a) The average power required to accelerate the car of 9000 N is 32798.57 W.
(b) The average power required to accelerate the car of 14,000 N is 51020.40 W.
Given data:
The initial velocity of car is, u = 0 m/s. (Since car was initially at rest)
The final velocity of car is, v = 20 m/s.
The time interval is, t = 5.6 s.
The given problem is based on the concept of average power. The average power is defined as the amount of work done during a time interval. Then,
P = W/t
Here, W is the work done and its value is obtained from the work - energy theorem as,
[tex]W = \Delta KE\\\\W = \dfrac{1}{2}m(v^{2}-u^{2})[/tex]
Here, m is the mass.
(a)
For the weight of 9000 N, the mass of car is,
[tex]w = mg\\\\9000 = m \times 9.8\\\\m =918.36 \;\rm kg[/tex]
So, the Work is obtained as,
[tex]W =\dfrac{1}{2} \times 918.36 \times (20^{2}-0^{2})\\\\W =183672\;\rm J[/tex]
Then, the average power required to accelerate the car is,
P = W/t
P = 183672 / 5.6
P = 32798.57 W
Thus, we can conclude that the average power required to accelerate the car of 9000 N is 32798.57 W.
(b)
For the weight of 14,000 N, the mass of car is,
[tex]w = mg\\\\14,000 = m \times 9.8\\\\m =1428.57 \;\rm kg[/tex]
So, the Work is obtained as,
[tex]W =\dfrac{1}{2} \times 1428.57.36 \times (20^{2}-0^{2})\\\\W =285714.28\;\rm J[/tex]
Then, the average power required to accelerate the car is,
P = W/t
P = 285714.28 / 5.6
P = 51020.40 W
Thus, we can conclude that the average power required to accelerate the car of 14,000 N is 51020.40 W.
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A piano tuner hears a beat every 1.80 s when listening to a 272.0 Hz tuning fork and a single piano string. What are the two possible frequencies (in Hz) of the string? (Give your answers to at least one decimal place.)
Answer:
272.56 or 271.44
Explanation:
[tex]Fbeat=\frac{1}{Tbeat} \\=\frac{1}{1.8} \\=0.56 Hz[/tex]
[tex]Fbeat=abs(f1-f2)\\0.56=abs(272-f2)\\\\f2=272.56\\f2=271.44[/tex]
A racetrack curve has radius 70.0 m and is banked at an angle of 12.0 ∘. The coefficient of static friction between the tires and the roadway is 0.400. A race car with mass 1200 kg rounds the curve with the maximum speed to avoid skidding. consider friction when solving for a, b, and c.
a) As the car rounds the curve, what is the normal force exerted on it by the road?
b) What is the car's radial acceleration?
c) What is the car's speed?
d) In the case of a banked curve with friction, which of the following forces contribute to the centripetal (inward) acceleration: the frictional force, the normal force, and/or the gravity? and why?
Answer:
See attachment below
Explanation:
The normal force exerted on the car by the curved road is 13,139.7 N.
The radial acceleration of the car is 6.56 m/s².
The speed of the car is 21.43 m/s.
In the case of a banked curve with friction, the centripetal acceleration is increased by normal force and frictional force since it prevents the car from skidding.
The given parameters;
radius of the curved path, r = 70 mbanking angle, θ = 12⁰coefficient of friction, μ = 0.4 mass of the race car, m = 1200 kgThe normal force exerted on the car by the curved road is calculated as follows;
[tex]\Sigma F_y = 0\\\\Ncos(\theta) - \mu_s Nsin(\theta) - mg = 0\\\\Ncos(\theta) - \mu_s Nsin(\theta) = mg\\\\N(cos\theta \ - \ \mu_s sin(\theta)) = mg\\\\N = \frac{mg }{cos\theta - \mu_s sin(\theta)} \\\\N = \frac{(1200 \times 9.8)}{cos (12) - \ 0.4\times sin(12)} \\\\N = 13,139.7 \ N[/tex]
The radial acceleration of the car is calculated as follows;
[tex]\Sigma F_x = ma_c\\\\ma_c = Nsin(\theta) + \mu_s N cos(\theta)\\\\a_c = \frac{Nsin(\theta) + \mu_s N cos(\theta)}{m} \\\\a_c = \frac{(13, 139.7)sin(12) \ + \ 0.4 \times 13,139.7cos(12)}{1200} \\\\a_c = 6.56 \ m/s^2[/tex]
The car's speed is calculated as follows;
[tex]a_c = \frac{v^2}{r} \\\\v^2 = a_c r\\\\v = \sqrt{a_c r} \\\\v = \sqrt{6.56 \times 70} \\\\v = 21.43 \ m/s[/tex]
In the case of a banked curve with friction, the centripetal acceleration is increased by the following;
[tex]ma_c = Nsin(\theta) + F_s\\\\ma_c = Nsin(\theta) + \mu_s Ncos (\theta)[/tex]
where;
[tex]F_s[/tex] is the frictional forceN is the normal forceThus, In the case of a banked curve with friction, the centripetal acceleration is increased by normal force and frictional force since it prevents the car from skidding.
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A physics student with a stopwatch drops a rock into a very deep well, and measures the time between when he drops the rocks and when he hears the sound of the rock hitting the water below. If the speed of sound is 343 m/s, and the student measures a time of 6.20 s, how deep is the well?
Answer:
h= 161.06 m
Explanation:
Given that
Speed of the sound ,C= 343 m/s
Total time ,t= 6.2 s
lets take the depth of the well = h
The time taken by stone before striking the water = t₁
we know that
[tex]h=\dfrac{1}{2}gt_1^2[/tex]
[tex]t_1=\sqrt{\dfrac{2h}{g}}[/tex]
The time taken by sound =t₂
[tex]t_2=\dfrac{h}{343}[/tex]
The total time
t = t₁+ t₂
[tex]6.2 = \sqrt{\dfrac{2h}{g}}+\dfrac{h}{343}[/tex]
[tex]6.2 = \sqrt{\dfrac{2h}{9.81}}+\dfrac{h}{343}[/tex]
Now by solving the above equation we get
h= 161.06 m
Therefore the depth of the well will be 161.06 m.
In 1996, NASA performed an experiment called the Tethered Satellite experiment. In this experiment a 2.30 104-m length of wire was let out by the space shuttle Atlantis to generate a motional emf. The shuttle had an orbital speed of 7.50 103 m/s, and the magnitude of the earth's magnetic field at the location of the wire was 5.40 10-5 T. If the wire had moved perpendicular to the earth's magnetic field, what would have been the motional emf generated between the ends of the wire?
emf = 9.3 x 10³
Explanation:
When a conductor moves in the magnetic field , the emf is generated across its ends . Which can be calculated by the relation
emf ξ = B x l x v
here B is the magnetic field strength , l is the length of conductor and v is its velocity .
In our question B = 5.4 x 10⁻⁵ T
l = 2.30 x 10⁴ m and v = 7.5 x 10³
Thus ξ = 5.4 x 10⁻⁵ x 2.30 x 10⁴ x 7.5 x 10³ = 9.3 x 10³ Volt
Technician A says current is the same in each branch of a parallel circuit. Technician B says current is the same everywhere in a series circuit. Who is right?
Answer:
Technician B
Explanation:
The current amp value at the battery position is the current value. The current is the same everywhere for a series circuit without any connecting points. The current at the position of the battery is the same as in each resistor position.
Final answer:
Technician B is correct: current is the same in all parts of a series circuit. Technician A is incorrect: current divides in the branches of a parallel circuit and is not the same. The sum of branch currents equals the total current in parallel circuits.
Explanation:
Understanding Current in Series and Parallel Circuits
Technician B is correct when they say current is the same everywhere in a series circuit. In a series circuit, there is only one path for the current to flow, thus the current (I) is identical through any component in the series, as stated: I = I1 = I2 = I3. This principle is known as the first principle of series circuits. On the other hand, Technician A is incorrect in suggesting current is the same in each branch of a parallel circuit. In parallel circuits, the current divides at each junction and the total supplied current is the sum of the currents in the various branches.
So, in summary, in a series circuit, the current is constant through all components, and the total voltage drop across all components adds up to the voltage supplied by the source. In a parallel circuit, meanwhile, each resistance experiences the same voltage difference, while the current through each branch may vary depending on the resistance of that branch.
What is the excess charge on a conducting sphere of radius r = 0.31 m if the potential of the sphere is 1375 V and V = 0 at infinity?
Given Information:
Radius = r = 0.31 m
Potential difference = V = 1375 Volts
Required Information:
charge = q = ?
Answer:
q = 4.741x10⁻⁸ C
Solution:
[tex]V = kq/r[/tex]
Where k = 8.99x10⁹ N.m²/C² is the Coulomb's constant
Re-arranging the equation to find the q
[tex]q = Vr/k[/tex]
q = (1375*0.31)/8.99x10⁹
q = 4.741x10⁻⁸ C = 47.41 nC
Under what circumstances can energy level transitions occur?
Energy level transitions in physics occur when an atom or molecule absorbs or emits electromagnetic radiation. The circumstances of these transitions depend on the system being studied.
Explanation:Energy level transitions in physics typically occur when an atom or a molecule absorbs or emits electromagnetic radiation. When an electron within an atom moves from a lower energy level to a higher energy level, it absorbs energy. Conversely, when an electron moves from a higher energy level to a lower energy level, it emits energy in the form of electromagnetic radiation.
These transitions can occur under various circumstances, such as when an atom is excited by heat or light, or when an electron interacts with another particle. For example, in the hydrogen atom, transitions between different energy levels give rise to the absorption and emission of specific wavelengths of light, leading to the existence of discrete spectral lines. It is important to note that the exact circumstances of energy level transitions depend on the specific system being studied, such as atoms, molecules, or subatomic particles.
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Energy level transitions in atoms occur when an electron absorbs or emits energy, typically through photon interactions, due to electronic rearrangements.
Energy level transitions in atoms happen when electrons change their energy states. These transitions can occur under several circumstances:
Absorption of Photons: Electrons can move to higher energy levels by absorbing photons with specific energy levels corresponding to the energy difference between the levels. This is observed in absorption spectra.
Emission of Photons: Electrons release energy in the form of photons when transitioning from higher to lower energy levels. These emitted photons produce emission spectra.
Electron Collisions: In certain situations, such as in a plasma, collisions between electrons and other particles can excite electrons to higher energy states, leading to subsequent de-excitation and photon emission.
External Energy Sources: External sources like electrical discharges or high-temperature environments can energize electrons, causing transitions between energy levels.
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After skiding down a snow-covered hill on an inner tube, Ashley is coasting across a level snowfield at a constant velocity of 2.7 m/s. Miranda runs after her at a velocity of 4.1 m/s and hops on the inner tube. How fast do the two of them slide across the snow together on the inner tube?
Answer:
3.3 m/s
Explanation:
The question is incomplete, here is the complete question:
After skiding down a snow-covered hill on an inner tube, Ashley is coasting across a level snowfield at a constant velocity of 2.7 m/s. Miranda runs after her at a velocity of 4.1 m/s and hops on the inner tube. How fast do the two of them slide across the snow together on the inner tube? Ashley's mass is 71 kg and Miranda's is 58 kg. Ignore the mass of the inner tube and any friction between the inner tube and the snow.
SOLUTION:
mass of Ashley (Ma) = 71 kg
mass of Miranda (Mm) = 58 kg
initial velocity of Ashley (Va) = 2.7 m/s
initial velocity of Miranda (Vm) = 4.1 m/s
Find the final velocity (Vf) at which they both slide together
from the conservation of momentum, initial momentum = final momentum
Ma.Va + Mm.Vm = (Ma + Mm) . Vf
[tex]Vf = \frac{Ma.Va + Mm.Vm}{Ma + Mm} \\ Vf=\frac{(71 x 2.7) + (58 x 4.1)}{71+58}[/tex]
Vf = (191.7 +237.8) / (129)
Vf = 3.3 m/s
A boy is sledding down a hill and has an initial speed of +12 m/s. He continues to speed up and reaches a final velocity of +18m/s after traveling for 12 seconds what distance does the boy travel?
Answer:
180m
Explanation:
V = Final velocity = 18
U = Initial Velocity = 12
a = Acceleration = (18-12)/12 = 6/12 = 0.5
d = ?
V² = U² + 2ad
18² = 12² + 2*0.5*d
324 = 144 +d
d = 324 - 144
d = 180m
Final answer:
The boy travels a distance of 180 meters. This calculation is based on the kinematic equations for uniformly accelerated motion using the given initial and final velocities and the time taken.
Explanation:
The subject of the question is Physics, and it appears to be at a High School level. The problem deals with kinematics, specifically the equations of motion for an object with constant acceleration. To find the distance traveled by the boy sledding down the hill, we can use the kinematic equation: s = ut + ½ at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. The initial speed u is given as +12 m/s, the final velocity v is +18 m/s, and the time t is 12 seconds. We first find the acceleration using the formula a = (v - u) / t. Substituting the known values, we get a = (18 m/s - 12 m/s) / 12 s = 0.5 m/s². Now we can calculate the distance s using the kinematic equation: s = (12 m/s)(12 s) + ½(0.5 m/s²)(12 s)² = 144 m + 36 m = 180 m. Therefore, the boy travels a distance of 180 meters.
A car went 60 km in 5/6 of an hour while a second car went 54 km in 2/3h. Which car was faster? How many times faster?
Answer:
The car that went 54 km in 2/3h was faster, 5/4 times faster than the other car
Explanation:
Average speed of a car is the ratio between the displacement [tex]\Delta x [/tex] and the time (t) it takes to do that displacement:
[tex]V_{avg}=\frac{\Delta x}{t} [/tex]
So, for the first car:
[tex]V_{avg1}=\frac{60km}{\frac{5h}{6}} =72\frac{km}{h} [/tex] (1)
for the second car we have:
[tex]V_{avg2}=\frac{60km}{\frac{2h}{3}} =90\frac{km}{h} [/tex] (2)
So, the second cart is faster than the first one. To find how many times divide speed 2 by speed 1:
[tex]\frac{90}{72}=\frac{5}{4} [/tex]
If you don't answer this in the next 5 seconds, tomorrow morning you will step on a LEGO trying to grab the milk from the fridge.
Answer:
d = 78 [km]
Explanation:
The displacement shift was from Tipton to Iowa city, so we have:
d1 = 27[km]
Then he travels back for 15[km]
d2 = 15[km]
Finally he returns back to Iowa city the displacement 36 [km]
d3 = 36[km]
d = d1 + d2 + d3
d = 27 + 15 + 36
d = 78 [km]
Which materials should Hannah choose for her pot and spoon?
A. Hannah should choose copper for her pot because it has the lowest specific heat, and
she should choose wood for her spoon because it has the highest specific heat.
B. Hannah should choose wood for her pot because it has the highest specific heat, and
she should choose copper for her spoon because it has the lowest specific heat.
C. Hannah should choose wood for her pot because it has the highest specific heat, and
she should choose rubber for her spoon because it has the second highest specific
heat.
D. Hannah should choose copper for her pot because it has the lowest specific heat, and
she should choose steel because it has the second lowest specific heat.
A. Hannah should choose copper for her pot because it has the lowest specific heat, and she should choose wood for her spoon because it has the highest specific heat
Explanation:
The specific heat capacity of a substance is the amount of heat required to increase the temperature of 1 kg of the substance by 1 degree.
Mathematically:
[tex]C=\frac{Q}{m\Delta T}[/tex]
where
Q is the amount of heat supplied to the substance
m is the mass of the subtsance
[tex]\Delta T[/tex] is the increase in temperature
This means that:
A substance with higher specific heat needs a lot of energy to raise its temperatureA substance with higher specific heat needs a small amount of energy to increase its temperatureIn this situation, we want:
A material with low specific heat for the pot, because the pot must be able to transmit heat efficiently, so it must be able to become hot (=increase its temperature) with a small amount of heatA material with high specific heat for the spoon, so that the spoon does not become too hot too fast (because it takes a lot of energy to increase its temperature)Therefore, the most reasonable choice in this situation is:
A. Hannah should choose copper for her pot because it has the lowest specific heat, and she should choose wood for her spoon because it has the highest specific heat
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to prevent a manual-type torch from damage, it is made of a special plastic that resists what?
Answer:
heat
Explanation:
If the voltage of the secondary side of the transformer is less than ? and the voltage of the primary side of the transformer exceeds ? to ground, the secondary side shall be grounded.?
Answer:
50 V / 150 V
Explanation:
Final answer:
The question discusses the grounding of a transformer based on voltage levels and defines the characteristics of step-down and step-up transformers in terms of voltage and current. Transforming the power efficiently is crucial, and the number of windings plays a significant role in determining the output voltage and current.
Explanation:
The question pertains to transformers and their operating principles regarding voltage, current, and the number of windings. If the voltage at the secondary side of a transformer is less than 50 volts and the primary side exceeds 150 volts to ground, the secondary side shall be grounded.
When a transformer has fewer windings in the secondary coil compared to the primary coil, it is a step-down transformer, meaning the secondary voltage is less than the primary voltage. Conversely, a step-up transformer has more windings in the secondary coil, resulting in a higher secondary voltage compared to the primary voltage.
In the context of primary and secondary properties of transformers, option (a) the Primary voltage is higher than secondary voltage defines a step-down transformer. Additionally, (c) the Primary current is higher than secondary current is not typically true for a step-down transformer.
Instead, the primary current would be lower than the secondary current since power (P = IV) is conserved across the transformer. Therefore, if the voltage decreases, the current must increase to maintain the equivalent power (assuming ideal conditions with no power losses).
A vertical steel beam in a building supports a load of 6.0×10⁴. If the length of the beam is 4.0m and it's cross-sectional area is 8.0×10^-3m². Find the diameter of the beam which is compressed along its length
Answer:
DL = 1.5*10^-4[m]
Explanation:
First we will determine the initial values of the problem, in this way we have:
F = 60000[N]
L = 4 [m]
A = 0.008 [m^2]
DL = distance of the beam compressed along its length [m]
With the following equation we can find DL
[tex]\frac{F}{A} = Y*\frac{DL}{L} \\where:\\Y = young's modulus = 2*10^{11} [Pa]\\DL=\frac{F*L}{Y*A} \\DL=\frac{60000*4}{2*10^{11} *0.008} \\DL= 1.5*10^{-4} [m][/tex]
Note: The question should be related with the distance, not with the diameter, since the diameter can be found very easily using the equation for a circular area.
[tex]A=\frac{\pi}{4} *D^{2} \\D = \sqrt{\frac{A*4}{\pi} } \\D = \sqrt{\frac{0.008*4}{\\pi } \\\\D = 0.1[m][/tex]
Fiora starts riding her bike at 18 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the trip. The whole trip of 69 miles takes her 4.5 hours. For how long did she travel at 18 mi/h?
Answer:
t = 2.5 hours
Explanation:
given,
speed of the bike for t time= 18 mi/h
final speed of the bike after t time = 12 mi/h
total distance, D = 69 miles
total time, T= 4.5 hour
time for which speed of the bike is 18 mi/h = ?
we know distance = speed x time
now,
18 x t + 12 (4.5 - t) = 69
6 t + 54 = 69
6 t = 15
t = 2.5 hours
The bike was at the speed of 18 mi/h for 2.5 hours.
Which motion maps show an object in uniform circular motion? Check all that apply. V W X Y
The correct options are V, W, and Y
Uniform Circular Motion is the motion when an object moves in a circular path with constant speed (uniform speed). In such a motion,
1- The position keeps on changing,
2- The speed remains constant
Now checking all the option
V- It shows the uniform circular motion since all the direction is uniform,
W-It shows the uniform circular motion since all the direction is uniform,
X-Direction are different,
Y-It shows the uniform circular motion since all the direction is uniform,
Hence, all the above options,
The correct options are V, W, and Y
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A 56.0kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.50 turns each second. The distance from one hand to the other is 1.5m. Biometric measurements indicate that each hand typically makes up about 1.25 % of body weight.
Part A) horizontal force must her wrist exert on her hand F=150N
Express the force in part (a) as a multiple of the weight of her hand?
Answer:
Net force = 129.4, Force as multiple of weight of her hand = 18.84
Explanation:
Given Data:
Total body weight = 56.0 kg ;
no. of turns = 2.5/second ;
hand to hand distance = 1.5m ;
weight of hand = 1.25% of body weight ;
Solution:
mass of hand = [tex]\frac{1.25}{100}[/tex]*56 = 0.7kg ;
radius = d/2 = 1.5/2 = 0.75m ;
Now we need to find velocity, as we know that velocity can be calculated by dividing distance by time
v = d/t = [tex]\frac{2.5*2*3.14*0.75}{1}[/tex] = 11.775 m/s or 12 m/s;
a.
The formula to calculate force is given as
F = mv²/r = (0.7*11.775²)/0.75 = 129.4 N
b.
To calculate force as multiple of weight on her hand, we need to calculate the gravitational force W on her hand first.
W = gm = 9.81 * 0.7 = 6.867 N
Now the wieght on her hand can be represented by
[tex]\frac{Force_{net} }{weight of hand}[/tex] = 129.4 / 6.867 = 18.84
The force that the ice skater exerts on her hand (150N) is roughly 22 times the weight of her hand (6.86N). She achieves this by spinning with her arms outstretched.
Explanation:The skater's hand, as given by biometric measurements, makes up about 1.25% of her body weight. Since her body weight is 56.0kg, her hand weight would be 0.0125 * 56.0kg = 0.7kg. In terms of newtons (N), when we consider the acceleration due to gravity as approximately 9.8 m/s², the weight of her hand becomes 0.7kg * 9.8 m/s² = 6.86N. Thus, if we're asked to express the force that her wrist exerts on her hand (150N) as a multiple of the weight of her hand, we'll divide the two forces. Namely, F_hand/F_weight = 150N/6.86N which is approximately 21.86. Therefore, the force her wrist exerts on her hand is roughly 22 times the weight of her hand.
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the phrase positive to positive negative to ground is correct when jump starting a car true or false
Answer: TRUE
Explanation:Jump starting a car is a process of starting a car whose battery is Dead,which means it doesn't have enough battery life to start a Car.
The process involves bringing both cars close to each other and Connecting the positive side of a good battery to the positive side of the Dead battery using a JUMPER CABLES,
Connect the positive cable to the positive side of the working battery,
Connect the Negative cable to the Negative side of the working battery,
Finally,connect the Negative cable to a grounded meta part of the car we want to JUMP START.
The phrase "positive to positive, negative to ground" is true when jump starting a car.
Explanation:The phrase "positive to positive, negative to ground" is true when jump starting a car.
When jump starting a car, it is important to connect the positive terminals of both the donor and recipient vehicles using jumper cables. Then, the negative terminal of the recipient vehicle should be connected to a grounded metal surface, such as the engine block or a bolt on the frame.
By following this phrase, you ensure that the electrical current flows in the correct direction, providing a safe and effective way to start a car with a dead battery.
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Newton's Law of Cooling says that the rate of cooling of an object is proportional to the difference between its own temperature and the temperature of its surrounding. Write a differential equation that expresses Newton's Law of Cooling for this particular situation.
Answer:
[tex]\frac{dQ}{dt} =-hA\Delta T(t)[/tex]
Explanation:Newton.s law of cooling states that the rate of cooling of an object is proportional to the difference between its own temperatures and temperature of its surroundings. Mathematically,
[tex]\frac{dQ}{dt} =-hA [T(t)-T(s)]\\[/tex]
[tex]\frac{dQ}{dt} =-hA\Delta T(t)[/tex]
where [tex]Q[/tex] is the heat transfer
[tex]h[/tex] is heat transfer coefficient
[tex]A[/tex] is the heat transfer surface area
[tex]T[/tex] is the temperature of the object's surface
[tex]T(s)[/tex] is the temperature of the surroundings
Final answer:
Newton's Law of Cooling can be expressed using a differential equation: dQ/dt = -k(T - Ts).
Explanation:
Newtons's Law of Cooling states that the rate of cooling of an object is proportional to the difference between its temperature and the temperature of its surroundings. In this particular situation, we can express this law using a differential equation:
dQ/dt = -k(T - Ts)
Where:
dQ/dt represents the rate of heat transfer or coolingk is the proportionality constantT is the temperature of the objectTs is the temperature of the surroundingsA carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3.66 m/s and a centripetal acceleration :a of magnitude 1.83 m/s2. Position vector :r locates him relative to the rotation axis. (a) What is the magnitude of :r? What is the direction of :r when :a is di- rected (b) due east and (c) due south?
The magnitude of the position vector :r is approximately 7.32 m. When the centripetal acceleration is directed due east, :r will also be due east. When the centripetal acceleration is directed due south, :r will also be due south.
Explanation:The magnitude of the position vector :r can be determined using the equation :a = :v^2 / r where :v is the speed of the man and :a is the centripetal acceleration. Rearranging the equation, we get :r = :v^2 / :a. Substituting the given values, we find :r = (3.66^2) / 1.83 which is approximately 7.32 m.
When the centripetal acceleration is directed due east, the position vector :r will be due east. When the centripetal acceleration is directed due south, the position vector :r will be due south.
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The average speed of a nitrogen molecule in air is proportional to the square root of the temperature in kelvins (K). If the average speed is 475 m/s on a warm summer day (temperature=300.0 K), what is the average speed on a frigid winter day (250.0 K)?
Answer:
v₂ = 395.83 m/s
Explanation:
given,
Average speed of nitrogen molecule, v₁ = 475 m/s
Temperature in summer, T₁ = 300 K
Average speed of nitrogen molecule in winter, v₂ = ?
Temperature in winter, T₂ = 250 K
The relation of average speed with temperature
[tex]v\ \alpha\ \sqrt{T}[/tex]
now,
[tex]\dfrac{v_2}{v_1} = \dfrac{\sqrt{T_2}}{\sqrt{T_1}}[/tex]
[tex]\dfrac{v_2}{475} = \dfrac{\sqrt{250}}{\sqrt{300}}[/tex]
v₂ = 0.833 x 475
v₂ = 395.83 m/s
The average speed on a frigid winter day is equal to v₂ = 395.83 m/s
What is the magnitude of energy of a photon of light (in J) that is emitted when an excited electron in a hydrogen atom falls from n = 4 down 1 energy levels?
Answer:
[tex]2.044 * 10^{-18}[/tex] J
Explanation:
Parameters given:
[tex]n_{1} = 4\\\\n_{2} = 1[/tex]
We know that the electron fell from level 4 to level 1. We can use the Rydberg's equation to find the [tex]2.044 * 10^{-18}[/tex] that is emitted by the electron in the process. Rydberg's equation is given as:
1/λ = [tex]R * (\frac{1}{(n_{2})^2} - \frac{1}{(n_{1})^2})[/tex]
where R = Rydberg's constant = [tex]1.0973731568508 * 10^{7}[/tex]
1/λ = [tex]1.0973731568508 * 10^{7} (\frac{1}{1^{2}} - \frac{1}{4^{2}} } )[/tex]
1/λ = [tex]1.0973731568508 * 10^{7} (\frac{1}{1} - \frac{1}{16} } )[/tex]
1/λ = [tex]1.0973731568508 * 10^{7} * \frac{15}{16}[/tex]
1/λ = [tex]1.029 * 10^{7}[/tex] [tex]m^{-1}[/tex]
∴ λ = [tex]9.72 * 10^{-8}[/tex] m
Energy of the photon is given by:
E = (hc)/λ
where
h = Planck's constant = [tex]6.62607004 * 10^{-34} m^2 kg / s[/tex]
c = speed of light = 299792458 m / s
∴ E = [tex]\frac{6.62607004 * 10^{-34} * 299792458}{9.72 * 10^{-8}}[/tex]
E = [tex]2.044 * 10^{-18}[/tex] J
The energy of the photon is [tex]2.044 * 10^{-18}[/tex] J
Final answer:
The energy of a photon emitted when an electron in a hydrogen atom falls from n=4 to n=3 can be found using the Rydberg formula. The constant R (Rydberg constant) along with the energy levels n1 and n2 are used in the formula to calculate this energy in joules.
Explanation:
The magnitude of energy of a photon of light that is emitted when an excited electron in a hydrogen atom falls from n = 4 to n = 3 can be calculated using the Rydberg formula for the energy difference between two hydrogen energy levels. In this case, since the question only specifies a drop of one energy level, we'll correct the typo where it currently indicates a fall from n=4 to n=1, and we'll proceed with the fall from n=4 to n=3. The formula to use is:
E = R × (1/n1² - 1/n2²)
where:
R is the Rydberg constant (2.179 × 10⁻¹⁸ J),
n1 and n2 are the principal quantum numbers (integers) for the initial and final energy levels, respectively.
The calculation is then:
E = 2.179 × 10⁻¹⁸ J × (1/3² - 1/4²) = 2.179 × 10⁻¹⁸ J × (1/9 - 1/16) = 2.179 × 10⁻¹⁸ J × (7/144)
After calculating the above expression, you will get the energy in joules of the photon emitted due to the electron's transition.
6. A car, 1110 kg, is traveling down a horizontal road at 20.0 m/s when it locks up its brakes. The coefficient of friction between the tires and road is 0.901. How much distance will it take to bring the car to a stop?
To calculate the distance that it will take to bring the car to a stop, we can use the equations of motion. The force of kinetic friction can be calculated by multiplying the coefficient of friction by the normal force. Using the equations v = u + at and s = ut + 0.5at^2, we can find the distance.
Explanation:To calculate the distance that it will take to bring the car to a stop, we can use the equations of motion. The force of kinetic friction can be calculated by multiplying the coefficient of friction by the normal force. The normal force is equal to the weight of the car, which is given by the mass of the car multiplied by the acceleration due to gravity. The force of kinetic friction is equal to the mass of the car multiplied by the acceleration. Rearranging the equation F = ma to solve for acceleration gives us a = F/m. Substituting the force of kinetic friction for F and the mass of the car for m, we can find the acceleration. Using the equation v = u + at, where v is the final velocity (0 m/s), u is the initial velocity (20.0 m/s), a is the acceleration, and t is the time, we can solve for t. Finally, we can use the equation s = ut + 0.5at^2 to calculate the distance s that it will take to bring the car to a stop. Plugging in the values for u, t, and a will give us the answer.
What is the angular diameter (in arcseconds) of an object that has a linear diameter of 75 cm and a distance of 2 km?
Answer:
The angular diameter is 77.35 arc-seconds.
Explanation:
The angular diameter, as shown in the figure, is the angle [tex]x[/tex] subtended by the the diameter of the object.
Before we do the calculation, we first convert everything to meters.
The diameter of the of the object in meters is
75cm =0.75m,
and the distance to the object in meters is
2 km = 2000 m.
Now, from trigonometry we get:
[tex]tan (\frac{x}{2} )= \dfrac{radius}{length} = \dfrac{0.75/2}{2000} \\\\\dfrac{x}{2} = tan^{-1}(\dfrac{0.75/2}{2000})\\\\\dfrac{x}{2}= 0.0107\\\\\boxed{x= 0.0215^o}[/tex]
and since 1 degree = 3600 arc-seconds, [tex]x[/tex] in arc-seconds is
[tex]x= 0.0215*3600 \\\\\boxed{x= 77.35''}[/tex]
A protester carries his sign of protest, starting from the origin of an xyz coordinate system, with the xy plane horizontal. He moves 40 m in the negative direction of the x axis, then 20 m along a perpendicular path to his left, and then 25 m up a water tower.
(a) In unit-vector notation, what is the displacement of the sign from start to end?
(b) The sign then falls to the foot of the tower. What is the magnitude of the displacement of the sign from start to this new end?
Answer:
Part A:
[tex]Displacement\ vector= -40\hat i+20\hat j+25\hat k[/tex]
Part B:
Magnitude of displacement=44.721359 m
Explanation:
Given Data:
40 m in negative direction of the x axis
20 m perpendicular path to his left (considering +ve y direction)
25 m up a tower (Considering +ve z direction)
Required:
In unit-vector notation, what is the displacement of the sign from start to end.The magnitude of the displacement of the sign from start to this new end=?Solution:
Part A:
[tex]Displacement= (0-40)\hat i+(0+20)\hat j+(0+25)\hat k\\Displacement= -40\hat i+20\hat j+25\hat k[/tex]
where:
i,j,k are unit vectors
Part B:
Sign falls to foot of tower so z=0
[tex]Displacement= -40\hat i+20\hat j+0\hat k[/tex]
Magnitude of displacement:
[tex]Magnitude\ of\ displacement=\sqrt{(-40)^2+(20)^2+0^2} \\Magnitude\ of\ displacement=44.721359\ m[/tex]
A 5.67 gram coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of the static friction is 0.100, how far from the center of the record can the coin be placed without having it slip off?
Answer:
81 mm from the center
Explanation:
5.67g = 0.00567 kg
33.3 revolutions per minute = 33.3 (rev/min) * 2π (rad/rev) * (1/60) (min/sec) = 3.487 rad/s
The weight of the coin is product of mass and gravitational acceleration g = 9.81m/s2
W = mg = 0.00567 * 9.81 = 0.0556 N
Which is also the normal force of the record acting back on the coin to balance it.
Them the static friction is product of normal force and friction coefficient
[tex]F_f = N\mu = 0.0556*0.1 = 0.00556 N[/tex]
For the coin to NOT slip off, its centripetal force should at most be equal to the static friction
[tex]F_c = 0.00556[/tex]
[tex]a_cm = 0.00556[/tex] Newton's 2nd law
[tex]a_c = 0.00556 / 0.00567 = 0.981 m/s^2[/tex]
The centripetal acceleration is the product of squared angular velocity and radius of circular motion
[tex]a_c = \omega^2r[/tex]
[tex]r = \frac{a_c}{\omega^2} = \frac{0.981}{3.487^2} = 0.081m[/tex] or 81 mm
Which statement correctly compares sound and light waves?
A) Both light and sound waves need matter to carry energy from one place to another.
B) Neither light nor sound waves need matter to carry energy from one place to another.
C) Light waves carry energy paral
Answer:
D) Sound waves carry energy parallel to the motion of the wave, while light waves carry energy perpendicular to it.
Explanation:
A) This is incorrect because Light waves do not need a medium to pass through it while Sound waves need a medium to pass through it.
B) This is incorrect as explained above.
C) This is incorrect because Light waves do not carry energy parallel to the motion of the wave.
Light waves are transverse waves, as so they carry energy perpendicular to the motion of the wave while Sound waves are longitudinal waves and so, they carry energy parallel to the motion of the wave.
D) This is correct because Sound waves carry energy parallel to the motion of the wave.
Sound waves are transverse waves, as so they carry energy parallel to the motion of the wave while Light waves are longitudinal, as so they carry energy perpendicular to the motion of the wave.
Answer:
D-Sound waves carry energy parallel to the motion of the wave, while light waves carry energy perpendicular to it.
Explanation:
A rifle with a weight of 25 N fires a 4.0 g bullet with a speed of 290 m/s.
(a) Find the recoil speed of the rifle.
Your response is off by a multiple of ten. m/s
(b) If a 750 N man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle.
m/s
Answer:
Explanation:
Given
Weight of rifle [tex]W=25\ N[/tex]
mass of rifle [tex]M=\frac{25}{10}=2.5\ kg[/tex]
mass of bullet [tex]m=4\ gm[/tex]
speed of bullet [tex]u=290\ m/s[/tex]
As there is no net force on the bullet-rifle system therefore momentum is conserved
[tex]0=Mv+mu[/tex]
[tex]v=-\frac{mu}{M}[/tex]
[tex]v=-\frac{4\times 10^{-3}\times 290}{2.5}[/tex]
[tex]v=-0.464\ m/s[/tex]
i.e. opposite to the direction of bullet speed
(b)Weight of Man [tex]=750\ N[/tex]
Combined weight of man and rifle[tex]=750+25=775\ N[/tex]
mass of man-rifle system [tex]M'=\frac{775}{10}=77.5\ kg[/tex]
Now man and rifle combinedly recoil
therefore
[tex]0=(M')v '+mu [/tex]
[tex]v'=-\frac{mu}{M'}[/tex]
[tex]v'=-\frac{4\times 10^{-3}\times 290}{77.5}[/tex]
[tex]v'=0.0149\ m/s[/tex]
Halley's comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0.59 A.U. and its greatest distance being 35 A.U. (A.U. is the Earth-Sun distance). If the comet's speed at closest approach is 47 km/s, what is its speed when it is farthest from the Sun?
Answer:
Explanation:
Given
Halley's closest distance from sun is [tex]r_1=0.59\ A.U.[/tex]
Greatest distance is [tex]r_2=35\ A.U.[/tex]
Comet's speed at closest approach is [tex]v_1=47\ km/s[/tex]
As there is no external torque so angular momentum of comet about the sun is conserved
[tex]L_1=L_2[/tex]
[tex]mr_1^2\times \omega _1=mr_2^2\times \omega _2[/tex]
where [tex]\omega =angular\ velocity [/tex]
This can be written as [tex]\omega =\frac{v}{r}[/tex]
Therefore
[tex]mr_1^2\times \frac{v_1}{r_1}=mr_2^2\times \frac{v_2}{r_2}[/tex]
[tex]r_1\times v_1=r_2\times v_2[/tex]
[tex]0.59\times 47=35\times v_2[/tex]
[tex]v_2=0.79\ km/s[/tex]