Answer: perpendicular to it oscillations.
Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.
By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.
Examples of transverse waves includes wave in a string, water wave and light.
Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.
If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.
Since the oscillations is perpendicular to the direction of wave, it is a transverse wave
In a transverse wave, the vibrations move in a direction perpendicular to the direction of wave travel. This distinguishes them from other types of waves. An example is the movement of water waves.
Explanation:In a transverse wave, the vibrations or disturbances move in a direction perpendicular (at right angles) to the direction of the wave's travel. This is the key characteristic that separates transverse waves from other types of waves. A common example of transverse waves would be waves in a body of water where the water particles move up and down causing the wave to move forward.
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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).
Indicate whether each of the following statements is true or false. 1. The molecules in hot air move at the same speed as in cold air, but there are more molecules so they feel hotter. 2. In a coal-fired power plant, the types of energy from start-to-finish are: electrical, mechanical, thermal, and chemical. 3. In a coal-fired power plant, the approximate percentage of original energy in the coal lost to heat is 10%.
Answer:
1) False
2) False (chemical, thermal, mechanical and electrical)
3) False
Explanation:
1.
We have the expression for the root mean square velocity of the molecules of gases as:
[tex]v_{rms}=\sqrt{\frac{3.k_b.T}{M} }[/tex]
where:
[tex]k_b=[/tex] Boltzmann constant
[tex]T=[/tex] temperature of the gas
[tex]M=[/tex] molecular mass of the gas
2.
In a coal-fired powered the types of energy form start to finish can be given as;
At first the chemical energy of the coal gets converted into heat energy after the process of combustion.Then next, this heat is utilized for generating high pressure steam which drives the turbine converting the heat energy into the mechanical energy.Then this rotational motion of turbine shaft is coupled with the armature of the alternator to convert the mechanical energy into electrical energy.3.
In a coal fired power plant the approximate percentage of original energy in the coal lost to heat is approximately 60%.
Blood is accelerated from rest to 25.00 cm/s in a distance of 2.10 cm by the left ventricle of the heart. How long does the acceleration take? (To solve this problem, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units.)
Answer:
Time taken, t = 0.16 seconds
Explanation:
Given that,
Initial speed of blood, u = 0
Final speed of blood, v = 25 cm/s = 0.25 m/s
Distance, d = 2.1 cm = 0.021 m
Using first equation of motion as :
v = u + at
u = 0
[tex]v=at\\0.25 =at[/tex]
Let t is the time taken. Using second equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
[tex]d=\dfrac{1}{2}at\times t[/tex]
[tex]t=\dfrac{2d}{at}[/tex]
Since, at = 0.25
So,
[tex]t=\dfrac{2\times 0.021}{0.25}[/tex]
t = 0.16 seconds
So, the time taken by the blood to accelerate is 0.16 seconds.
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
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 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]
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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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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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.
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]
If an aircraft is loaded 90 pounds over maximum certificated gross weight and fuel (gasoline) is drained to bring the aircraft weight within limits, how much fuel should be drained?
Answer:
15 gallons.
Gallons = (Pounds) ÷ (Pounds ÷ Gallons)
Explanation:
To bring an aircraft's weight under the maximum certificated gross weight by 90 pounds, 15 gallons of gasoline should be drained.
If an aircraft is loaded with 90 pounds over the maximum certificated gross weight and needs to be brought within the weight limit by draining fuel, we need to calculate how much fuel, in pounds, should be drained. Gasoline's weight can vary slightly with temperature, but for general aviation purposes, it's usually taken to be about 6 pounds per gallon.
Hence, to remove 90 pounds of weight, the amount of fuel that should be drained would be 90 divided by 6.
Performing the calculation, 15 gallons of fuel should be drained to remove 90 pounds of weight from the aircraft and bring it back to or below the maximum certificated gross weight.
A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 37.6 vibrations in 27.9 s. Also, a given maximum travels 450 cm along the rope in 11.3 s. What is the wavelength? Answer in units of cm.
Answer:
[tex]\lambda = 25.79\ cm[/tex]
Explanation:
given,
Wave vibrates = 37.6
time = 27.9 s
maximum distance travel = 450 cm
time = 11.3 s
wavelength = ?
frequency of wave
[tex]f=\dfrac{37.6}{27.9}[/tex]
f = 1.35 Hz
Speed of wave
[tex]v = \dfrac{450}{11.3}[/tex]
v = 39.82 cm/s
wavelength of wave
v = fλ
[tex]\lambda =\dfrac{v}{f}[/tex]
[tex]\lambda =\dfrac{34.82}{1.35}[/tex]
[tex]\lambda = 25.79\ cm[/tex]
Hence, wavelength of the wave is equal to 25.79 cm.
Final answer:
The wavelength of a wave can be calculated using the speed of the wave and its frequency. In this case, the wavelength is 29.50 cm.
Explanation:
Wavelength: The wavelength of a wave is the distance between two similar points on the medium that have the same height and slope. In this case, the wavelength can be found using the formula: λ = v/f, where v is the speed of the wave and f is the frequency.
Given Data: Number of vibrations = 37.6, Time = 27.9 s, Distance traveled = 450 cm, Time = 11.3 s.
Calculation: Speed of the wave = Distance/Time = 450 cm / 11.3 s = 39.82 cm/s. Frequency = Number of vibrations / Time = 37.6 / 27.9 = 1.35 Hz. Therefore, Wavelength = 39.82 cm / 1.35 Hz = 29.50 cm.
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
The wavelength of green light from a traffic signal is centered at 5.20*10^-5cm. Calculate the frequency.
Answer:
5.8×10^11Hz
Explanation:
Frequency is the ratio of the speed of the light wave to its wavelength.
Frequency (f)= velocity(v)/Wavelength (¶)
Given wavelength = 5.2×10^-5cm = 5.2×10^-3m (converted to meters)
Velocity of light = 3×10^8
Substituting the values given in the formula we have
Frequency = 3×10^8/ 5.2×10^-3
Frequency = 5.8×10^11Hz
The frequency of the green light from the traffic signal is 5.77 × 10¹⁴ Hertz.
Given the data in the question;
Wavelength; [tex]\lambda = 5.20*10^{-5}cm = 5.20*10^{-7}m[/tex]
Frequency; [tex]f = \ ?[/tex]
To determine the frequency of the green light, we use the expression for the relations between wavelength, frequency and speed of light.
[tex]\lambda = \frac{c}{f} \\[/tex]
Where [tex]\lambda[/tex] is wavelength, f is frequency and c is speed constant ([tex]3* 10^8m/s[/tex])
We substitute the values into the equation
[tex]5.20*10^{-7}m = \frac{3*10^8m/s^}{f} \\\\f = \frac{3*10^8m/s^}{5.20*10^{-7}m}\\\\f = 5.77 * 10^{14}s^{-1}\\\\f = 5.77 * 10^{14}Hz[/tex]
Therefore, the frequency of the green light from the traffic signal is 5.77 × 10¹⁴ Hertz.
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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 75-kilogram hockey player is skating across the ice at a speed of 6.0 meters per second. What is the magnitude of the average force required to stop the player in 0.65 second?
(1) 120 N
(2) 290 N
(3) 690 N
(4) 920 N
Answer:
(3) 690 N
Explanation:
Force: This is the product of mass and acceleration of a body. The S.I unit of force is Newton(N).
The formula for force is given as,
F = ma..................... Equation 1
Where F = force, m = mass, a = acceleration.
Also,
a = (v-u)/t................... Equation 2
Where v = Final velocity, u = initial velocity, t = time.
Given: u = 6.0 m/s, v = 0 m/s (bring to stop), t = 0.65 s.
Substitute into equation 2
a = (0-6)/0.65
a = -6/0.65
a = -9.23 m/s²
Also given: m = 75 kg
Substitute again into equation 1
F = 75(-9.23)
F = -692.25 N
The negative sign tells that the force oppose the motion of the player
F ≈ 690 N
Hence the right option is (3) 690 N
The magnitude of the average force required to stop the player in 0.65 second is 690 N. And option (3) is correct.
Given data:
The mass of hockey player is, m = 75 kg.
The speed of hockey player is, v = 6.0 m/s.
The reaction time is, t = 0.65.
Use the concept of impulse - momentum theorem to obtain the value of average force acting on the hockey player as,
As per the impulse - momentum theorem,
[tex]F = \dfrac{mv}{t}[/tex]
Here, F is the average force exerted on the hockey player.
Solving as,
[tex]F = \dfrac{75 \times 6}{0.65 }\\\\F \approx 690 \;\rm N[/tex]
Thus, the magnitude of the average force required to stop the player in 0.65 second is 690 N. And option (3) is correct.
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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 surroundingsWhat 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.
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]
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.
A(n) _____ is a procedure for using ultrasonic sound waves to create a picture of an embryo or fetus.
Answer:
Sonogram
Explanation:
A sonogram is a medical diagnostic method that is commonly uses sound waves in order to generate images of any tissues, organs, or some other internal structures inside the human body. This sonogram does not use any kind of harmful radiations like the X-rays. These are sound waves of high frequencies.
This is widely used by doctors for health checkups. The other name for sonogram is ultrasound.
Two charges that are separated by one meter exert 1-N forces on each other. If the charges are pushed together so the separation is 25 centimeters, the force on each charge will be______
Final answer:
If the charges are pushed together so the separation is 25 centimeters, the force on each charge will be 16 N.
Explanation:
In electrostatics, the force between two charged particles is given by Coulomb's law. The formula to calculate the force is F = k * (q1 * q2) / r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the charges of the particles, and r is the distance between them.
Using this formula, if the distance between the charges is decreased from 1 meter to 25 centimeters, the force on each charge will increase. The force is inversely proportional to the square of the distance, so when the distance is reduced to 0.25 meters (25 centimeters), the force will be 16 times stronger. Therefore, the force on each charge will be 16 N.
A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular speed. Answer in units of rad/s.
Answer:
[tex]\omega=0.12\frac{rad}{s}[/tex]
Explanation:
In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:
[tex]\omega=\frac{2\pi}{T}(1)[/tex]
Here T is the period, that is, the time taken to complete onee revolution:
[tex]T=\frac{2\pi r}{v}(2)[/tex]
Replacing (2) in (1):
[tex]\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}[/tex]
The angular speed of the racing car is approximately 0.1208861 rad/s.
The angular speed of the racing car can be calculated using the formula that relates linear speed (v) to angular speed and the radius (r) of the circular path, which is:
[tex]\[ v = r \cdot \omega \][/tex]
Given that the linear speed (v) of the car is 19.1 m/s and the radius (r) of the circular track is 158 m, we can solve for the angular speed (Ï) as follows:
[tex]\[ \omega = \frac{v}{r} \][/tex]
[tex]\[ \omega = \frac{19.1 \text{ m/s}}{158 \text{ m}} \][/tex]
[tex]\[ \omega =\frac{19.1}{158} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = 0.1208861 \text{ rad/s} \][/tex]
Therefore, the angular speed of the racing car is approximately 0.1208861 rad/s.
To express this in a more simplified fraction, we can write:
[tex]\[ \omega \ = \frac{19.1}{158} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{191}{1580} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{191}{1580} \times \frac{10}{10} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{1910}{15800} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{191}{1580} \text{ rad/s} \][/tex]
Since 191 and 1580 do not have any common factors other than 1, this fraction is already in its simplest form. Thus, the final answer for the angular speed of the racing car is:
[tex]\[ \ {0.1208861 \text{ rad/s}} \][/tex]
A(n) _____ satellite system employs many satellites, each in an orbit at an altitude of less than 1,000 miles.
Answer:
The correct answer is a Low earth orbit.
Explanation:
A low earth orbit can be understood as an earth orbit with an altitude of 1,000 miles or less. It is a satellite sustem that employs many satelites, in fact, most man-made objects that are currently in outer-space are part of this low earth orbit. (LEO).
The most famous LEO satellite system is the one from planet earth. Almost every space flight that human beings have ever done are done in LEO, and every spacial station is located in this zone.
In conclusion, A low earth orbit satellite system employs many satellites, each in an orbit at an altitude of less than 1,000 miles.
Answer:A(n) low Earth Orbit (LEO) satellite system employs many satellites, each in an orbit at an altitude of less than 1,000 miles.
Explanation:A low Earth orbit (LEO) is an Earth-centered orbit with an altitude of 2,000 km (1,200 mi) or less (approximately one-third of the radius of Earth), or with at least 11.25 periods per day (an orbital period of 128 minutes or less) and an eccentricity less than 0.25. Most of the manmade objects in outer space are in LEO.
There is a large variety of other sources that define LEO in terms of altitude. The altitude of an object in an elliptic orbit can vary significantly along the orbit. Even for circular orbits, the altitude above ground can vary by as much as 30 km (19 mi) (especially for polar orbits) due to the oblateness of Earth's spheroid figure and local topography
In the video, the torque due to the mass of the plank is used in the calculations. For this question, ignore the mass of the board. Rank, from largest to smallest, the mass m needed to keep the board from tipping over. To rank items as equivalent, overlap them.
Answer:
D, A & B, C, E
Explanation:
In each problem, sum the torques about the edge of the table.
A) ∑τ = Iα
mg (-1.5 m) + (100 kg) g (1.5 m) = 0
m = 100 kg
B) ∑τ = Iα
mg (-0.75 m) + (100 kg) g (0.75 m) = 0
m = 100 kg
C) ∑τ = Iα
mg (-1.5 m) + (100 kg) g (0.75 m) = 0
m = 50 kg
D) ∑τ = Iα
mg (-1.5 m) + (200 kg) g (1.5 m) = 0
m = 200 kg
E) ∑τ = Iα
mg (-1.5 m) + (100 kg) g (0.5 m) = 0
m = 33.3 kg
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 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]
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 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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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