Answer:22,833.12 N
Explanation:
Given
Velocity of Jet [tex]v=201 m/s[/tex]
Velocity of Exhaust gases [tex]u=669 m/s[/tex]
Rate of intake of air [tex]\frac{\mathrm{d} M_a}{\mathrm{d} t}=44.1 kg/s[/tex]
Rate at which Fuel is burned is [tex]\frac{\mathrm{d} M_f}{\mathrm{d} t}=3.28 kg/s[/tex]
Rate of change of mass in Rocket [tex]\frac{\mathrm{d} M}{\mathrm{d} t}=\frac{\mathrm{d} M_a}{\mathrm{d} t}+\frac{\mathrm{d} M_f}{\mathrm{d} t}=44.1+3.28=47.38 kg/s[/tex]
Thrust on Rocket is given by
[tex]F=\frac{\mathrm{d} M}{\mathrm{d} t}u-\frac{\mathrm{d} M_a}{\mathrm{d} t}v[/tex]
[tex]F=47.38\times 669-44.1\times 201[/tex]
[tex]F=31,697.22-8864.1=22,833.12 N[/tex]
When output is below potential and the policy rate has hit the floor of zero, the resulting fall in inflation leads to ________ real interest rates, which ________ output further, which causes inflation to fall further.
Answer:
high real rate of interest, depressing
Explanation:
If output is less than potential and the rate of policy has fall to the ground of zero, the subsequent decline in inflation results in high real rate of interest which further depressing output, causing inflation to even further decline.
If the actual output is less than potential output, demand or supply has fallen, resulting in job and output declines
) The sprocket assembly on a bicycle consists of a chain and two sprockets, one on the pedal and the other on the rear wheel. If the sprocket on the pedal is 6 inches in diameter, the sprocket on the rear wheel is 4 inches in diameter, and the rear wheel is 26 inches in diameter, how fast is the bicycle traveling in miles per hour when the cyclist is pedaling at the rate of 1.1 revolutions per second? Round your answer to the nearest tenth mph.
Answer:
7.8 mph
Explanation:
Rate of cycling = 1.1 rev/s
Rear wheel diameter = 26 inches
Diameter of sprocket on pedal = 6 inches
Diameter of sprocket on rear wheel = 4 inches
Circumference of rear wheel = [tex]\pi d=26\pi[/tex]
Speed would be
[tex]\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s[/tex]
Converting to mph
[tex]1\ inch/s=\frac{1}{63360}\times 3600\ mph[/tex]
[tex]134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph[/tex]
The Speed of the bicycle is 7.8 mph
To determine the bicycle speed, calculate the rear wheel's circumference, use the gear ratio to find the rear wheel's rotation rate, and then convert the distance traveled per second into miles per hour. The resulting speed is 17.3 mph when pedaling at 1.1 revolutions per second.
Explanation:The student's question involves calculating the speed of a bicycle based on the diameters of the sprockets on the pedal and rear wheel, the diameter of the rear wheel, and the pedaling rate in revolutions per second. To find the speed in miles per hour, we first calculate the circumference of the rear wheel, which gives us the distance traveled per revolution. We then use the gear ratio to find the effective revolutions of the rear wheel based on the pedaling speed. The total distance traveled per second is then converted to miles per hour.
The diameter of the rear wheel is 26 inches, so its circumference is π(26 inches) ≈ 81.68 inches. The gear ratio is the diameter of the pedal sprocket (6 inches) over the diameter of the rear wheel sprocket (4 inches), giving us a 1.5:1 gear ratio. For every pedal revolution, the rear wheel rotates 1.5 times. Pedaling at 1.1 revolutions per second, the rear wheel rotates at 1.5 * 1.1 = 1.65 revolutions per second. The distance traveled per second is then 81.68 inches * 1.65 ≈ 134.77 inches per second.
To convert inches per second to miles per hour, we use the conversion factors 12 inches per foot and 5280 feet per mile, and 3600 seconds per hour. Thus, the bike speed is (134.77 inches/second) * (1 foot/12 inches) * (1 mile/5280 feet) * (3600 seconds/hour) ≈ 17.26 miles per hour. Rounded to the nearest tenth, the bicycle is traveling at 17.3 mph.
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume that the photons spread out spherically. The area of a sphere is 4????????2.
Answer:
The distance from the radio station is 0.28 light years away.
Solution:
As per the question:
Distance, d = 4 ly
Frequency of the radio station, f = 854 kHz = [tex]854\times 10^{3}\ Hz[/tex]
Power, P = 50 kW = [tex]50\times 10^{3}\ W[/tex]
[tex]I_{p} = 1\ photon/s/m^{2}[/tex]
Now,
From the relation:
P = nhf
where
n = no. of photons/second
h = Planck's constant
f = frequency
Now,
[tex]n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s[/tex]
Area of the sphere, A = [tex]4\pi r^{2}[/tex]
Now,
Suppose the distance from the radio station be 'r' from where the intensity of the photon is [tex]1\ photon/s/m^{2}[/tex]
[tex]I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}[/tex]
[tex]1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}[/tex]
[tex]r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m[/tex]
Now,
We know that:
1 ly = [tex]9.4607\times 10^{15}\ m[/tex]
Thus
[tex]r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly[/tex]
What is the magnitude φinitial of the magnetic flux through one turn of the coil before it is rotated?
Answer:
Explanation:
Facts about magnetic flux;
(1). Change in magnetic flux is equal to the current generated.
(2). The magnetic field moves from the north to the south.
(3). Magnetic flux is equal to how much magnetic field goes through a direction.
(4). Magnetlc direction can be found found by using Lenz's law.
Faraday's law= emf= N(BA/ ∆t.
Where N= number of loop
∆B/∆T = rate at which flux changes.
The magnitude φ(initial) of the magnetic flux through one turn of the coil before it is rotated is
φ(initial) = BA while φ(final)= zero(0).
The initial magnetic flux through one turn of the coil, with the coil initially perpendicular to Earth's magnetic field, is 6.10 × 10⁻⁹ T×m².
To calculate the magnitude of the initial magnetic flux (φinitial) through one turn of the coil, we use the formula:
φ = B × A × cos(θ)
where:
B is the magnetic field (5.00 × 10⁻⁵ T)A is the area the coil encloses (12.2 cm²)θ is the angle between the magnetic field and the normal to the plane of the coilSince initially, the plane of the coil is perpendicular to Earth's magnetic field, θ = 0°, and hence cos(θ) = 1. First, we must convert the area from cm² to m²:
A = 12.2 cm² × (1 m² / 10,000 cm²) = 12.2 × 10⁻⁴ m²
Now, we can calculate the magnetic flux:
φinitial = 5.00 × 10⁻⁵ T × 12.2 × 10⁻⁴ m² × cos(0°)
φinitial = 5.00 × 10⁻⁵ T × 12.2 × 10⁻⁴ m²
φinitial = 6.10 × 10⁻⁹ T×m²
This is the magnitude of the initial magnetic flux through one turn of the coil before it is rotated.
Complete Question:
A coil has 230 turns enclosing an area of 12.2 cm³. In a physics laboratory experiment, the coil is rotated during the time interval 3.00 × 10⁻² s from a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.00 × 10⁻⁵ T.
What is the magnitude φinitial of the magnetic flux through one turn of the coil before it is rotated?
The filament in an incandescent light bulb is made from tungsten. The resistivity of tungsten = 5.6e-8 Ω*m. The radius of the tungsten wire is 0.045 mm. If the bulb is plugged into a 120 V outlet and is to draw a current of 1.24 A, how long must the wire be?
Answer:
11m
Explanation:
Given:
Resistivity ρ = 5.6e-8 Ωm
Radius r = 0.045 mm [tex]=\frac{0.045}{1000}[/tex] = 4.5 x 10⁻⁵ m
Voltage V = 120V
Current I = 1.24A
From Ohm's law, [tex]R = \frac{V}{I}[/tex]
[tex]R = \frac{120}{1.24}[/tex]
R = 96.77 Ω
Resistivity = (Resistance × Area)/ length
ρ = (RA)/L
Therefore, the length of a wire is given by;
L = (RA)/ρ
Calculating the area A of the wire;
A = πr²
A = π × (4.5 x 10⁻⁵)²
A = 6.36 x 10⁻⁹ m²
Substituting area of the wire A = 6.36 x 10⁻⁹ m² into the equation of the length of wire
L = (96.77 × 6.36×10⁻⁹ ) / 5.6×10⁻⁸
L = 10.9977m
L = 11m (approximately)
Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 2,000 N/C. If the voltage is doubled and the distance between the plates is reduced to 1/5 the original distance, the magnitude of the new electric field is
A) 800 N/C
B) 1600 N/C
C) 2400 N/C
D) 5000 N/C
E) 20000 N/C
Answer:
E) E₂= 20000 N/C
Explanation:
The electric field between two parallel conductive plates is thus calculated
E= V/d
Where:
E: Electric field (N/C)
V: voltage (V)
d : distance between the plates (m)
Problem development
E₁= V₁/d₁ = 2000 N/C
E₂= V₂/d₂
[tex]E_{2} = \frac{2V_{1} }{\frac{d_{1} }{5} }[/tex]
[tex]E_{2} = 10(\frac{V_{1} }{d_{1} })[/tex]
E₂= 10* (2000)
E₂= 20000 N/C
Final answer:
When the voltage between two parallel conducting plates is doubled and the distance between them is reduced to 1/5 of the original distance, the magnitude of the new electric field is 20,000 N/C.
Explanation:
The question relates to the electric field between two parallel conducting plates connected to a voltage source. The original electric field's magnitude is given as 2,000 N/C, and we're asked to find the new electric field magnitude when the voltage is doubled and the distance between the plates is reduced to 1/5 of the original distance. The electric field (E) between two parallel plates is given by the equation E = V/d, where V is the potential difference (voltage) between the plates, and d is the distance between them. When the voltage is doubled, V becomes 2V, and when the distance is reduced to 1/5, d becomes d/5. Thus, the new electric field E' = (2V)/(d/5) = 10(V/d) = 10E. So, the new electric field strength is 10 * 2,000 N/C = 20,000 N/C, which corresponds to option E).
Two balls are thrown horizontally. Ball C is thrown with a force of 20 N, and ball D is thrown with a force of 40 N. Assuming all other factors are equal, ball D will fall toward the ground
Answer:
ball D will fall toward the ground at the same time as ball C
Explanation:
both balls experience the same downward (vertical) force of gravity as such they will both fall down at the same time, given that all other factors are equal.
although the ball were through with different forces,
those forces where in the horizontal direction but the force of gravity (downward force) will act on them equally to bring them down at the same time
Technician A says that magnetic lines of force can be seen by placing iron filings on a piece of paper and then holding them over a magnet. Technician B says that the effects of magnetic lines of force can be seen using a compass. Which technician is correct?
Answer:
Both A and B are correct
Explanation:
Magnetic line of force can be seen by placing iron fillings on a piece of paper and then hold a magnet over them, the iron fillings align themselves in the magnetic field of the magnetic along the line of forces thus showing the loops of the magnetic line of forces.Also, we know that unlike poles attract where as like poles repel each other. This fact can be proved when we hang a bar magnet it aligns itself in the North-South direction because of the the magnetic field due to the alignment of the geographical North-South poles and the poles of the bar magnet and the compass works the same way.Thus both the technicians are correct.
Answer:
Technician A is correct
Magnetic lines of force can be seen by placing iron filings on a piece of paper and then holding them over a magnet.
Explanation:
As we know that the magnetic field lines are always originated out from the North pole of magnet and then terminate at south pole of the magnet
Here we know that when magnetic material aligns itself in the direction of external magnetic field
So here when we put the iron filings on the cardboard and then place a magnet near it then all the iron filing will align itself in the direction of the magnetic field
Now if we place a compass near the magnet then it will aligns itself in the tangential direction of the magnetic field at the same position
So here we can say that Technician A is correct
An object is acted upon by a constant force which gives it a constant acceleration a. At a certain time t1, having started from rest at t = 0, it has kinetic energy K1. At what time t2 has its kinetic energy doubled?
Answer:
[tex]t_2=\sqrt{2}(t_1)[/tex]
Explanation:
The kinetic energy of a body is that energy it possesses due to its motion. In classical mechanics, this energy depends only on its mass and speed, as follows:
[tex]K=\frac{mv^2}{2}[/tex]
The speed in an uniformly accelerated motion is given by:
[tex]v=at[/tex]
Replacing this expression in the formula for the kinetic energy, we have:
[tex]K=\frac{ma^2t^2}{2}\\[/tex]
So, if we have [tex]K_2=2K_1[/tex]:
[tex]K_1=\frac{ma^2t_1^2}{2}(1)\\K_2=\frac{ma^2t_2^2}{2}\\2K_1=\frac{ma^2t_2^2}{2}\\K_1=\frac{ma^2t_2^2}{4}(2)\\[/tex]
Equaling (1) and (2) and solving for [tex]t_2[/tex]:
[tex]\frac{ma^2t_1^2}{2}=\frac{ma^2t_2^2}{4}\\t_2=\frac{4t_1^2}{2}\\t_2=\sqrt{2t_1^2}\\t_2=\sqrt{2}(t_1)[/tex]
A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx3 , where c = 8.0 N/m3 . The particle’s speed at A, where xA = 1.0 m, is 6.0 m/s. What is its speed at B, where xB = −2.0 m?
Answer:4.58 m/s
Explanation:
Given
mass of Particle [tex]m=4 kg[/tex]
[tex]F=-cx^3[/tex]
[tex]a=\frac{F}{m}[/tex]
[tex]a=-\frac{cx^3}{m}[/tex]
[tex]a=-\frac{8x^3}{4}[/tex]
[tex]a=-2x^3[/tex]
[tex]v\frac{\mathrm{d} v}{\mathrm{d} x}=-2x^3[/tex]
[tex]vdv=-2x^3dx[/tex]
integrating
[tex]\int_{6}^{v_b}vdv=\int_{1}^{-2}-2x^3dx[/tex]
[tex]\frac{v_b^2-6^2}{2}=-\frac{1}{2}\left [ \left ( -2\right )^4-\left ( 1\right )^4\right ][/tex]
[tex]\frac{v_b^2-36}{2}=-0.5\times 15[/tex]
[tex]v_b^2=36-15[/tex]
[tex]v_b=\sqrt{21}[/tex]
[tex]v_b=4.58 m/s[/tex]
A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx³ The speed of the particle at point B is 5.3 m/s.
The potential energy at any point x is given by:
F(x) = -cx³
U(x) = -∫F(x) dx
U(x)= ∫cx³dx
U(x) = c/4 × x⁴ + C
The total mechanical energy E remains constant:
E = K + U
E_A = K(A) + U(A)
E_A = (1/2)mv(A)²+ c÷4 × x(A)⁴ + C
At point B, the potential energy U(B) is:
U(B) = c/4 × x(B)⁴ + C
The total mechanical energy at point B is:
E(B) = K(B) + U(B)
Since the total mechanical energy is conserved, we have:
E(A) = E(B)
(1/2)mv(A)² + c÷4 × x(A)⁴ + C = (1/2)mv(B)² + c÷4 × x(B)⁴ + C
(1/2)(4.0)(6.0)² + ((8.0) ÷ 4) × (1.0)⁴ = (1/2)(4.0)v(B)² + ((8.0) ÷ 4) × (-2.0)⁴
72.0 + 2.0= 2.0 × v(B)² + 16.0
56.0 J = 2.0× v(B)²
v(B)² = 28.0
v(B) = ±5.3 m/s
Since speed cannot be negative, the speed of the particle at point B is 5.3 m/s.
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A 58.1g sample of quartz is put into a calorimeter that contains 250.0g of water. The quartz sample starts off at 98.4°C and the temperature of the water starts off at 25.0°C. When the temperature of the water stops changing it's 27.9°C. The pressure remains constant at 1 atm. Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to the correct number of significant digits.
Answer:
[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]
Explanation:
Given:
mass of sample quartz, [tex]m_q=58.1\ g[/tex]mass of water in calorie-meter, [tex]m_w=250\ g[/tex]initial temperature of quartz, [tex]T_{qi}=98.4^{\circ}C[/tex]initial temperature of water, [tex]T_{wi}=25^{\circ}C[/tex]final temperature of the mixture, [tex]T_{f}=27.9^{\circ}C[/tex]We have:
specific heat capacity of water, [tex]m_c=4.186\ J.g^{-1}.^{\circ}C^{-1}[/tex]Assuming that heat loss is neither from any of the components before mixing nor from the mixture just after mixing. Also the container does not absorbs any heat.
Therefore,
heat gained by the water = heat lost by the quartz
[tex]Q_w=Q_q[/tex]
[tex]m_w.c_w. (T_f-T_{wi})=m_q.c_q.(T_{qi}-T_f)[/tex]
where: [tex]c_q=[/tex] specific heat capacity of quartz
[tex]250\times 4.186\times (27.9-25)=58.1\times c_q\times (98.4-27.9)[/tex]
[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]
To calculate the specific heat capacity of quartz, use the equation q = mcΔT and set it equal to the equation for the heat transferred to the water. Solve for c to find the specific heat capacity of quartz.
Explanation:The specific heat capacity of quartz can be calculated using the equation:
q = mcΔT
where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
The heat transferred to the water can be calculated using the equation:
q = mcΔT
where q is the heat transferred, m is the mass, c is the specific heat capacity of water (4.184 J/g °C), and ΔT is the change in temperature.
By setting the two equations equal to each other and solving for c, the specific heat capacity of quartz can be determined.
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An aircraft is loaded 110 pounds over maximum certificated gross weight. If fuel (gasoline) is drained to bring the aircraft weight within limits, how much fuel should be drained?
Answer:
242 lts de fuel
Explanation:
We know density of fuel (is variable) but we can say that is close to 1 Kg/lts.
If we have 110 pounds over maximun certificated gross weight, we need to get rid of that 110 pounds or 110 * 2.2 = 242 Kgs
Now by rule of three
If 1 lt weight 1 kg.
x 242 kgs
x = 242 lts de fuel
To bring the aircraft within the maximum gross weight limits, approximately 18.33 gallons of fuel need to be drained, calculated by dividing the excess weight (110 pounds) by the weight of fuel per gallon (6 pounds/gallon).
To calculate how much fuel needs to be drained to bring an aircraft weight within the maximum certificated gross weight limits, we need to know two pieces of information: the weight by which the aircraft is over its maximum gross weight and the weight of the fuel that needs to be drained.
In this scenario, the aircraft is 110 pounds over the maximum certificated gross weight. In general, aviation gasoline (avgas) weighs about 6 pounds per gallon. To solve this problem, we will divide the excess weight by the weight of the fuel per gallon to find out how many gallons need to be drained.
So, the calculation is as follows:
Excess weight: 110 pounds.
Weight of fuel per gallon: 6 pounds/gallon.
Number of gallons to be drained = Excess weight / Weight of fuel per gallon.
Number of gallons to be drained = 110 pounds / 6 pounds/gallon.
Number of gallons to be drained = 18.33 gallons (approximately).
Therefore, to bring the aircraft back within weight limits, approximately 18.33 gallons of fuel need to be drained.
In the experiment, you will determine the wavelength of a laser by shining the laser beam on a diffraction grating, a set of regularly spaced lines. If the screen is 10 m away, the spots are 5.4 cm apart, and the lines of the grating are separated by 0.1 mm,
what is (in nm)?
Answer:
The wavelength is 539 nm.
Explanation:
Given that,
Distance of screen L= 10 m
Width s= 5.4 cm
Grating separation d= 0.1 mm
Suppose the pattern is displayed on a screen a distance L from the grating and the spots are separated by s. we need to find the wavelength.
The angle between the central and first maximum is given as
[tex]\tan\theta=\dfrac{s}{L}[/tex]
[tex]\theta=\tan^{-1}\dfrac{s}{L}[/tex]
[tex]\theta=\tan^{-1}\dfrac{5.4\times10^{-2}}{10}[/tex]
[tex]\theta=0.309[/tex]
We need to calculate the wavelength
The condition for maximum of a diffraction grating is
[tex]d\sin\theta=m\lambda[/tex]
[tex]\lambda=\dfrac{d\sin\theta}{m}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{0.1\times10^{-3}\times\sin0.309}{1}[/tex]
[tex]\lambda=539\times10^{-9}\ m[/tex]
[tex]\lambda=539\ nm[/tex]
Hence, The wavelength is 539 nm.
The wavelength of a laser can be determined using the diffraction grating equation. Calculate the θ value with θ = tan⁻¹(y/L), then substitute the known values into the equation λ = (d sin θ) / m. Convert the resultant wavelength to nanometers.
Explanation:Your question relates to the physics of light and diffraction grating. We can find the wavelength of the laser by using the diffraction grating equation. The equation is: d sin θ = mλ
Where:
d is the grating spacing (distance between the lines represented in meters)θ is the angle to the m'th order maximumm is the order of maximum (in your case, it's likely 1)λ is the wavelength we are trying to findIn your case, we need to calculate θ first: θ = tan⁻¹(y/L) where y is the distance between the spots (5.4cm) and L is the distance from the grating to the screen (10m).
Substitute the known values and solve for λ: λ = (d sin θ) / m
Remember, the resultant wavelength should be converted to nanometers (1m = 1 x 10⁹ nm).
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Tech A says the first external component to install when assembling an engine is the intake manifold. Tech B says the exhaust manifolds should be installed after the intake manifold. Who is correct?
Answer:
both technicians are correct
Explanation:
both technicians are correct because technician A says the first external component to be installed is the intake manifold and technician B says the exhaust manifold should be installed after the intake manifold has been installed, this implies the same thing because installing the exhaust manifold after installing the intake manifold means you are installing the intake manifold first.
Amy is out with her friends. Misfortune occurs and Amy and her friends find themselves getting a workout. They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel station. Determine the work done on the car.A. 2.35 x 105 JB. 2.05 x 105 JC. 1.35 x 105 JD. 3.25 x 105 J
Answer:
2.35 x 10^{5} J
Explanation:
force (f) = 1080 N
distance (d) = 218 m
find the work done
work is said to be done when force applied to an object moves the object, therefore work done = force x distance
work done = 1080 x 218 = 235,440 J = 2.35 x 10^{5} J
Suppose you are asked to find the amount of time t, in seconds, it takes for the turntable to reach its final rotational speed. Which of the following equations could you use to directly solve for the numerical value of t?
A) θ=θ0+ω0t+(1/2)αt2
B) ω=ω0+αt
C) ω2=ω02+2α(θ−θ0)
D) More information is needed before t can be found.
Answer:
option (D)
Explanation:
Here initial rotation speed is given, final rotation speed is given and asking for time.
If we use
A) θ=θ0+ω0t+(1/2)αt2
For this equation, we don't have any information about the value of angular displacement and angular acceleration, so it is not useful.
B) ω=ω0+αt
For this equation, we don't have any information about angular acceleration, so it is not useful.
C) ω2=ω02+2α(θ−θ0)
In this equation, time is not included, so it is not useful.
D) So, more information is needed.
Thus, option (D) is true.
The equation to directly solve for time t in this context is ω = ω₀ + αt, as it includes all necessary variables except for t, making it easy to solve algebraically. Hence, option(b) is correct.
The correct equation to use in this problem is ω = ω₀ + αt.
By rearranging this equation, you can directly find the value of t using given angular velocities and acceleration. This is applicable in analyzing rotational motion in physics.
To solve for the time t, we can rearrange the equation algebraically:
[tex]\\t = \frac{\omega - \omega_0}{\alpha}\\[/tex]
This equation directly relates
the final angular velocity (ω), initial angular velocity (ω₀), and angular acceleration (α) to time t.Given the initial and final angular velocities and the angular acceleration, you can plug these values in to find t.
A very loud train whistle has an acoustic power output of 100 W. If the sound energy spreads out spherically, what is the intensity level in dB at a distance of 100 meters from the train?
The intensity level in dB at a distance of 100 meters from the train is 98.46 dB.
Explanation:To calculate the intensity level in dB at a distance of 100 meters from the train, we need to first calculate the intensity of sound at that distance.
Since the sound energy spreads out spherically, the intensity decreases with the square of the distance from the source.
Using the formula for intensity, I = P/4πr², where P is the acoustic power output and r is the distance from the source, we can calculate the intensity at 100 meters:
I = 100W / (4π * (100m)²) = 0.0796 W/m².
Now, we can use the formula for the intensity level in dB, β = 10log(I/I₀), where I₀ is the reference intensity (10⁻¹² W/m²).
Substituting the values, β = 10log(0.0796 / 10⁻¹²) = 98.46 dB.
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A type of cuckoo clock keeps time by having a mass bouncing on a spring, usually something cute like a cherub in a chair. What force constant (in N/m) is needed to produce a period of 0.660 s for a 0.0100 kg mass?
Answer:0.906 N/m
Explanation:
Given
time period [tex]T=0.66 s[/tex]
mass [tex]m=0.01 kg[/tex]
System can be considered as spring mass system
Time Period of spring mass system is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
squaring
[tex]k=\frac{4\pi ^2m}{T^2}[/tex]
[tex]k=\frac{4\times \pi ^2\times 0.01}{0.66^2}[/tex]
[tex]k=0.906 N/m[/tex]
A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a surface area of 7.70 m2. The total power it absorbs from the star is 3800 W.
Assuming the surface is an ideal absorber and radiator, calculate the equilibrium temperature of the asteroid (in K)
Answer:
Temperature will be 305 K
Explanation:
We have given The asteroid has a surface area [tex]A=7.70m^2[/tex]
Power absorbed P = 3800 watt
Boltzmann constant [tex]\sigma =5.67\times 10^{-8}Wm/K^4[/tex]
According to Boltzmann rule power radiated is given by
[tex]P=\sigma AT^4[/tex]
[tex]3800=5.67\times 10^{-8}\times 7.70\times T^4[/tex]
[tex]T^4=87.0381\times 10^8[/tex]
[tex]T=305K[/tex]
So temperature will be 305 K
A uniform meter stick of mass $M$ has a half-filled can of fruit juice of mass $m$ attached to one end. The meter stick and the can balance at a point 21.2 cm from the end of the stick where the can is attached. When the balanced stick-can system is suspended from a scale, the reading on the scale is 2.58 N. Calculate the mass of the meter stick.
Answer:
111.6 g
Explanation:
Given
m + M = 2.58 / 9.8
= 0.2632 kg
When the can of fruit juice is balance from scale , we get following relation
M x ( 50 - 21.2 ) = m x 21.2 ( balancing the torque due to weight of scale and can about the balancing point )
M x 28.8 = 21.2 m
= 21.2 ( 0.2632 - M )
= 5.58 - 21.2 M
M ( 28.8 + 21.2 ) = 5.58
M = .1116 kg
= 111.6 g
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg · m2. The father exerts a force on the merry-go-round perpendicular to its 1.50 m radius to achieve a torque of 375 N · m.(a) Calculate the rotational kinetic energy (in J) in the merry-go-round plus child when they have an angular velocity of 29.2 rpm. J(b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. revolutions(c) Again, using energy considerations, calculate the force (in N) the father must exert to stop the merry-go-round in seven revolutions. N
In this part, we discuss the expected decrease in angular acceleration when the child is on the merry-go-round due to the greater moment of inertia. We calculate the child's moment of inertia by approximating the child as a point mass at a specific distance from the axis.
Explanation:b. We expect the angular acceleration for the system to be less in this part because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia I, we first find the child's moment of inertia Ic by approximating the child as a point mass at a distance of 1.25 m from the axis. Then
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Final answer:
The rotational kinetic energy of the merry-go-round plus child at 29.2 rpm is 394.5834 J. The father must push for approximately 0.167 revolutions to reach this velocity from rest. A force of -8.975 N (in the opposite direction of motion) must be exerted to stop the merry-go-round in seven revolutions.
Explanation:
The problem addresses the concepts of angular velocity, rotational kinetic energy, and the application of work-energy principles within a physics context, specifically relating to a merry-go-round system.
(a) To calculate the rotational kinetic energy (KErot) of the merry-go-round plus child, we use the formula KErot = (1/2)Iω^2 where I is the moment of inertia and ω is the angular velocity in radians per second. First, convert the angular velocity from rpm to rad/s using the conversion factor 1 rpm = (2π/60) rad/s.
KErot = (1/2)(84.4 kg · m^2)(29.2 rpm × 2π/60 rad/s)^2
= (1/2)(84.4)(3.058 rad/s)^2
= (1/2)(84.4)(9.351)
= (1/2)(789.1668) J
= 394.5834 J
(b) The number of revolutions needed to achieve this angular velocity from rest can be found using the work-energy principle, where work done (W) by the father's force equals the rotational kinetic energy of the merry-go-round system.
W = τr(2πN)
394.5834 J = 375 N · m × (2πN)
N = 394.5834 J / (375 N · m × 2π)
N = 0.167 revolutions
(c) The force to stop the merry-go-round in seven revolutions can be calculated using the concept that the work done by the father's force (W) to stop the merry-go-round equals the negative change in the rotational kinetic energy. Thus, W = -KErot, and the rotational kinetic energy is initially computed in part (a). This gives:
W = -394.5834 J
Force (F) = W / (2π×7)
F = -394.5834 J / (2π × 7)
F = -8.975 N (negative sign indicates that the force is applied in the opposite direction of motion)
A weather balloon is designed to expand to a maximum radius of 21 m at its working altitude, where the air pressure is 0.030 atm and the temperature is 200 K. If the balloon is filled at atmospheric pressure and 323 K, what is its radius at liftoff?
Answer:
7.65 m
Explanation:
[tex]P_1[/tex] = Initial pressure = 0.03 atm
[tex]P_2[/tex] = Final pressure = 1 atm
[tex]r_1[/tex] = Inital radius = 21 m
[tex]V_1[/tex] = Intial volume of gas = [tex]\frac{4}{3}\pi r_1^3[/tex]
[tex]V_2[/tex] = Final volume of gas = [tex]\frac{4}{3}\pi r_2^3[/tex]
[tex]T_1[/tex] = Initial temperature = 200 K
[tex]T_2[/tex] = Final temperature = 323 K
From ideal gas law we have
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{P_1\frac{4}{3}\pi r_1^3}{T_1}=\frac{P_2\frac{4}{3}\pi r_2^3}{T_2}\\\Rightarrow \frac{P_1 r_1^3}{T_1}=\frac{P_2r_2^3}{T_2}\\\Rightarrow r_2=\frac{P_1r_1^3T_2}{T_1P_2}\\\Rightarrow r_2=\left(\frac{0.03\times 21^3\times 323}{200\times 1}\right)^{\frac{1}{3}}\\\Rightarrow r_2=7.65\ m[/tex]
The radius at liftoff is 7.65 m
The initial radius of the weather balloon at liftoff can be calculated using the ideal gas law by comparing the conditions at the ground and at the balloon's working altitude.
Explanation:The subject of this question deals with the physics principle of the behavior of gases, namely the ideal gas law. The ideal gas law links the pressure, temperature and volume of a gas, represented by the equation PV=nRT. Here, 'P' denotes pressure, 'V' is volume, 'n' is the number of moles, 'R' is the universal gas constant, and 'T' is temperature.
In the case of the weather balloon, when it is filled at atmospheric pressure and 323 K on the ground, and then expands to a maximum radius of 21m at a working altitude where the air pressure is 0.030 atm and the temperature is 200K, the principle of the ideal gas law can be applied. Assuming the same quantity of gas (n) we have P1V1/T1=P2V2/T2 where indices 1 and 2 refer to conditions at lift off and at altitude respectively.
We know that the volume of the balloon (V) is related to its radius (r) by the equation for volume of a sphere V=(4/3)*pi*r^3. So we can substitute the volumes in above equation with their expressions in terms of radius and solve for r1 (radius at liftoff).
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The electric motor of a model train accelerates the train from rest to 0.740 m/s in 30.0 ms. The total mass of the train is 560 g. Find the average power delivered to the train during its acceleration.
Answer:
= 5.1 W
Explanation:
time (t) = 30 ms = 0.03 s
mass (m) = 560 g = 0.56 kg
initial velocity (U) = 0 m/s
final velocity (V) = 0.74 m/s
power = \frac{work done}{t} = \frac{f x d}{t} = f x v = m x a x v
m x a x v = m x \frac{V-U}{t} x \frac{V + U}{2}
m x \frac{V-U}{t} x \frac{V + U}{2} = 0.56 x \frac{0.74 - 0}{0.03} x \frac{0.74+0}{2}
= 5.1 W
An unknown mass is hung from a very light spring with a spring constant of 13.2 N/m. Neglect the mass of the spring. If the period of oscillation of the mass-spring system is 2.60 s, then the unknown mass is kg.
Amplitude is the maximum displacement of the mass from the _______.
end point equilibrium point top point vibrational point
A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of +14.8 cm at time, t = 0.00 s. The displacement from equilibrium of the mass at time, t = 1.25 s is cm.
A ball of mass 1.55 kg is hung from a spring which stretches a distance of 0.3800 m. The spring constant of the spring (a scalar quantity) is N/m
A mass m on a spring with a spring constant k is in simple harmonic motion with a period T. If the same mass is hung from a spring with a spring constant of 3k, the period of the motion will be _______.
increased by a factor of the square root of 3 decreased by a factor of the square root of 3 decreased by a factor of 3 increased by a factor of 3
Answer:
Explanation:
Time period of oscillation of spring mass system
T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]
m = [tex]\frac{T^2\times k}{4\pi^2}[/tex]
= [tex]\frac{(2.6)^2\times 13.2}{4\pi^2}[/tex]
= 2.26 kg
Amplitude is the maximum displacement of the mass from the _equilibrium point .
maximum displacement from equilibrium of +14.8 cm at time, t = 0.00 s
x(t) = 14.8cosωt
ω = 2πn = 2π x 9.73 = 61.1 rad /s
x(t) = 14.8cos61.1 t
when t = 1.25
x(t) = 14.8cos61.1 x 1.25
= 14.8cos61.1 x 1.25
= 14.8cos76.38
= 8.22 cm
A ball of mass 1.55 kg is hung from a spring which stretches a distance of 0.3800 m
spring constant
k = mg / x
= 1.55 x 9.8 / .38
= 40 N / m
T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]
so if mass is hung from a spring with a spring constant of 3k
T decreases by a factor of the square root of 3
For the first question, the unknown mass is approximately 0.71 kg. For the second question, the term 'equilibrium point' completes the statement. For the third question, the displacement at time 1.25 seconds is -14.2 cm. For the fourth question, the spring constant k is approximately 40.0 N/m. For the fifth question, the period of the motion will be decreased by a factor of the square root of 3.
Explanation:The unknown mass hung from a spring can be calculated using the formula for the period of a mass-spring system, T = 2π√(m / k), where T is the period, m is the mass, and k is the spring constant. Rearranging for m gives us m = (T²k) / (4π²). Given that T = 2.60 seconds and k = 13.2 N/m, we find m = (2.60^2 * 13.2) / (4*π^2), which gives a mass of approximately 0.71 kg.
The amplitude is the maximum displacement of the mass from the equilibrium point.
The displacement of a harmonically oscillating mass can be found using the equation: x(t) = A*cos(2πft + φ), where A is the amplitude, f is the frequency, and φ is the phase angle. Given the information we have a displacement at 1.25 seconds of -14.2 cm.
The spring constant can be calculated using the equation k = mg / x, where m is the mass, g is gravity (9.81 m/s²), and x is the distance stretched. Given that the mass is 1.55 kg and the distance stretched is 0.3800 m, we find that the spring constant k is approximately 40.0 N/m.
If the same mass is hung from a spring with a spring constant of 3k, the period of the motion will be decreased by a factor of the square root of 3.
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A freight car with a mass of 10 metric tons is rolling at 108 km/h along a level track when it collides with another freight car, which is initially at rest. If the speed of the cars after they couple together is 36 km/h, what is the mass of the second car?
Answer:
20 metric tons
Explanation:
Using the Principle of Conservation of Linear Momentum which states that the total momentum of a system is constant in any direction in which no external force acts.
momentum before collision= momentum after collision
mathematically expressed as
m1u1 + m2u2 = m1v1+m2v2 =0
where:
m1= mass of rolling freight car = 10 metric tons = 10Mg
1 ton = 1000kg= 1Mg
u1 =initial velocity of freight car = 108km/h,
m2 = mass of stationary freight car = ?, this the unknown we are finding
u2 = initial velocity of
m2 = 0, because it is at rest,
v1= final velocity of m1,
v2= final velocity of m2
since they move together in the same direction they share a common final velocity as such
v1=v2= 36km/h
substituting valves in expression give
m1u1 + m2u2 = m1v1 + m2v2= 0
10Mgx 108km/hr + m2 x 0 = 10Mg x 36km/h + m2x36km/h
1080( Mg x km/hr) + 0 = 360 (Mg x Km/h) + 36m2 (km/h)
1080 (Mgx km/hr) - 360(Mg x km/h) = 36m2 (km/h)
720(Mg x km/h) = 36m2 (km/h)
divide both side by 36 (km/h) gives
m2 = 20Mg =20 metric tons
Answer: B). 20 metric tons
Explanation: If a freight car with a mass of 10 metric tons is going at 108 km/h collides with another freight car, that is parked. If the speed of the cars, after they crash together, is 36 km/h, the mass of the second car is 20 metric tons.
I AM 100% CONFIDENT IN THIS ANSWER!Listed following are several objects in the solar system. Rank these objects from left to right based on the distance from the Sun at which they are presumed to have formed, from nearest to farthest.
a typical asteroid in the asteroid belt, a typical oort cloud object, a typical kuiper belt object
Although Oort cloud comets are now located far beyond the Kuiper belt, they are thought to have formed in the region of the jovian planets. They were then "kicked out" to their current orbits by gravitational encounters with the jovian planets.
The objects in the solar system can be ranked from nearest to farthest based on their formation distance from the Sun: Mercury, Venus, Earth, Mars, and Jupiter.
Explanation:The objects in the solar system can be ranked based on their distance from the Sun as follows:
Mercury: This planet is the closest to the Sun and is presumed to have formed at its current location.Venus: Venus is the second planet from the Sun and also formed relatively close to it.Earth: Our home planet is the third planet from the Sun and formed within the habitable zone.Mars: Mars is the fourth planet from the Sun and formed farther away than Earth.Jupiter: As the largest planet in the solar system, Jupiter formed even farther away from the Sun than Mars.Therefore, the ranking from nearest to farthest would be Mercury, Venus, Earth, Mars, and Jupiter.
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According to the principle of conservation of momentum, A. the amount of momentum of all the objects in the universe is constant. B. in an open system, the total amount of momentum of the objects is conserved. C. in a closed system, the speed of any two colliding objects will remain constant. D. in a closed system, an object's momentum before a collision will equal its momentum after the collision.
The principle of conservation of momentum states that in a closed system, an object's momentum before a collision will equal its momentum after the collision.
Explanation:The principle of conservation of momentum states that in a closed system, the total momentum of the objects before a collision will be equal to the total momentum of the objects after the collision. This means that in a closed system, an object's momentum before a collision will equal its momentum after the collision.
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then released from rest, and whose period is 0.66 s?
Assume that the displacement at the start of the motion is positive.
What will be its displacement after 2.3 s ?
Answer:
[tex]0.836cm[/tex] in the positive direction.
Explanation:
The equation that describes the motion of this mass-spring system is given by;
[tex]y=Asin\omega t...................(1)[/tex]
Where A is the amplitude, which defined as the maximum displacement from the equilibrium position for a body in simple harmonic motion.
[tex]\omega[/tex] is the angular velocity measured in [tex]rads^{-1[/tex], this is the angle turned through per unit time.
[tex]y[/tex] is the displacement along the axis of the amplitude, and [tex]t[/tex] is any instant of time in the motion.
Given; A = 8.8cm = 0.088m
The angular velocity is given by the following relationship also;
[tex]\omega=2\pi/T[/tex]
Where T is the period, which is defined as the time taken for a body in simple harmonic motion to make one complete oscillation.
Given; T=0.66s
Therefore;
[tex]\omega=2\pi/0.66\\\omega=3.03\pi rads^{-1[/tex]
Substituting all values into equation (1) we obtain the following;
[tex]y=0.088sin3.03\pi t................(2)[/tex]
Equation (2) is the equation describing the motion of the mass on the spring.
At an instant of time t = 2.3s, the displacement is therefore given as follows;
[tex]y=0.088sin[3.03\pi(2.3)]\\y=0.088sin6.97\pi\\[/tex]
By conversion, [tex]6.97\pi rad=6.97*180^o=1254.55^o[/tex]
Therefore
[tex]y=0.088sin1254.54=0.00836m\\[/tex]
[tex]y=0.00836m=0.836cm[/tex]
A race car travels 99 m/s around a circular track of radius 52 m. What is the magnitude of the resultant force on the 1900 kg driver and his car if the car does not slip? Answer in units of kN.
The magnitude of the resultant force on the race car and driver is 357.375 kN.
Explanation:First, we need to find the centripetal acceleration of the race car traveling around the circular track. The centripetal acceleration can be found using the formula:
a = v² / r
where v is the velocity of the car and r is the radius of the circular track. Plugging in the values, we get:
a = (99 m/s)² / 52 m = 188.25 m/s²
Next, we can use Newton's second law of motion to find the magnitude of the resultant force acting on the car. The formula is:
F = m * a
where F is the force, m is the mass of the car, and a is the centripetal acceleration. Plugging in the values, we get:
F = (1900 kg) * (188.25 m/s²) = 357,375 N
Finally, we can convert the force to kilonewtons (kN) by dividing by 1000:
F = 357,375 N / 1000 = 357.375 kN
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An engine does 19900 J of work and rejects 5300 J of heat into a cold reservoir whose temperature is 285 K. What would be the smallest possible temperature of the hot reservoir?
Answer:
[tex]T_H = 1335\ J[/tex]
Explanation:
Given,
Work done by engine = 19900 J
rejected energy of heat = 5300 J
temperature = 285 K
efficiency of the engine
[tex]\eta = \dfrac{W}{Q_H}[/tex]
where
[tex]W = Q_H - Q_C[/tex]
[tex]19900 = Q_H - 5300[/tex]
[tex]Q_H = 25200\ J[/tex]
efficiency in terms of temperature
[tex]\eta = \dfrac{T_H-T_C}{T_H}[/tex]
from the above two equation
[tex]\dfrac{Q_H-Q_C}{Q_H}= \dfrac{T_H-T_C}{T_H}[/tex]
[tex]1-\dfrac{Q_C}{Q_H}= 1- \dfrac{T_C}{T_H}[/tex]
[tex]\dfrac{Q_C}{Q_H}= \dfrac{T_C}{T_H}[/tex]
[tex]T_H=T_C\ \dfrac{Q_H}{Q_C}[/tex]
[tex]T_H=285\times \dfrac{25200}{5300}[/tex]
[tex]T_H = 1335\ J[/tex]
The smallest possible temperature of the hot reservoir, given the work done by the engine, heat rejected into the cold reservoir, and temperature of the cold reservoir, is approximately 1357.14 Kelvin.
Explanation:The problem is based on the Carnot heat engine, a theoretical model used for understanding the thermodynamics of heat conversion into work. The efficiency of Carnot's engine depends on the temperatures of the hot and cold reservoirs (Th and Tc respectively). The efficiency, represented as e, is calculated as e = (Th-Tc) / Th.
Before we do the calculation, it's important to note that temperatures must be expressed in Kelvin for this formula to be accurate. Tc, the temperature of the cold reservoir, is given as 285 K. The work done by the engine, represented as W, is 19900 J, and the heat rejected by the engine into the cold reservoir, represented as Qc, is 5300J. The heat absorbed by the engine from the hot reservoir, represented as Qh, can be calculated using the first law of thermodynamics, which states Qh = W + Qc. Therefore, Qh = 19900 J + 5300 J = 25200 J.
Now, to find Th, we rearrange the efficiency formula to Th = Tc / (1 - e), where e is now W / Qh = 19900 J / 25200 J ≈ 0.79. Substituting Tc = 285 K and e ≈ 0.79 into the formula, Th = 285 K / (1 - 0.79) ≈ 1357.14 K. This is the smallest possible temperature of the hot reservoir for the given conditions.
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