Answer:
the aquifer’s hydraulic conductivity is K = 10.039 m/day
transmissivity T = 331.287 m/day
the drawdown 100 m away from the well is s = 1.452 m
Explanation:
Given that :
The constant pumps rate Q = 2000 m³/day
R₁ =160 m → H₁ = 249 m
R₂ = 453 m → H₂ = 250 m
The confined aquifer B is 33 m thick
The hydraulic conductivity K = [tex]\frac{Q*In (\frac{R_1}{R_2}) }{2 \pi B(H_2-H_1)}[/tex]
K = [tex]\frac{2000*In (\frac{160}{453}) }{2 \pi *33(250-249)}[/tex]
K = [tex]\frac{2081.43662}{207.3451151}[/tex]
K = 10.039 m/day
Transmissivity T = K × B
T = 10.039×33
T = 331.287 m/day
TO find the drawdown 100 m away from the well; we have:
K = [tex]\frac{Q* In(\frac{R_2}{R_1} )}{2 \pi B (H_2-H_1)} =\frac{Q* In(\frac{R_2}{R_3} )}{2 \pi B (H_2-H_3)}[/tex]
[tex]\frac{ In(\frac{453}{160} )}{(250-249)} =\frac{ In(\frac{453}{100} )}{ (250-H_3)}[/tex]
H₃ = 248.548 m
Drawdown (s) = H₂ - H₃
s = (250 - 248.548)m
s = 1.452 m
In order for a current to be maintained through a conductor a difference in ________________________ must be maintained from one end of the conductor to the other end.
Answer:
Potential
Explanation:
In order for a current to be maintained through a conductor a difference in potential must be maintained from one end of the conductor to the other end. Due to potential difference and emf is induced in it. As a result current starts to flow.
Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example, one study found that drivers of trains had a higher incidence of blood cancer than other railway workers, possibly due to long exposure to mechanical devices in the train engine cab. Consider a magnetic field of magnitude 1.00 10-3 T, oscillating sinusoidally at 72.5 Hz. If the diameter of a red blood cell is 7.60 µm, determine the maximum emf that can be generated around the perimeter of a cell in this field.
Given Information:
Magnetic field = B = 1×10⁻³ T
Frequency = f = 72.5 Hz
Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m
Required Information:
Maximum Emf = ?
Answer:
Maximum Emf = 20.66×10⁻¹² volts
Explanation:
The maximum emf generated around the perimeter of a cell in a field is given by
Emf = BAωcos(ωt)
Where A is the area, B is the magnetic field and ω is frequency in rad/sec
For maximum emf cos(ωt) = 1
Emf = BAω
Area is given by
A = πr²
A = π(d/2)²
A = π(7.60×10⁻⁶/2)²
A = 45.36×10⁻¹² m²
We know that,
ω = 2πf
ω = 2π(72.5)
ω = 455.53 rad/sec
Finally, the emf is,
Emf = BAω
Emf = 1×10⁻³*45.36×10⁻¹²*455.53
Emf = 20.66×10⁻¹² volts
Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts
A small car of mass m and a large car of mass 4m drive along a highway at constant speeds VS and VL. They approach a curve of radius R. The small and large cars have accelerations as and aL respectively, as they travel around the curve. The magnitude of as is twice of that of aL. How does the speed of the small car VS compare to the speed of the large car VL as they move around the curve
Answer:
[tex]v_S=\sqrt{2}v_L[/tex]
Explanation:
The acceleration experimented while taking a curve is the centripetal acceleration [tex]a=\frac{v^2}{r}[/tex]. Since [tex]a_S=2a_L[/tex], we have that: [tex]\frac{v_S^2}{r_S}=\frac{2v_L^2}{r_L}[/tex]
They take the same curve, so we have: [tex]r_S=r_L=R[/tex]
Which means: [tex]v_S^2=2v_L^2[/tex]
And finally we obtain: [tex]v_S=\sqrt{2}v_L[/tex]
There are several types of drag on a car other than air resistance. Effects having to do with the squeezing of the tires (rolling resistance) and frictional forces in the drivetrain (the system that transfers energy from the engine to the rotation of the wheels) also must be taken into account. Engineers use the following equation to model the total force due to these different effects Fdrag=A+Bv+Cv2 For a Accord, these coefficients are estimated to be A=220.500 N, B=−5.930 N s/m, and C=0.611 N s2/m2. Suppose that the driver steadily accelerates the car from 0 km/hr to 100 km/hr over a 3.5 s. What is the magnitude of the work done by the drag forces?]
Answer:[tex]W=16.837\ kJ[/tex]
Explanation:
Given
Drag force is given by
[tex]F_{drag}=A+Bv+Cv^2[/tex]
for [tex]A=220.5\ N[/tex]
[tex]B=-5.93\ N-s/m[/tex]
[tex]C=0.611N-s/m^2[/tex]
car accelerate from 0 to [tex]100\ km/hr[/tex] in [tex]3.5\ s[/tex]
so acceleration is given by
[tex]v=u+at[/tex]
here u=initial velocity is zero
[tex]v=100\km/hr\approx 27.78\ m/s[/tex]
[tex]27.78=0+a(3.5)[/tex]
[tex]a=7.936\ m/s^2[/tex]
Now work done is given by
[tex]dW=F\cdot vdt[/tex]
[tex]\int_{0}^{W}dW=\int_{0}^{3.5}F\cdot vdt[/tex]
[tex]W=\int_{0}^{3.5}[Av+Bv^2+Cv^3]dt[/tex]
[tex]W=\int_{0}^{3.5}[220.5at-5.93a^2t^2+0.611a^3t^3]dt[/tex]
[tex]W=\int_{0}^{3.5}220.5\times 7.936tdt-\int_{0}^{3.5}5.93\times (7.936)^2t^2dt+\int_{0}^{3.5}0.611\times (7.936)^3t^3dt[/tex]
[tex]W=1749.88[\frac{t^2}{2}]_0^{3.5}-373.47[\frac{t^3}{3}]_0^{3.5}+305.389[\frac{t^4}{4}]_0^{3.5}[/tex]
[tex]W=10,718.015-5337.508+11,456.85[/tex]
[tex]W=16.837\ kJ[/tex]
The work done by drag forces on a car accelerating from 0 to 100 km/hr can be calculated using the equation for total force and its given coefficients, by first converting the velocity to m/s, then finding the average velocity to use it to find the drag force, and finally using the net force and distance covered to calculate the work done.
Explanation:The work done by the drag forces on a car accelerating steadily from 0 km/hr to 100 km/hr over 3.5 s can be calculated using the given equation for total force due to different effects, Fdrag=A+Bv+Cv2, and the coefficients for an Accord car. Please note that before calculating, the velocity needs to be converted from km/hr to m/s as SI units should be consistent.
First, find the average velocity of the car during its acceleration period, which is (0 + final velocity)/2. Then, substitute the values in the given equation Fdrag=A+Bv+Cv2 to find out the Net Force acting on the car.
Next, use this Net Force to find the Work done against the drag forces during the acceleration. The Work done (W) against a force is calculated by the equation W = F * d, where F is the net force (drag force) acting on the body and d is the total distance covered during the acceleration. To find the distance covered by the car, replace 's' in the equation s = ut + 1/2*a*t^2 with the average velocity of the car times the total time (t) it took to accelerate.
This will give you the total work done against drag forces on accelerating the car from 0 km/hr to 100 km/hr over 3.5 s.
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Two speakers produce waves of the same wavelength that are in phase. 1) At a point midway between the speakers, you would expect to hear: Louder sound Softer sound Alternating louder and softer sounds Louder or softer sounds depending on the wavelength No interference You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. (Survey Question) 2) Briefly explain your answer to the previous questio
Answer:
louder
Explanation:
Since the two speakers producing same wavelength that are in phase,at the midpoint, the waves travel the same distance and hence path difference is zero
hence constructive interference takes place , due to this a louder sound is observed .
hence the answer is a) louder
When two speakers produce waves of the same wavelength that are in phase at a point midway between the speakers, constructive interference occurs. This results in the overlapping and combination of the waves to form a wave with higher amplitude, creating a louder sound.
Explanation:The subject of this question involves the principle of wave interference in physics. This is a phenomenon that occurs when two waves come together while traveling through the same medium. At a point midway between the speakers, when two speakers produce waves of the same wavelength that are in phase, taking into account the path lengths traveled by the individual waves, you would expect to hear a louder sound.
This is a case of constructive interference, where the two sound waves, being in phase and of the same wavelength, will overlap and combine to form a wave with a greater amplitude, leading to a louder sound. This is explained in Figure 17.17 and 16.36, where the difference in the path lengths is one wavelength, resulting in total constructive interference and a resulting amplitude equal to twice the original amplitude.
However, it is worthy to note that in the real world recognition of this increased amplitude or louder sound will depend on the specific frequency of the sound, as sonic perception can vary with frequencies. This explanation is in reference to a single tone or frequency. When discussing music which is composed of many frequencies, the actual perception might be a bit more complex.
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what is the the wavelength if a wave cycles up and down three times per second and the distance between each wave is 1.7m?
Answer:5.1 meters
Explanation:
wavelength=3 x1.7
Wavelength=5.1 meters
Two football players collide head-on in midair, moving along the same horizontal direction, while trying to catch a thrown football. The first player is 89.5 kg and has an initial velocity of 6.05 m/s (in the positive direction), while the second player is 111 kg and has an initial velocity of –3.55 m/s. What is their velocity just after impact if they cling together?
Final answer:
The final velocity of the two football players after the collision is 0.738 m/s in the positive direction, calculated using the conservation of momentum.
Explanation:
The scenario described involves a conservation of momentum problem, where two football players collide and cling together. We can solve for the final velocity by using the principle of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces are acting on it.
To find the final combined velocity of the players after the collision, we use the formula:
Momentum before collision = Momentum after collision
(m1 × v1) + (m2 × v2) = (m1 + m2) × v_final
Plugging in the given values:
(89.5 kg × 6.05 m/s) + (111 kg × (-3.55 m/s)) = (89.5 kg + 111 kg) × v_final
After calculating both sides,
541.475 kg·m/s - 393.55 kg·m/s = 200.5 kg · v_final
We get the final combined velocity,
v_final = 147.925 kg·m/s / 200.5 kg
v_final = 0.7377 m/s (to four significant figures)
The players will be moving with a velocity of 0.738 m/s in the positive direction.
Identify three pollutants released into the air when fossil fuels are burned.
Answer:
Carbon Monoxide (CO), Carbon Dioxide (CO2), and Sulfur Dioxide (SO2).
Answer:Carbon dioxide (CO2), Carbon monoxide (CO), Methane (CH4)
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A square loop of wire with a small resistance is moved with constant speed from a field free region into a region of uniform B field (B is constant in time) and then back into a field free region to the right. The self inductance of the loop is negligible
True/False:
a. When leaving the field the coil experiences a magnetic force to the left.
b. Upon entering the field, a clockwise current flows in the loop.
c. Upon leaving the field, a clockwise current flows in the loop.
d. When entering the field the coil experiences a magnetic force to the right.
The loop experiences a magnetic force to the left when leaving the field, a counterclockwise current flows when entering, a clockwise current flows when leaving, and a magnetic force to the left when entering the field.
Explanation:The statement a is true. When leaving the magnetic field, the loop experiences a magnetic force to the left. This is because the magnetic field lines are directed from right to left in the field and the loop opposes the change in magnetic flux by generating a current that creates its own magnetic field, according to Lenz's law.
The statement b is false. Upon entering the field, a counterclockwise current flows in the loop. This is because the increasing magnetic flux induces a current that opposes the increase, as stated by Lenz's law.
The statement c is true. Upon leaving the field, a clockwise current flows in the loop. This is because the decreasing magnetic flux induces a current that opposes the decrease, according to Lenz's law.
The statement d is false. When entering the field, the coil experiences a magnetic force to the left. This is because the magnetic field lines are directed from left to right in the field and the loop opposes the change in magnetic flux by generating a current that creates its own magnetic field, according to Lenz's law.
The portion of a cello string between the bridge and upper end of the fingerboard (that part of the string that is free to vibrate) is of length 60.0 cm, and this length of the string has mass 2.00 g. The string sounds an A4 note (440 Hz) when played.
Where must the cellist put a finger (what distance x from bridge) to play a D5 note (587 Hz)? For both notes, the string vibrates in its fundamental mode.
Answer:
Explanation:
length of vibration l = .6 m
mass per unit length m = 2 x 10⁻³ / .6
= 3.33 x 10⁻³ kg/ m
n = [tex]\frac{1}{2l} \sqrt{\frac{T}{m} }[/tex]
n is frequency of vibration , l is length , T is tension in the string .
Apply this formula in the first case
440 = [tex]\frac{1}{2\times.6} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]
Apply this formula for second case
n = [tex]\frac{1}{2\times.6} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]
587 = [tex]\frac{1}{2\times l} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]
Dividing
[tex]\frac{440}{587}[/tex] = [tex]\frac{l}{.6}[/tex]
l = .45 m .
The distance from the bridge where the cellist must put his finger to play the D5 note is 45 cm.
The given parameters;
length o the string at A4, l = 60 cmmass, m = 2.0 gfrequency of A4 note, f = 440 Hzfrequency of D5 note, f = 5587 HzThe frequency of a sound wave in a stretched string is calculated as;
[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]
where;
T is the tension in the stringμ is the mass per unit length[tex]f_1(2l_1) = f_2(2l_2)\\\\f_1l_1 = f_2l_2\\\\l_2 = \frac{f_1l_1}{f_2} \\\\l_2 = \frac{440 \times 0.6}{587} \\\\l_2 = 0.45 \ m\\\\l_2 = 45 \ cm[/tex]
Thus, the distance from the bridge where the cellist must put his finger to play the D5 note is 45 cm.
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A coil is wrapped with 140 turns of wire on the perimeter of a square frame of sides 35 cm. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 1.46 . A uniform magnetic field is turned on perpendicular to the plane of the coil. If the field changes linearly from 0 to −0.426 Wb/m2 in a time of 1.19 s, find the magnitude of the induced emf in the coil while the field is changing. Answer in units of V.
The magnitude of the induced emf in the coil, calculated using Faraday's Law of Electromagnetic Induction, is 6.18 V when a uniform magnetic field changes linearly over a time of 1.19 seconds.
To find the magnitude of the induced emf in the coil while the magnetic field is changing, we will apply Faraday's Law of Electromagnetic Induction which states that the emf induced in a circuit is proportional to the rate of change of the magnetic flux through the circuit. Given that the coil has 140 turns (N) and each side of the square frame is 35 cm (which we need to convert to meters to maintain SI units, 0.35 m), we can calculate the area (A) of the square frame as:
A = side x side = 0.35 m x 0.35 m = 0.1225 m2
The magnetic field (B) changes from 0 to -0.426 Wb/m2 over a time (t) of 1.19 s. The rate of change of magnetic field (dB/dt) is:
dB/dt = - ( 0 - (-0.426) ) / 1.19 s = 0.358 Wb/m2/s
Now using Faraday’s Law, the induced emf (ε) is given by:
ε = -N x (dΦ/dt) = -N x A x (dB/dt)
Plugging in the values we get:
ε = -140 x 0.1225 m2 x 0.358 Wb/m2/s = -6.18 V
The magnitude of the induced emf is therefore 6.18 V (taking the absolute value as the question asks for magnitude).
A large, 36.0 kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell’s center of mass is 0.55 m below the pivot. The bell’s moment of inertia about an axis at the pivot is 36.0 kg ·m2 . The clapper is a small, 2.8 kg mass attached to one end of a slender rod of length L and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length L of the clapper rod for the bell to ring silently — that is, for the period of oscillation for the bell to equal that of the clapper?
Answer:
Length of the rod is 0.043m
Explanation:
Mass of the bell (M) = 36kg
Distance of the center of mass from pivot = 0.55m
Moment of inertia (I) = 36.0kgm²
Mass of clipper (m) = 2.8kg
Length of the bell to ring = L
The period of the pendulum with small amplitude of oscillation is
T = 2π √(I / mgd)
Where g = acceleration due to gravity = 9.8 m/s²
T = 2π √(36 / 36*9.8*0.55)
T = √(0.1855)
T = 0.43s
The period of the pendulum is 0.43s
To find the length of the Clapper rod for which the bell ring slightly,
T = 2π√(L / g)
T² = 4π²L / g
T² *g = 4π²L
L = (T² * g) / 4π²
L = (9.8* 0.43²) / 4π²
L = 1.7287 / 39.478
L = 0.043m
The length of the Clapper rod for the bell to ring slightly is 0.043m
our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three on the right (total mass mp=10.1 kgmp=10.1 kg). The distance from the fulcrum to the center of the pile of rocks is rp=0.360 m.rp=0.360 m. Answer the two questions below, using three significant digits. Part A: What is the value of the torque (????pτp) produced by the pile of rocks? (Enter a positive value.)
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The torque produced by the pile of rocks is [tex]\tau = 35.63\ N \cdot m[/tex]
b
The distance of the single for equilibrium to occur is [tex]r_s =1.62 \ m[/tex]
Explanation:
From the question we are told that
The mass of the left rock is [tex]m_s = 2.25 \ kg[/tex]
The mass of the rock on the right [tex]m_p = 10.1 kg[/tex]
The distance from fulcrum to the center of the pile of rocks is [tex]r_p = 0.360 \ m[/tex]
Generally the torque produced by the pile of rock is mathematically represented as
[tex]\tau = m_p * g * r_p[/tex]
Substituting values
[tex]\tau = 10.1 * 9.8 * 0.360[/tex]
[tex]\tau = 35.63\ N \cdot m[/tex]
Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows
The torque due to the single rock is
[tex]\tau = m_s * g * r_s[/tex]
At equilibrium the both torque are equal
[tex]35.63 = m_s * r_s * g[/tex]
Making [tex]r_s[/tex] the subject of the formula
[tex]r_s = \frac{35.63 }{m_s * g}[/tex]
Substituting values
[tex]r_s = \frac{35.63 }{2.25 * 9.8}[/tex]
[tex]r_s =1.62 \ m[/tex]
Two students are holding opposite ends of a spring in a classroom. One student stands on the left end of the classroom and the other stands at the right end. They shake the spring so that a longitudinal wave travels along the spring. In which directions will the longitudinal wave oscillate?
Answer:
Up And Down
Explanation:
In this case, the particles of the medium move parallel to the direction that the pulse moves. This type of wave is a longitudinal wave. Longitudinal waves are always characterized by particle motion being parallel to wave motion.
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Final answer:
In a longitudinal wave on a horizontal spring, the medium oscillates back and forth horizontally, parallel to the direction of the wave's travel.
Explanation:
The question pertains to the behavior of longitudinal waves in a spring. When one student shakes the spring to create a longitudinal wave, the oscillation of the spring occurs in the same direction as the wave's propagation. This means that in a longitudinal wave, the medium—the spring in this case—oscillates parallel to the wave's direction of motion. Therefore, if the spring is held horizontally and the wave travels from left to right, the individual coils of the spring will move left and right along the same horizontal line, compressing and expanding as the wave passes through.
In contrast, with transverse waves, the medium moves perpendicularly to the direction of the wave's travel, such as in the motion of a rope being moved up and down while the wave travels horizontally.
Submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1 400 Hz. The speed of sound in the water is 1 533 m/s. A second submarine (sub B) is located such that both submarines are traveling directly toward each other. The second submarine is moving at 9.00 m/s. What frequency is detected by an observer riding on sub B as the subs approach each other?
Answer:
1385 hz
Explanation:
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Which side of the electromagnetic spectrum has shorter wavelengths?
Which side of the electromagnetic spectrum has higher energy?)
Which side of the electromagnetic spectrum has higher frequencies?
Gamma rays have the smallest wavelengths and the most energy of any wave in the electromagnetic spectrum. They are produced by the hottest and most energetic objects in the universe, such as neutron stars and pulsars, supernova explosions, and regions around black holes.
What is wavelength ?
"Distance between corresponding points of two consecutive waves." “Corresponding points” refers to two points or particles in the same phase—i.e. points that have completed identical fractions of their periodic motion.
What is frequency?
"Tthe number of waves that pass a fixed point in unit time. "It also describes the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.
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Answer:
1.blue
2.blue
3.blue
Explanation:
I took the quiz
QUESTIONS
A 800 kg person sprints up a 700 meter high stairway in 4.00 seconds flat. What is their power output in watts?
Answer:
Power;P = 1372000 W
Explanation:
We are given;
Mass of person;m = 800 kg
Distance sprinted;h = 700m
Time taken;t = 4 seconds
Now, the formula for Power is;
Power = Work done/Time taken
Where work done = Force x Distance
Now, Force is expressed as;
F = mg
Where m is mass and g is acceleration due to gravity = 9.8 m/s²
Thus, plugging in the relevant values into power equation gives;
Power;P = 800 x 9.8 x 700/4
Power;P = 1372000 W
A long uniform wooden board (a playground see-saw) has a pivot point at its center. An older child of mass M=32 kg is sitting a distance L from the pivot. On the other side of the pivot point are two smaller children each of mass M/2. One is sitting a distance L/6 from the pivot. How far from the pivot must the other small child be sitting in order for the system to be balanced?
Answer:[tex]\frac{5L}{6}[/tex]
Explanation:
Given
Wooden board is pivoted at center and
Older child of mass [tex]M=32\ kg[/tex] is sitting at a distance of L from center
if two child of mass [tex]\frac{M}{2}[/tex] is sitting at a distance [tex]\frac{L}{6}[/tex] and [tex]x[/tex](say) from pivot then net torque about pivot is zero
i.e.
[tex]\Rightarrow \tau_{net}=MgL-\frac{M}{2}g\frac{L}{6}-\frac{M}{2}gx[/tex]
as [tex]\tau_{net}=0[/tex]
Therefore
[tex]MgL=\frac{M}{2}g\frac{L}{6}+\frac{M}{2}gx[/tex]
[tex]L-\frac{L}{6}=x[/tex]
[tex]x=\frac{5L}{6}[/tex]
Therefore another child is sitting at a distance of [tex]\frac{5L}{6}[/tex]
What is the most common and powerful agent of erosion? wind ice water animal behavior
Answer:
water
Explanation:
Answer:water
Explanation:because it shaped the grand canyon
A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.42). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad>s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle.
a. Is angular momentum conserved?
b. Find the change in kinetic energy of the block, in J.
c. How much work was done in pulling the cord? in J.
Answer:
W= [tex]K_2-K_1==9.12\times10^{-3} J[/tex]
Explanation:
a) Yes, In the absence of external torques acting on the system, the angular momentum is conserved.
b) By the law of conservation of energy angular momentum
[tex]L_1=L_2[/tex]
[tex]I_1\omega_1=I_2\omega_2[/tex]
[tex]mr_1^2\omega_1=mr_2^2\omega_2\\\omega_2=(\frac{r_1}{r_2} )^2\omega_1[/tex]
[tex]\omega_2=(\frac{0.3}{0.15})^2\times2.85[/tex]
[tex]\omega_2=5.7\text{ rad/sec}[/tex]
c) work done in pulling the chord W= Final kinetic energy(K_2)-Initial Kinetic energy(K_1)
[tex]K_1=\frac{1}{2} mr_1^2\omega_1^2[/tex]
[tex]K_1=\frac{1}{2} \times0.025\times0.3^2\times2.85^2[/tex]
[tex]=9.12\times10^{-3} J[/tex]
Now,
[tex]K_2=\frac{1}{2} mr_2^2\omega_2^2[/tex]
[tex]K_2=\frac{1}{2} \times0.025\times0.15^2\times5.7^2[/tex]
[tex]K_2=18.24\times10^{-3}[/tex] J
Therefore, Work done W= [tex]K_2-K_1==9.12\times10^{-3} J[/tex]
a. Angular momentum is conserved in this system. b. The change in kinetic energy of the block can be calculated. c. No work is done in pulling the cord.
Explanation:a. Angular momentum is conserved when there are no external torques acting on the system. In this case, since the block is revolving on a frictionless surface, and there is no mention of any external torques, we can assume that angular momentum is conserved.
b. The change in kinetic energy of the block can be calculated using the equation ΔKE = KE_final - KE_initial. Since the block is modeled as a particle, its kinetic energy is given by KE = 1/2 * m * v^2, where m is the mass and v is the linear velocity. As the radius is changed, the linear velocity changes, and we can calculate the change in kinetic energy.
c. The work done in pulling the cord can be calculated using the equation W = ΔKE + ΔPE, where W is the work done, ΔKE is the change in kinetic energy, and ΔPE is the change in potential energy. In this case, since the block is on a frictionless surface, there is no change in potential energy, and we only need to calculate the change in kinetic energy.
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A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the dry steel frying pan is only 0.100 N. Knowing the coefficient of kinetic friction between the two materials (0.3), he quickly calculates the normal force in newtons. What is it
Answer:
Normal force = 3N
Explanation:
We are given;
Coefficient of kinetic friction friction; μ_k = 0.3
Frictional force;F_f = 0.1N
Now,the formula for frictional force is;
F_f = μ_k*N
Where μ_k is coefficient of friction and N is the normal force.
So, making N the subject, we have;
N = μ_k/F_f
Plugging in the relevant values to obtain;
N = 0.3/0.1
N = 3N
The noise level coming from a pig pen with
199 pigs is 74.3 dB. The intensity of the sound
coming from the pig pen is proportional to the
number of pigs.
Assuming each of the remaining pigs squeal
at their original level after 61 of their compan-
ions have been removed, what is the decibel
level of the remaining pigs?
Answer in units of dB.
The decibel level of the remaining pigs is 34.02dB.
To solve this problem, we need to understand the relationship between the number of pigs and the intensity of sound. Since the intensity is proportional to the number of pigs, we can use the formula:
[tex]\[ \text{Intensity} = k \times \text{Number of pigs} \][/tex]
where ( k ) is a constant of proportionality.
Given that the noise level from 199 pigs is 74.3 dB, we can set up the equation:
[tex]\[ 74.3 = k \times 199 \][/tex]
First, let's solve for ( k ):
[tex]\[ k = \frac{74.3}{199} \][/tex]
Now, we can use this value of ( k ) to find the intensity when only 138 pigs remain:
[tex]\[ \text{Intensity}_{\text{new}} = k \times 138 \][/tex]
Finally, we can convert this intensity back into decibels using the formula:
[tex]\[ \text{Noise level (dB)} = 10 \log_{10}(\text{Intensity}) \][/tex]
Now, let's calculate:
[tex]\[ k = \frac{74.3}{199} \][/tex]
[tex]\[ k \approx 0.3726 \][/tex]
[tex]\[ \text{Intensity}_{\text{new}} = 0.3726 \times 138 \][/tex]
[tex]\[ \text{Intensity}_{\text{new}} = 51.36 \][/tex]
[tex]\[ \text{Noise level (dB)} = 10 \log_{10}(51.36) \][/tex]
[tex]\[ \text{Noise level (dB)} = 34.02 \][/tex]
So, the decibel level of the remaining pigs is 34.02dB.
A researcher with the Ministry of Transportation is commissioned to study the drive times to work (one-way) for U.S. cities. The underlying hypothesis is that average commute times are different across cities. To test the hypothesis, the researcher randomly selects six people from each of the four cities and records their one-way commute times to work. Refer to the below data on one-way commute times (in minutes) to work. Note that the grand mean is 36.625. Houston Charlotte Tucson Akron 45 25 25 10 65 30 30 15 105 35 19 15 55 10 30 10 85 50 10 5 90 70 35 10 x¯i 74.167 36.667 24.833 10.833 s2i 524.167 436.667 82.167 14.167 The competing hypotheses about the mean commute times are ______________ a) H0: μ1 = μ2 = μ3, HA: Not all population means are equal H0: b) Not all population means are equal, HA: μ1 = μ2 = μ3 H0: μ1 = μ2 = μ3 = μ4, HA: c) Not all population means are equal H0: d) Not all population means are equal, HA: μ1 = μ2 = μ3 = μ4
Answer:
d) H0: Not all population means are equal,
HA: μ1 = μ2 = μ3 = μ4
Explanation:
The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.
Therefore, for the case above;
Since the underlying hypothesis is that average commute times are different across cities.
Null hypothesis H0: is that Not all population means are equal.
Alternative hypothesis Ha: μ1 = μ2 = μ3 = μ4
g I can test a new wheel design by rolling it down a test ramp. I release a wheel of mass m=1.6 kg and radius r=0.37 m from rest at an initial height of h=6.7 m at the top of a test ramp. It rolls smoothly to the bottom without sliding. I measure the linear speed of the wheel at the bottom of the test ramp to be v=4.7 m/s. What is the rotational inertia of my wheel?
Answer:
The rotational inertia of my wheel is [tex]I =1.083 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The mass of the wheel is [tex]m = 1.6 \ kg[/tex]
The radius of the wheel is [tex]r = 0.37 \ m[/tex]
The height is [tex]h = 6.7 m[/tex]
The linear speed is [tex]v = 4.7 m/s[/tex]
According to the law of energy conservation
[tex]PE = KE + KE_R[/tex]
Where PE is the potential energy at the height h which is mathematically represented as
[tex]PE = mgh[/tex]
While KE is the kinetic energy at the bottom of height h
[tex]KE = \frac{1}{2} mv^2[/tex]
Where [tex]KE_R[/tex] is the rotational kinetic energy which is mathematically represented as
[tex]KE_R = \frac{1}{2} * I * \frac{v^2}{r^2}[/tex]
Where [tex]I[/tex] is the rotational inertia
So substituting this formula into the equation of energy conservation
[tex]mgh = \frac{1}{2} mv^2 + \frac{1}{2} * I * \frac{v^2}{r^2}[/tex]
=> [tex]I =[ \ mgh - \frac{1}{2} mv^2 \ ]* \frac{2 r^2}{v^2}[/tex]
substituting values
[tex]I =[ \ 1.6 * 9.8 * 6.7 - \frac{1}{2} * 1.6 *4.7^2 \ ]* \frac{2 * 0.37^2}{4.7^2}[/tex]
[tex]I =1.083 \ kg \cdot m^2[/tex]
Sarah is 14 years old and skips school twice a week without any written explanation. What can she be charged with?
Answer:
Too many people are unaware or indifferent to that.” Fines can cost up to $500 per truancy, due within 30 days unless a judge gives an extension. For many students and families, it's another debt they can't pay. And if fines aren't paid, they can convert into an arrest warrant when a student turns 17.
Explanation:
A 0.415-kg mass suspended from a spring undergoes simpleharmonic oscillations with a period of 1.4 s. How much mass, inkilograms, must be added to the object to change the period to2.2 s?
Answer:
m=0.893kg
Explanation:
time period of oscillations is given by= 2π√(m/k)
m: mass of the object
k: spring constant
when T=1.5 and m=0.415
1.5= 2π√(0.415/k)
k= 7.27 N/m
when T= 2.2s
2.2= 2π√(m/7.27)
m=0.893kg
The mass in kilograms that must be added to the object to change the period to 2.2 s is 1.025kg
The formula for calculating the period of a simple pendulum is expressed as:
[tex]T= 2\pi \sqrt{\frac{m}{k} }[/tex]
m is the mass of the spring
k is the spring constant
Given the following parameters
m = 0.415kg
T = 1.4 secs
Get the spring constant
[tex]1.4=2(3.14)\sqrt{\frac{0.415}{k} } \\1.4=6.28\sqrt{\frac{0.415}{k} }\\ 0.2229=\sqrt{\frac{0.415}{k} }\\ \frac{0.415}{k} =0.0497\\0.0497k=0.415\\k=\frac{0.415}{0.0497}\\k= 8.35N/m[/tex]
Given the period is 2.2secs, the mass of the spring will be expressed as:
[tex]2.2=2 (3.14)\sqrt{\frac{m}{8.35} } \\2.2=6.28\sqrt{\frac{m}{8.35} }\\\sqrt{\frac{m}{8.35} }=0.3503\\\frac{m}{8.35} =0.1227\\m=1.025kg[/tex]
Hence the mass in kilograms that must be added to the object to change the period to 2.2 s is 1.025kg
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Albert and Emmy purchase identical state-of-the-art atomic watches to take on their interplanetary entomology expedition. After a busy day collecting spacebug specimens, Albert waits in the main spaceship while Emmy flies a shuttle to the nearest food court to pick up dinner. As Emmy flies by at a very high but constant velocity, Albert measures the rate at which Emmy's watch is ticking and compares it with his watch.
Albert observe that Emmy's watch is ticking ________ his own watch.
A. more slowly than
B. more quickly than
C. at the same rate as
Answer:
Option is A, Albert observes that Emmy's watch is ticking more slowly than his watch.
Explanation:
According to Einstein's theory of relativity, the time between two observers, as one of them is moving respect to the other, is different, occurring the known time dilation. This theory indicates that, for the observer that is in an inertial frame of reference, he will measure a clock that is moving relative to him, to tick slower than his clock that is at rest. The faster the relative velocity, the greater the time dilation between the two of them.
In our case, the person that is at rest (Albert) will measure the clock of the observer that is moving (Emmy) as ticking slower than his own watch, because Emmy is moving at high speed in relation to Albert (that is at rest), thus, time dilation is occurring for Emmy.
Therefore, the correct option is A, Albert observes that Emmy's watch is ticking more slowly than his watch.
I hope it helps you!
Nitrogen is stored in a large chamber under conditions of 450 K and 1.5 × 105 N/ m2 . The gas leaves the chamber through a convergent-only nozzle whose outlet area is 30 cm2 . The ambient room pressure is 1 × 105 N/m2 and there are no losses. (a) What is the velocity of the nitrogen at the nozzle exit? (b) What is the mass flow rate? (c) What is the maximum flow rate that could be obtained by lowering the ambi- ent pressure?
Answer:
a) 319.56 m/s
b) 0.8057 kg /s
c) 0.840 kg /s
Explanation:
Check the pictures attached for the whole explanation
Final answer:
The velocity of nitrogen at the nozzle exit and the mass flow rate depend on the conditions and properties of the gas, using principles such as the Bernoulli equation and the ideal gas law. Without specific constants, these values cannot be determined precisely. The maximum flow rate would be obtained by reducing the ambient pressure below the critical pressure ratio, causing choked flow.
Explanation:
To answer the student's question about the velocity of the nitrogen gas at the nozzle exit, we need to apply the principles of fluid dynamics. Under the provided conditions, without any losses and assuming adiabatic expansion, the velocity can be calculated using the Bernoulli equation and the equation of continuity. However, given that we only have the inlet pressure, outlet pressure, and temperature, we need additional equations like the ideal gas law to find the density at the exit. Unfortunately, without the specific heat ratio or gas constant for nitrogen, a precise calculation cannot be provided.
The mass flow rate can then be calculated by multiplying the outlet velocity by the outlet area and the density of nitrogen under the given conditions. An approximation can be made, but with the given data, it is not possible to provide an exact value since the outlet velocity is not known.
Regarding the maximum flow rate, it is related to the critical pressure ratio of the gas. If the ambient pressure is decreased below a certain critical point, the flow becomes choked, and the mass flow rate reaches a maximum. This maximum is a function of the chamber conditions and the specific properties of nitrogen.
While Dr. Chesnutt is making breakfast, she turns on her 1060 W toaster, 500 W coffee pot, and 1230 W microwave at the same time. Part Description Answer Save Status A. If all three appliances are connected in parallel across a 120 V power source, what total current would they draw from the source? (include units with answer) Format Check Click here to check your answer 4 pts.100% 13% try penalty 8 try limit # tries: 0
Answer:23.25 A
Explanation:
Given
Rating of toaster [tex]P_1=1060\ W[/tex]
Coffee pot [tex]P_2=500\ W[/tex]
microwave [tex]P_3=1230\ W[/tex]
Voltage applied [tex]V=120\ V[/tex]
if they are connected in parallel then all three operates at same voltage
so their resistance are
[tex]P=\frac{V^2}{R}[/tex]
thus [tex]R_1=\frac{V^2}{P_1}=\frac{120^2}{1060}[/tex]
[tex]R_1=13.58\ \Omega[/tex]
[tex]R_2=\frac{V^2}{P_2}=\frac{120^2}{500}[/tex]
[tex]R_2=28.8\ \Omega[/tex]
[tex]R_3=\frac{V^2}{P_3}=\frac{120^2}{1230}[/tex]
[tex]R_3=11.707\ \Omega[/tex]
and [tex]V=IR[/tex]
where I=current
thus [tex]I_1=\frac{V}{R_1}=\frac{120}{13.58}[/tex]
[tex]I_1=8.83\ A[/tex]
[tex]I_2=\frac{V}{R_2}=\frac{120}{28.8}[/tex]
[tex]I_2=4.16\ A[/tex]
[tex]I_3=\frac{V}{R_3}=\frac{120}{11.707}[/tex]
[tex]I_3=10.25\ A[/tex]
Total current [tex]I=I_1+I_2+I_3=23.24\ A[/tex]
Which resistors in the circuit must always have the same current?
Answer:
Resistors in series in the circuit must always have the same current
Explanation:
Resistors are said to be connected in series if they are connected one after another.
The total resistance in the circuit with resistors connected in series is equal to the sum of individual resistances.
Individual resistors in series do not get the total source voltage. Total source voltage divide among them.
Answer:
it's a and d
Explanation:
because it all goes around if that makes sense.. and if u guys are here from apex