Answer:
[tex]A = 1.056\,m^{2}[/tex]
Explanation:
The portion of solar energy converted into electric energy is given by the following equation:
[tex]\dot E = \eta \cdot I\cdot A[/tex]
The area needed to produce energy is derived by clearing the corresponding variable:
[tex]A = \frac{\dot E}{\eta \cdot I}[/tex]
[tex]A = \frac{75\,W }{(0.1)\cdot (710\,\frac{W}{m^{2}} )}[/tex]
[tex]A = 1.056\,m^{2}[/tex]
What pollutant forms when automobile emissions react with oxygen gas and ultraviolet rays?
Ozone is a molecule that consists of three oxygen atoms. When high-energy ultraviolet rays strike ordinary oxygen molecules (O2), they split the molecule into two single oxygen atoms, known as atomic oxygen. A freed oxygen atom then combines with another oxygen molecule to form a molecule of ozone. Hydrocarbons and nitrogen oxides come from great variety of industrial and combustion processes. Motor vehicle exhaust and industrial emissions, gasoline vapors, and chemical solvents are some of the major sources of NOx and VOC that acts as a precursor of ozone. In urban areas, the number of automobiles are more and therefore, more production of such harmful gases. These gases in presence of sunlight leads to the formation of bad ozone.
What are the strengths and limitations of the doppler method?
Answer:
Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on
Limitations: yield only planet's mass and orbital properties
Explanation:
Strengths: yield planet's size
Limitations: few planets have the necessary orbital alignment to be detectable
The Doppler method is a valuable tool for detecting and characterizing exoplanets, particularly those with significant mass and short orbital periods. However, it has limitations in terms of its sensitivity to certain types of planets and the limited information it provides about exoplanet properties. So, option A is correct.
The Doppler method is particularly effective in detecting massive exoplanets that are relatively close to their host stars. It has been successful in identifying hundreds of exoplanets since its inception.The Doppler method is more sensitive to massive exoplanets that are closer to their host stars. This means that it may miss smaller, more distant planets or those with longer orbital periods. The method also has difficulty detecting exoplanets with orbits that are nearly aligned with the line of sight, resulting in underrepresentation of such planets in detection statistics.
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Complete question is below
What are the strengths and limitations of the doppler method?
A.it has limitations in terms of its sensitivity to certain types of planets and the limited information it provides about exoplanet properties.
B.it has NO limitations in terms of its sensitivity to certain types of planets and the limited information it provides about exoplanet properties
A 66-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.90 s. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.6 s, and then comes to rest. What does the spring scale register in each of the following time intervals?
Answer:735.46N, 626.34N
Explanation: when an elevator ascend, it means it moving against gravity
F= m(a+g)
M=66kg
a=v/t
g=9.81m/s^2
For the first a= 1.2/0.9
a=1.2/0.9
a=1.33m/s2
F=66(1.33+9.81)
F=66(11.143)
F=735.46N
F=m(a+g)
a=v/t
a=-1.6/5
a=-0.32m/s2
F=66(-0.32+9.8)
F=66(9.49)
F=626.34N
Find the magnitude of the torque produced by a 4.5 N force applied to a door at a perpendicular distance of 0.26 m from the hinge. Answer in units of N · m.
Answer:
The required torque is 1.17 N-m.
Explanation:
The given data :-
The magnitude of force ( F ) = 4.5 N.
The length of arm ( r ) = 0.26 m.
Here given that force is applied at perpendicular means ( ∅ ) = 90°.
The torque ( T ) is given by
T = F * r * sin∅
T = 4.5 * 0.26 * sin 90°
T = 4.5 * 0.26 * 1
T = 1.17 N-m.
The magnitude of the torque produced by the given force perpendicular to the lever is 1.17N.m.
Given the data in the question;
Force; [tex]F = 4.5N[/tex]Perpendicular distance or radius; [tex]r = 0.26m[/tex]Since the force is perpendicular to the lever, Angle; [tex]\theta = 90^o[/tex]Torque; [tex]T = \ ?[/tex]
Torque simply the measure of the force that can cause an object to rotate about an axis. It is expressed as:
[tex]T = rFsin\theta[/tex]
Where r is radius, F is force applied and θ is the angle between the force and the lever arm.
We substitute our given values into the equation;
[tex]T = 0.26m * 4.5N * sin90^o\\\\T = 0.26m * 4.5N * 1\\\\T = 1.17N.m[/tex]
Therefore, the magnitude of the torque produced by the given force perpendicular to the lever is 1.17N.m.
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An aminoacyl-trna that enters the a site of the ribosome will next occupy which site?
Answer:
The P site
Explanation:
After the formation of peptide linkage, the trna of P site becomes free, more to the E site and then slip away. The A site trna has dipeptide instead of a single aminoacid.
In the prescence of translocase and energy from GTP, the ribosome move in such a way the peptidyl bearing trna of A site come to lie with P sit. This exposes next codon to A site. A new aminoacid trna complex reaches the fresh codon exposed at A site.
An aminoacyl-tRNA molecule that enters the A site of the ribosome next moves to the P site during protein synthesis. This is part of the process known as translocation. After the P site, it moves to the E site, then exits the ribosome.
Explanation:In protein synthesis, an aminoacyl-tRNA that enters the A site of the ribosome will next occupy the P site. This transition between the A site (aminoacyl site) and the P site (peptidyl site) happens during the process known as translocation, which is driven by elongation factors. Once in the P site, the tRNA molecule will be linked with the growing polypeptide chain. After this the tRNA will move to the E site (exit site) and leave the ribosome.
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During a certain comet’s orbit around the Sun, its closest distance to the Sun is 0.6 AU, and its farthest distance from the Sun is 35 AU. At what distance will the comet’s orbital velocity be the largest?
Answer:
At the closest point
Explanation:
We can simply answer this question by applying Kepler's 2nd law of planetary motion.
It states that:
"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"
In this problem, we have a comet orbiting around the Sun:
- Its closest distance from the Sun is 0.6 AU
- Its farthest distance from the Sun is 35 AU
In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: [tex]A\propto vr[/tex], therefore if r is larger, then v (velocity) must be lower).
On the other hand, when the the comet is closer to the Sun the line must move faster ([tex]A\propto vr[/tex], if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.
A voltage source of 10 V is connected to a series RC circuit where R = 2.0 ´ 106 W , and C = 3.0 µF. Find the amount of time required for the current in the circuit to decay to 5% of its original value. Hint: This is the same amount of time for the capacitor to reach 95% of its maximum charge.
Answer:
0.5 s
Explanation:
The discharge current in a series RC circuit is given by
[tex]i = \frac{V_{0} }{R} e^{-\frac{t}{RC} }[/tex] where i₀ = V₀/R
For the current i to decay to 5%i₀, i/i₀ = 0.05. Where R = resistance = 2.0 × 10⁶ Ω and C = capacitance = 3.0 μF = 3.0 10⁻⁶ F
So,
[tex]i = {i_{0}e^{-\frac{t}{RC} }[/tex]
[tex]{\frac{i}{i_{0} } = e^{-\frac{t}{RC}[/tex]
[tex]0.05 = e^{-\frac{t}{2 X 10^{6} X 3 X 10^{-6}}\\[/tex]
[tex]0.05 = e^{-\frac{t}{6}[/tex]
taking natural logarithm of both sides
-t /6 = ln(0.05)
t = -ln(0.05)/6 = 0.5 s
So it takes the current 0.5 s to reach 5% of its original value
In an RC circuit, the time required for the current to decay to 5% of its original value, or for the capacitor to reach 95% of its maximum charge, is approximately three times the circuit's time constant. Here, the time constant is 6 seconds, so the time required is approximately 3 * 6 = 18 seconds.
Explanation:In an RC circuit, the time required for the current to decay to a certain percentage of its original value is connected to the circuit's time constant. The time constant is calculated by the formula τ = R*C. In your case, R equals 2.0 * 106 ohms and C equals 3.0 * 10-6 farads, so τ = 2.0 * 106 * 3 * 10-6 = 6 seconds.
The time required for the current to decay to 5% of its original value or for the capacitor to reach 95% of its maximum charge in an RC circuit is roughly three time constants, or 3τ.From the calculation above, τ equals 6 seconds, so the time required is approximately 3 * 6 = 18 seconds.
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A hollow aluminum cylinder 19.0 cm deep has an internal capacity of 2.000 L at 23.0°C. It is completely filled with turpentine at 23.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 91.0°C. (The average linear expansion coefficient for aluminum is 2.4 x10^-5/°C, and the average volume expansion coefficient for turpentine is 9.0 x10^-4/°C.)
(a) How much turpentine overflows?
in cm^3?
(b) What is the volume of turpentine remaining in the cylinder at 91.0°C? (Give you answer to four significant figures.)
in cm^3?
(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder's rim does the turpentine's surface recede?
in cm?
Answer:
Explanation:
The approach of Linear expansivity was used in solving the problem and the detailed steps and appropriate calculation is carefully shown in the attached file.
Answer:
a. 112.608 cm³ b. 2010 cm³ c. 0.032 cm
Explanation:
Given
Volume of aluminium cylinder = 2.0 L= 2.0 × 10³ cm³
Length of aluminium cylinder = 19.0 cm
Volume of turpentine = 2.0 L= 2.0 10³ cm
Initial temperature of aluminium cylinder = 23.0 °C
Initial temperature of turpentine = 23.0 °C
Average linear expansion coefficient of aluminium α = 2.4 × 10⁻⁵ °C
Average volume expansion coefficient of turpentine γ₁ = 9.0 × 10⁻⁴ °C
(a) How much turpentine overflows in cm³?
We calculate the volume expansion of the aluminium cylinder and that of the turpentine and then subtract the difference.
The volume expansion of material V = V₀(1 + γΔθ)
where V = final volume, V₀ = initial volume, γ = volume expansion and Δθ = temperature change.
For aluminium V₀ = 2000 cm³, γ = 3α = 3 × 2.4 × 10⁻⁵ /°C = 7.2 × 10⁻⁵ /°C
Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C
V₁ = 2000(1 + 7.2 × 10⁻⁵ /°C × 68°C) = 2000(1 + 0.004896) = 2000(1.004896) = 2009.792 cm³
For turpentine V₀ = 2000 cm³, γ = 9.0 × 10⁻⁴ /°C
Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C
V₂= 2000(1 + 9.0 × 10⁻⁴ /°C × 68°C) = 2000(1 + 0.0612) = 2000(1.0612) = 2122.4 cm³.
The volume of turpentine that overflows = V₂ - V₁ = 2122.4 cm³ - 2009.792 cm³ = 112.608 cm³
(b) What is the volume of turpentine remaining in the cylinder at 91.0°C? (Give you answer to four significant figures.) in cm³?
The volume of turpentine remaining is the final volume of the turpentine minus overflow = 2122.4 cm³ - 112.608 cm³ = 2009.792 cm³ = 2010 cm³ to four significant figures.
(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder's rim does the turpentine's surface recede in cm?
The linear expansion for the aluminium to 91 C from 23 C is L = L₀(1 + Δθ)
For aluminium L₀ = 19 cm, α = 2.4 × 10⁻⁵ /°C =
Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C
L = 19(1 + 2.4 × 10⁻⁵ /°C × 68°C) = 19(1 + 0.001632) = 19(1.001632) = 19.031 cm
The linear expansion for the aluminium from 91 C to 23 C is L = L₀(1 + Δθ)
For aluminium L₀ = 19 cm, α = 2.4 × 10⁻⁵ /°C =
Δθ = θ₂ - θ₁ = 23 °C - 91 °C = -68 °C
L₁ = 19.031(1 + 2.4 × 10⁻⁵ /°C × -68°C) = 19.031(1 - 0.001632) = 19.031(0.998368) = 18.999 cm
How far below the rim it recedes is L - L₁ = 19.031 cm - 18.999 cm = 0.032 cm.
Define electronegativity:
a. an atoms ability to attract electrons that are shared in a chemical bond
b. an atoms ability to form a cation an atoms ability to form double and triple bonds
c. an atoms ability to form an ionic bond with another atom
d. an atoms ability to donate valence electrons to another atom
Answer: a
Explanation: ur mom
Answer:
A) an atom's ability to attract electrons that are shared in a chemical bond
Explanation:
I took the test and got the answer right.
A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 10 sin(t/2)N (newtons) and moves in a medium that imparts a viscous force of 2 N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass.
Answer:
Initial value problem is:
u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0
u'(0) = 0.03m/s
Explanation:
The directions of Fd(t*) and U'(t*) are not specified in the question, so we'll take Fd(t*) to be negative and U'(t*) to be positive. This is due to the fact that the damping factor acts in the direction opposite the direction of the motion of the mass.
M = 5kg; L= 10cm or 0.1m;
F(t) = 10 sin(t/2) N ; Fd(t*) = - 2N
U'(t*) = 4cm/s or 0.04m/s
u(0) = 0
u'(0) = 3cm/s or 0.03m/s
Now, we know that W = KL.
Where K is the spring constant.
And L is the length of extension.
So, k = W/L
W= mg = 5 x 9.81 = 49.05N
So,k = 49.05/0.1 = 490.5kg/s^(2)
Now from spring damping, we know that; Fd(t*) = - γu'(t*)
Where,γ = damping coefficient
So, γ = - Fd(t*)/u'(t*)
So, γ = 2/0.04 = 50 Ns/m
Therefore, the initial value problem which describes the motion of the mass is;
5u'' + 50u' + 490u = (10 sin(t/2) N
Divide each term by 5 to give;
u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0
u'(0) = 0.03m/s
Strategic planning is long-range, formulated by top management, and made as if the company operated in a vacuum Group of answer choices True False
Answer:
THE ANSWER IS: TRUE
Explanation:
Strategic planning is indeed long-range and formulated by top management but not under the assumption that the company operates in a vacuum. Effective strategic planning acknowledges and incorporates external factors.
Explanation:The statement is partially true and partially false. It is accurate that strategic planning is long-range and formulated by top management. These types of plans typically look several years into the future to set comprehensive goals for the company. However, the statement is false in suggesting that strategic planning is done as if a company operates in a vacuum. In truth, effective strategic planning must incorporate external influences like market trends, competitive activities, and regulatory changes.
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The ΔG of a reaction would be at the minimum when... Group of answer choices the equilibrium constant is equal to 1 (i.e., the reactant and product concentrations are always equal). the reactant and product concentrations don't change over time (the system is at equilibrium). the entropy has reached its maximum positive value. the reaction goes to completion. the reaction is very slow
Answer:
the equilibrium constant is equal to 1 (i.e., the reactant and product concentrations are always equal).
Explanation:
ΔG is a symbol related to Gibbs free energy, which is a physical quantity related to thermodynamics. ΔG refers to the difference between the change in enthalpy (and sometimes entropy) and the temperature of a chemical reaction.
Gibbs free energy is very useful for measuring the work done between the reactants in a reaction. It is calculated using the formula: ΔG = change in enthalpy - (temperature x change in entropy).
The ΔG of a reaction would have a minimum value (zero), if the equilibrium constant is equal to 1 (that is, the concentrations of the reagent and the product are always equal).
the equilibrium constant should be equal to 1 (i.e., the reactant and product concentrations are always equal).
Gibbs free energy:ΔG refers to a symbol that should be related to Gibbs free energy, i.e. physical quantity related to thermodynamics. ΔG means the difference between the change in enthalpy and the temperature of a chemical reaction. Now it is determined using the formula:
ΔG = change in enthalpy - (temperature x change in entropy).
In the case when the ΔG of a reaction would have a minimum value (zero), so if the equilibrium constant is equivalent to 1 i.e. the concentrations of the reagent and the product should always equal.
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A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m>s. A 1.0-kg stone is thrown from the basket with an initial velocity of 15.0 m>s per- pendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m>s. (a) How high is the balloon when the rock is thrown
Answer:
a) 242.5 m
Explanation:
Given :
Vi= 20m/s
t = 5 seconds
According to kinematic equation, the displacement is given by:
d = Vit + 1/2 gt^2
d = 20 × 6 ) + (1/2 × 9.8 × 5^2)
d = 120 + 122.5
d = 242.5m
Answer:
Explanation:
Given:
Mass = 124 kg
U = -20 m/s
t = 5 s
Using the equation of motion,
S = U×t + 1/2 × gt^2
S = -20 × 5 - 1/2 × 9.8 × 5^2
= -222.5 m
= 222.5 m (the sign is showing the direction of the motion from its origin)
slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?
Answer: 800N
Explanation:
Given :
Mass of ball =0.8kg
Contact time = 0.05 sec
Final velocity = initial velocity = 25m/s
Magnitude of the average force exerted on the wall by the ball is can be calculated using the relation;
Force(F) = mass(m) * average acceleration(a)
a= (initial velocity(u) + final velocity(v))/t
m = 0.8kg
u = v = 25m/s
t = contact time of the ball = 0.05s
Therefore,
a = (25 + 25) ÷ 0.05 = 1000m/s^2
Therefore,
Magnitude of average force (F)
F=ma
m = mass of ball = 0.8
a = 1000m/s^2
F = 0.8 * 1000
F = 800N
A proton moves 10.0 cm on a path parallel to the direction of a uniform electric field of strength 3.0 N/C. What is the change in electrical potential energy?
Answer:
ΔPE= -4.8×10⁻²⁰J
Explanation:
Given data
Electric field of strength E=3.0 N/C
Charge of proton q=1.60×10⁻¹⁹C
Proton moves distance d=10 cm=0.10 m
To find
Change in electrical potential energy ΔPE
Solution
As we know that:
ΔPE= -qEd
[tex]=-(1.60*10^{-19}C )(3.0N/C)(0.10m)\\=-4.8*10^{-20}J[/tex]
ΔPE= -4.8×10⁻²⁰J
Answer:
Explanation:
Given:
D = 10 cm
= 0.1 m
E = 3.0 N/C
Qp = 1.602 × 10^-19 C
U = Q × V
But,
V = E × D
= 3 × 0.1
= 0.3 V
U = 1.602 × 10-19 × 0.3
= 4.806 × 10^-20 J.
Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets. Which of the following best explains why finding 1 planet with such a moon is consistent with the nebular theory, while finding 6 planets with such moons is not consistent?
Answer:
Unusually large moons form in giant impacts, which are relatively rare events
Explanation:
Solution:
- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.
- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible. The transition from an undifferentiated cloud to a star system complete with planets and moons takes about 100 million years.
The neurotransmitter dopamine is most closely associated with
Since there's no answer choices available, I would say "reward mechanisms in the brain", "pleasure", "Schizophrenia" or "appetite".
Answer:
A neurotransmitter used in the parts of the brain involved in regulating movement and experiencing pleasure.
Explanation:
Dopamine is one of the chemicals in our brain which controls mood Interacting with the pleasure and reward center of our brain, dopamine, along with other chemicals like oxytocin and endorphins plays a very crucial role in how we feel
The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At 15 degrees Celsius the pressure is 101.3 kPa at sea level and 87.14 kPa at h = 1000m. What is the pressure at an altitude of 3000 m?
Answer:
64.59kpa
Explanation:
See attachment
Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 62 kg, find the magnitudes of the impulse on him from the water.
Answer:
823.46 kgm/s
Explanation:
At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.
So, mgh = 1/2mv²
From here, his velocity just as he reaches the surface of the water is
v = √2gh
h = 9 m and g = 9.8 m/s²
v = √(2 × 9 × 9.8) m/s
v = √176.4 m/s
v₁ = 13.28 m/s
So his velocity just as he reaches the surface of the water is 13.28 m/s.
Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.
So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.
His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,
J = (62 × 0 + 62 × 13.28) kgm/s = 0 + 823.46 kgm/s = 823.46 kgm/s
So the magnitude of the impulse J of the water on him is 823.46 kgm/s
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.50 m in front of one of the speakers perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible? Express your answer with the appropriate units.
Answer:
242.85 Hz
Explanation:
For maximum intensity of sound, the path difference,ΔL = (n + 1/2)λ/2 where n = 0,1,2...
Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √(2.00² + 5.50²) = √(4.00 + 30.25) = √34.25 = 5.85 m.
The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.
Since ΔL = (n + 1/2)λ/2 and for lowest frequency n = 0,
ΔL = (n + 1/2)λ/2 = (0 + 1/2)λ/2 = λ/4
ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.
f = 340/(4 × 0.35) = 242.85 Hz
The condition of constructive interference and the speed of a wave allows finding the frequency for the first constructive interference that Abby hears is:
f = 971.4 Hz
The interference of the coherent sound wave occurs when two wave fronts are added at a point and the intensity depends on the path difference of these waves, we have two extreme cases:
Destructive. In this case the sum of the intensity gives a resultant of zero. Constructive. The sum of the intensities gives a maximum and is described by the expression.Δr = [tex]2n \ \frac{\lambda}{2}[/tex]
Where Δr is the path difference, λ is the wavelength and n is an integer.
In the attachment we can see the distance from Abby is to the speakers for the closest y = 5.50 m, the distance to the furthest speaker.
Let's use the Pythagoras' theorem,
r =[tex]\sqrt{x^2 + y^2}[/tex]
Let's calculate.
r = [tex]\sqrt{2^2 + 5.5^2}[/tex]
r = 5.85 m
The difference in path is:
Δr = r-y
Δr = 5.85 - 5.5
Δr = 0.35 m
let's find the wavelength.
Δr = 2n [tex]\frac{\lambda}{2}[/tex]
λ = [tex]\frac{\Delta r}{n}[/tex]
The first constructive interference occurs for n = 1
λ= Δr
λ = 0.35 m
Wave speed is proportional to wavelength and frequency.
v = λ f
f = [tex]\frac{v}{\lambda }[/tex]
Let's calculate.
f = [tex]\frac{340}{0.35}[/tex]
f = 971.4 Hz
In conclusion using the condition of constructive interference and the speed of a wave we can find the frequency for the first constructive interference that Abby hears is:
f = 971.4 Hz
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A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (unwisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you.
Answer:
1.9 s
Explanation:
We are given that
Length of cable=l=15 m
We have to find the time you have to move out of the way.
We know that
Time period,T=[tex]2\pi\sqrt{\frac{l}{g}}[/tex]
Where g=[tex]9.8m/s^2[/tex]
By using the formula
[tex]T=2\pi\sqrt{\frac{15}{9.8}}[/tex]
[tex]T=2\times 3.14\times \sqrt{\frac{15}{9.8}}=7.77 s[/tex]
Time you have to move out
[tex]t=\frac{T}{4}=\frac{7.77}{4}=1.9 s[/tex]
Hence,time you have to move out of the way=7.77 s
Which prediction of weather is the most accurate for the next two days if there is an occluded front over the area?
Question 2 options:
constant rain
dry warm weather
severe thunderstorms
cold but clear skies
The most accurate weather for the next two days would be dry warm weather or severe thunderstorms if there is an occluded front over the area.
Explanation:
The weather front that is created during the cyclogenesis process is an occluded front in meteorology. Cyclogenesis process is the development of extra-tropical cyclone and its intensification.
During the occurrence of this, the warm air is occluded (separated) from the center of cyclone at the surface of the earth.
The cold front rotates the storm as it intensifies and holds up the warm front. This develops an occluded front, that is the boundary which separates the new cold air mass and the older cool air mass that is already in warm front's north.
A solid cylinder and a hollow cylinder of equal mass and radius are at rest at the top of an inclined plane. They are released simultaneously and roll down the plane without slipping. Which object reaches the bottom of the incline first?
Answer:
Solid cylinder
Explanation:
Solid cylinder will reach first
If radius of each cylinder is r
mass is m
then,
Moment of inertia I= dm[tex]r^{2}[/tex]
Here d = measure as how close the mass is to the edge
Velocity of rolling cylinder is given by
v= [tex]\sqrt} \frac{2gh}{1+d}[/tex]
where,
g= 9.8 m/s2
h= height from ground
So from formula of velocity we can say that velocity will be maximum if denominator is minimum i-e if k is of small value -- or in other word if mass is away from edge i-e if mass is closer to the center !
As all the mass in hollow cylinder is near the edge so k value will be higher for it and hence it will have low velocity value. So it will reach later as compared to the solid cylinder in which mass is closer to the center and hence k is greater for solid cylinder.
A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the local ocean current is 1.50 m/s in a direction 40.0º north of east and changes the ship's intended motion. What is the velocity of the ship relative to the Earth?
Answer:
Explanation:
velocity of ship with respect to water = 6.5 m/s due north
[tex]\overrightarrow{v}_{s,w}=6.5 \widehat{j}[/tex]
velocity of water with respect to earth = 1.5 m/s at 40° north of east
[tex]\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)[/tex]
velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth
[tex]\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}[/tex]
[tex]\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j} \right )[/tex]
[tex]\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}[/tex]
The magnitude of the velocity of ship relative to earth is [tex]\sqrt{1.15^{2}+5.54^{2}}[/tex] = 5.66 m/s
A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95.0kg) stands on a platform 45.0m above the ground , and one endof the cord is tied securley to his ankle and the other end to theplatform. You have promised him that when he steps off he will falla maximum of 41m before the cord stops him. You had several bungeecords to select from and tested them by stretching them out, tyingone end to a tree and pulling the other end with a force of 380.0N.When you did this, what distance will the bungee cord you shouldselect have stretched?
Answer:
The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.
Explanation:
As from the given data
the length of the rope is given as l=30 m
the stretched length is given as l'=41m
the stretched length required is give as y=l'-l=41-30=11m
the mass is m=95 kg
the force is F=380 N
the gravitational acceleration is g=9.8 m/s2
The equation of k is given by equating the energy at the equilibrium point which is given as
[tex]U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}[/tex]
Here
m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so
[tex]k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m[/tex]
Now the force is
[tex]F=kx\\[/tex] or
[tex]x=\dfrac{F}{k}\\[/tex]
So here F=380 N, k=630.92 N/m
[tex]x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m[/tex]
So the distance is 0.602 m
The student requires help to calculate the stretch distance of a bungee cord when a force is applied, which is determined using Hooke's Law, F = kx, where F is the applied force and k is the spring constant.
Explanation:The student is asking about the properties of a bungee cord that would be suitable for their father-in-law's jump so that he will fall safely without hitting the ground. Specifically, the question involves calculating the amount by which the bungee cord would stretch when a force of 380.0 N is applied to it. The restoring force that the bungee cord exerts is given as kx, where k is the spring constant and x is the stretch distance.
To find the correct bungee cord, one must solve for x in the equation F = kx given that F (the force applied) is 380.0 N. The spring constant k is the unknown in this problem and must be determined based on the requirement that the maximum fall before the cord stops the father-in-law is 41 m, factoring in his weight and the initial unstretched length of the bungee cord (30 m).
This problem requires knowledge in the area of Hooke's Law and the work-energy principle, as it deals with forces and the properties of elastic materials. The actual calculation is not provided because it would require additional information about the bungee cords the student has access to, such as their specific spring constants.
And electromagnet is in temporary magnet made by Coiling wire around an iron core which becomes a magnet win and electric current flows through the wire how could a strength of an electromagnet be increase
Answer:
We can increase the strength by increasing current in the wire or increasing the number of turns of the coil around an iron core .
Explanation:
The magnetic strength due to current carrying conductor carrying current I and having N number of turns is given by
B = [tex]\mu _{0}\times N\times I[/tex]
[tex]\mu_{O}[/tex] is vacuum permeability .and its value is equal to [tex]4\pi\times10^{-7} \frac{H}{m}[/tex]
.so from the above equation we can see that magnetic strength is directly proportional to the current through wire and no. of turns .
Final answer:
An electromagnet's strength can be increased by passing an electric current through a coil of wire wrapped around an iron core and using a ferromagnetic material to intensify the magnetic field.
Explanation:
An electromagnet increases its strength when an electric current passes through a coil of wire wrapped around an iron core. The strength of the magnetic field generated by the electromagnet is proportional to the amount of current flowing through the wire.
By using a ferromagnetic core within the coil, such as soft iron, the magnetic field can be intensified to thousands of times stronger than the field produced by the coil alone. This combination creates a ferromagnetic-core electromagnet with significantly increased magnetic strength.
What is its maximum altitude above the ground? The answer is the maximum height above the ground
Answer:
Maximum altitude above the ground = 1,540,224 m = 1540.2 km
Explanation:
Using the equations of motion
u = initial velocity of the projectile = 5.5 km/s = 5500 m/s
v = final velocity of the projectile at maximum height reached = 0 m/s
g = acceleration due to gravity = (GM/R²) (from the gravitational law)
g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)
g = -9.82 m/s² (minus because of the direction in which it is directed)
y = vertical distance covered by the projectile = ?
v² = u² + 2gy
0² = 5500² + 2(-9.82)(y)
19.64y = 5500²
y = 1,540,224 m = 1540.2 km
Hope this Helps!!!
A grandfather clock is controlled by a swinging brass pendulum that is 1.6 m long at a temperature of 28°C. (a) What is the length of the pendulum rod when the temperature drops to 0.0°C? (Give your answer to at least four significant figures.)
Answer:
1.599 m.
Explanation:
Using
α = ΔL/(LΔT)............ Equation 1
Where
α = Coefficient of linear expansion of brass,
ΔL = Change in length,
ΔT = Change in temperature
L = Original length.
make ΔL the subject of the equation
ΔL = α(LΔT)................. Equation 2
Given: L = 1.6 m, ΔT = T₂-T₁ = 0-28 = -28 ⁰C
Constant: α = 1.8×10⁻⁵/°C
Substitute into equation 2
ΔL = (1.8×10⁻⁵)(1.6)(-28)
ΔL = -8.064×10⁻⁴ m
But,
ΔL = L₁-L
Where L = final length of the pendulum rod.
make L₁ the subject of the equation,
L₁ = ΔL+L......... Equation 2
L₁ = 1.6+( -8.064×10⁻⁴ )
L₁ = 1.599 m.
Explanation:
Below is an attachment containing the solution.
A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction between the hockey puck and the metal ramp are μs = 0.40 and μk = 0.30, respectively. The puck's initial speed is 26 m/s. What vertical height does the puck reach above its starting point?
Answer:
71.76 m
Explanation:
We will solve this question using the work energy theorem.
The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.
ΔK.E = W
In the attached free body diagram for the question, the forces acting on the puck are given.
ΔK.E = (final kinetic energy) - (initial kinetic energy)
Final kinetic energy = 0 J (since the puck comes to a stop)
Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J
ΔK.E = 0 - 67.6 = - 67.6 J
W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)
Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h
Workdone by the frictional force = F × d
F = μ N
μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)
N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N
F = μ N = 0.3 × 1.697 = 0.509 N
where d = distance along the incline that the puck travels.
d = h/sin 30° = 2h (from trigonometric relations)
Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h
ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)
- 67.6 = - 1.96h + 1.02h
-0.942h = - 67.6
h = 71.76 m
A block pushed along the floor with velocity v0xv0x slides a distance dd after the pushing force is removed. If the mass of the block is doubled but its initial velocity is not changed, what distance does the block slide before stopping?
Answer:
There is no change in distance
Explanation:
Given that block slides a distance dd with initial velocity [tex]v_{ox}[/tex][tex]v_{ox}[/tex]
Work energy theorem states that Work done W is the difference between final kinetic energy [tex]KE_{f}[/tex] and initial kinetic energy [tex]KE_{i}[/tex]
W = [tex]KE_{f}[/tex] - [tex]KE_{i}[/tex] ...........(equation 1)
In the given problem, the work is done by frictional force
Hence the work done is given by
[tex]W_{Friction}[/tex] = -[tex]F_{Friction}[/tex] x dd
[tex]F_{Friction}[/tex] is negative since it is resistive force and dd is the distance
[tex]W_{Friction}[/tex] = - μ mg X dd
where μ is coefficient of friction.
Also we know that Kinetic energy KE = [tex]\frac{1}{2}[/tex] [tex]mv^{2}[/tex]
[tex]KE_{f}[/tex] = 0 since final velocity is zero
[tex]KE_{i}[/tex] = [tex]\frac{1}{2}[/tex] [tex]m(v_{ox} v_{ox}) ^{2}[/tex]
Substituting the corresponding values equation 1 becomes
-μ mg x dd = 0 - [tex]\frac{1}{2}[/tex] [tex]m(v_{ox} v_{ox}) ^{2}[/tex]
dd= [tex]\frac{(v_{ox}v_{ox} )^{2} }{2g}[/tex]x (1/μ) ........... (equation 2)
To find distance when mass of block is doubled and initial velocity is not changed:
Equation 2 shows that the distance dd is independent of mass, therefore there is no change in distance.
The block will slide only half the distance before stopping when its mass is doubled while the initial velocity remains the same.
When a block is pushed along the floor with an initial velocity [tex]\(v_{0x}\)[/tex] and slides a distance [tex]\(d\)[/tex] before stopping, its stopping distance is determined by the initial velocity and the frictional force opposing its motion.
The force of friction is given by [tex]\(F_{\text{friction}} = \mu \cdot N\)[/tex], where [tex]\(\mu\)[/tex] is the coefficient of friction and [tex]\(N\)[/tex] is the normal force. In this case, we are assuming a constant coefficient of friction.
The deceleration of the block can be calculated using Newton's second law: [tex]\(F = m \cdot a\)[/tex]. When the mass of the block is doubled, the force required to decelerate it to a stop is also doubled.
Let's assume the initial velocity [tex]\(v_{0x}\)[/tex] remains the same. The deceleration [tex]\(a\)[/tex] is proportional to the force, so [tex]\(a\)[/tex] is also doubled when the mass is doubled. This is because [tex]\(F\)[/tex] is directly proportional to [tex]\(m\)[/tex].
The stopping distance [tex]\(d'\)[/tex] can be calculated using the equation of motion:
[tex]\[v_f^2 = v_i^2 + 2as\][/tex]
Where:
[tex]\(v_f\)[/tex] is the final velocity (0 m/s, as the block stops),
[tex]\(v_i\)[/tex] is the initial velocity [tex](\(v_{0x}\)),[/tex]
[tex]\(a\)[/tex] is the deceleration (doubled),
[tex]\(s\)[/tex] is the stopping distance [tex](\(d'\)).[/tex]
Plugging in the values:
[tex]\[0 = v_{0x}^2 + 2(2a)d'\][/tex]
Now, we can solve for [tex]\(d'\):[/tex]
[tex]\[2ad' = -v_{0x}^2\][/tex]
[tex]\[d' = \frac{-v_{0x}^2}{2a}\][/tex]
Since [tex]\(a\)[/tex] is doubled, the stopping distance [tex]\(d'\)[/tex] will be halved compared to the initial distance [tex]\(d\).[/tex] So, the block will slide only half the distance before stopping when its mass is doubled while the initial velocity remains the same.
For more such questions on initial velocity:
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