Answer:
Height will be 5.127 m
Explanation:
We have given mass m = 49 kg
Speed over the bar v = 10 m /sec
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
Kinetic energy on the ground [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 49\times 10^2=2450J[/tex]
Potential energy on the ground [tex]=mgh=0[/tex] (as height will be zero )
Speed above the bar = 0.7 m /sec
So kinetic energy above the bar [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 49\times 0.7^2=12J[/tex]
Potential energy above the bar = mgh
From energy conservation
Total kinetic energy = total potential energy
So [tex]2450+12=0+49\times 9.8\times h[/tex]
[tex]h=5.127m[/tex]
Using the principle of conservation of energy, we can calculate the pole vaulter's height above the bar as being approximately 5.02 meters.
In this question, we're dealing with conservation of energy. The energy of the pole vaulter is kinetic energy when she is running (1/2*mass*speed^2), and as she goes over the bar, her energy becomes potential energy (mass*gravity*height) and a little kinetic energy (1/2*mass*speed^2).
Let's use the following formula: Kinetic Energy initial + Potential Energy initial = Kinetic Energy final + Potential Energy final.
Initially, her kinetic energy is 1/2 * 49 kg * (10 m/s)^2 = 2450 J. She has no potential energy, because she's on the ground (height=0). When she's above the bar, her kinetic energy is 1/2 * 49 kg * (0.7 m/s)^2 = 12.075 J. We don't yet know her final potential energy, because we're trying to find her height. So, we'll call her final potential energy m*g*h, or 49 kg * 9.8 m/s^2 * h.
We then plug these values into our energy equation: 2450 J + 0 J = 12.075 J + 49 kg * 9.8 m/s^2 * h. Solving for h gives us h = (2450 J - 12.075 J) / (49 kg * 9.8 m/s^2) = 5.02 meters, so this would be her altitude as she crosses the bar.
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The angular speed of an automobile engine is increased at a constant rate from 1200 rev/min to 3000 rev/min in12 s.(a) What is its angular acceleration in revolutions per minutesquared?(b) How many revolutions does theengine make during this 12 s interval?
Answer:
The angular acceleration is
=
15.71
r a d s − 2 and the number of revolutions is = 419.9
Explanation:
a) The angular acceleration of the automobile is 9000 rev/min².
b) The engine makes 420 revolution during this 12 s interval.
What is angular acceleration?The temporal rate at which angular velocity changes is known as angular acceleration. The standard unit of measurement is radians per second per second.
Rotational acceleration is another name for angular acceleration. It is a numerical representation of the variation in angular velocity over time.
Initial angular speed = 1200 rev/min.
Final angular speed = 3000 rev/min.
Time taken = 12 second = 0.2 minute.
a) its angular acceleration is = (final angular speed - Initial angular speed )/ Time taken
= ( 3000 rev/min - 1200 rev/min)/0.2 minute
= 9000 rev/min²
b) The engine makes during this 12 s interval = (Initial angular speed + Final angular speed) × time interval/2
= (1200 + 3000)× 0.2/2 revolution
= 420 revolution.
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The Pony Express was a mail delivery system in the Old West that used a series of men on horseback to deliver mail from St. Joseph, MO to Sacramento, CA along a trail that was 2000 miles long. True or false: If each rider traveled 100 miles, got a fresh horse every 10 miles, and maintained an average speed of 10 mi/hr, it took 150 horses for each delivery.A. TrueB. False
Answer:
the given statement is False
Explanation:
given,
distance of the trail = 2000 miles long
each rider traveled = 100 miles
every fresh horse travel = 10 miles
to maintain speed of = 10 mile/hr
the given statement is
150 horses is used for each delivery.
if each horse is allowed to travel 10 miles to travel
distance traveled using 150 horses = 150 x 10
= 1500 miles
to travel 2000 miles horse required is equal to 200.
so, the given statement is False
Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence of dark energy. Rank each model from left to right based on the ratio of its actual mass density to the critical density, from smallest ratio (mass density much smaller than critical density) to largest ratio (mass density much greater than critical density).1. coasting universe2. recollapsing universe3. critical universe
Answer:
do your best
Explanation:
babyboiiii∛√
Express the angular velocity ω of the wheel in terms of the displacement d, the magnitude F of the applied force, and the moment of inertia of the wheel Iw, if you've found such a solution. Otherwise, following the hints for this part should lead you to express the angular velocity ω of the wheel in terms of the displacement d, the wheel's radius r, and α.
The angular velocity (ω) of a wheel can be expressed in terms of the force applied (F), wheel's radius (r), and its moment of inertia (I), all of which factor into the wheel's torque and resulting angular acceleration. Through the rotational analog of Newton's second law and kinematic equations, one can derive a formula for ω that connects these physical quantities
To express the angular velocity (ω) of the wheel in terms of displacement (d), magnitude of the force (F), and the moment of inertia of the wheel (Iw), we use physical principles of rotational motion. We can derive this using the definition of torque and the rotational version of Newton's second law, which states that torque (τ) is equal to the moment of inertia (I) times the angular acceleration (α): τ = Iα. For a wheel, torque is also equal to the force applied (F) times the radius (r): τ = Fr.
By equating the two expressions for torque, we get Fr = Iα, and therefore α = F × r/I. With the angular acceleration and knowing the time the force is applied, we can find the final angular velocity using the kinematic equation for rotational motion: ωfinal = ωinitial + α × t. Assuming that initial angular velocity (ωinitial) is zero and time (t) can be deduced from the linear displacement (d) and the linear velocity (v), given that v = ω × r, we can express the final angular velocity in terms of displacement (d), wheel's radius (r), and angular acceleration (α).
To further involve the moment of inertia in this expression, we would substitute the earlier derived expression for α into our final angular velocity equation: ω = (F × r/I) × t. Now, if we relate time (t) to the linear displacement (d) such that the linear velocity (v) = d/t, and since v = ω×r, we can isolate ω to derive a relationship that includes the moment of inertia.
A passenger on a balloon drops a baseball over the side of the gondola. As the baseball falls faster, the drag force from air resistance increases.
Which of these describes what happens to the motion of the ball from the time the ball is dropped to the time when the drag force becomes equal to the force of gravity?
A The acceleration of the ball remains constant.
B The speed of the ball decreases.
C The acceleration of the ball decreases.
D The speed of the ball remains constant.
Answer:
The answer is C.
Explanation:
For acceleration to be achieved so to speak there must be a force acting on it. The only force on the ball before the air drag increases is gravity. As the air resistance increases the force resisting gravity increases. This means that the forces start to cancel out. Therefore the acceleration must get smaller.
These forces will continue to cancel until it reaches terminal velocity.
What happens to the motion of the ball is, the acceleration of the ball decreases and will become zero when drag force on the ball equals the force of gravity.
The downward motion of the ball is reduced by frictional force opposing the motion. The frictional force opposing the motion is the drag force of the air or air resistance.
The net downward force on the ball is given as;
[tex]W -F_D = ma\\\\[/tex]
when the drag force on the ball equals force of gravity, the acceleration of the ball will be zero.
[tex]W - W = ma\\\\(W = F_D)\\\\0 = ma\\\\a = 0[/tex]
Thus, we can conclude that what happens to the motion of the ball is, the acceleration of the ball decreases and will become zero when drag force on the ball equals the force of gravity.
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(a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp equals 746 W) to reach a speed of 15.0 m/s, neglecting friction? (b) How long will this acceleration take if the car also climbs a 3.00-m high hill in the process?
Answer:
(A) time = 3.205 s
(B)time =4.04 s
Explanation:
mass (m) = 850 kg
power (P) = 40 hp = 40 x 746 = 29,840 W
final velocity (Vf) = 15 m/s
final height (Hf) = 3 m
since the car is starting from rest at the bottom of the hill, its initial velocity and initial height are both 0
(A) from the work energy theorem
work = 0.5 x m x ([tex](Vf)^{2} - (Vi)^{2}[/tex]) (change in kinetic energy)work = power x timethereforepower x time = 0.5 x m x ([tex](Vf)^{2} - (Vi)^{2}[/tex])
time = [tex]\frac{0.5 x m x ([tex](Vf)^{2} - (Vi)^{2}[/tex])}{power}[/tex]
time = [tex]\frac{0.5 x 850 x ([tex](15)^{2} - (0)^{2}[/tex])}{29,840}[/tex]
time = 3.205 s
(B) from the work energy theorem
work = change in potential energy + change in kinetic energywork = (mg (Hf - Hi)) + (0.5m([tex](Vf)^{2} - (Vi)^{2}[/tex])work = power x timethereforepower x time = (mg (Hf - Hi)) + (0.5m([tex](Vf)^{2} - (Vi)^{2}[/tex])
time = [tex]\frac{(mg (Hf - Hi)) + (0.5m([tex](Vf)^{2} - (Vi)^{2}[/tex])}[/tex])}{power}[/tex]
time = [tex]\frac{(850 x 9.8 x (3 - 0)) + (0.5 x 850 x [tex](15)^{2} - (0)^{2}[/tex])}[/tex])}{29,840}[/tex]
time =4.04 s
Answer:
a) [tex]\Delta t = 3.205\,s[/tex], b) [tex]\Delta t = 4.043\,s[/tex]
Explanation:
a) The time needed is determined by the Work-Energy Theorem and the Principle of Energy Conservation:
[tex]K_{1} + \Delta E = K_{2}[/tex]
[tex]\Delta E = K_{2} - K_{1}[/tex]
[tex]\dot W \cdot \Delta t = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]\Delta t = \frac{m\cdot v^{2}}{2\cdot \dot W}[/tex]
[tex]\Delta t = \frac{(850\,kg)\cdot \left(15\,\frac{m}{s} \right)^{2}}{2\cdot (40\,hp)\cdot \left(\frac{746\,W}{1\,hp} \right)}[/tex]
[tex]\Delta t = 3.205\,s[/tex]
b) The time is found by using the same approach of the previous point:
[tex]U_{1} + K_{1} + \Delta E = U_{2} + K_{2}[/tex]
[tex]\Delta E = (U_{2}-U_{1})+(K_{2} - K_{1})[/tex]
[tex]\dot W \cdot \Delta t = m\cdot \left(g\cdot \Delta h + \frac{1}{2}\cdot v^{2} \right)[/tex]
[tex]\Delta t = \frac{m\cdot\left(g\cdot \Delta h + \frac{1}{2}\cdot v^{2}\right)}{\dot W}[/tex]
[tex]\Delta t = \frac{(850\,kg)\cdot \left[\left(9.807\,\frac{m}{s^{2}} \right)\cdot (3\,m) + \frac{1}{2}\cdot \left(15\,\frac{m}{s} \right)^{2}\right]}{(40\,hp)\cdot \left(\frac{746\,W}{1\,hp} \right)}[/tex]
[tex]\Delta t = 4.043\,s[/tex]
A balloon filled with helium gas has an average density of rhob = 0.27 kg/m3. The density of the air is about rhoa = 1.23 kg/m3. The volume of the balloon is Vb = 0.084 m3. The balloon is floating upward with acceleration a.
The student is asking about a helium-filled balloon's behavior in relation to its density and the density of air. To determine whether the balloon rises or falls, we compare the buoyant force with the gravitational force. The buoyant force is calculated using the formula (rhoa - rhob) * Vb * g.
Explanation:The subject of this question is Physics.
The student is inquiring about a balloon filled with helium gas that has an average density of 0.27 kg/m3. The density of air is approximately 1.23 kg/m3. The volume of the balloon is 0.084 m3. The balloon is floating upwards with an acceleration.
To determine if the balloon is floating upwards or downwards, we need to compare the buoyant force with the gravitational force. If the buoyant force is greater, the balloon will rise; if it is less, the balloon will descend.
The buoyant force can be calculated using the formula:
Buoyant force = (rhoa - rhob) * Vb * g
where rhoa is the density of air, rhob is the density of the balloon, Vb is the volume of the balloon, and g is the acceleration due to gravity.
If the buoyant force is greater than the gravitational force (given by the formula mg, where m is the mass of the balloon and g is the acceleration due to gravity), then the balloon will float upwards.
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The balloon's acceleration is about 34.84 m/s².
First, let's calculate the buoyant force (F[tex]_b[/tex]) acting on the balloon using Archimedes' principle:
F[tex]_b[/tex] = ρ[tex]_a[/tex] × g × V[tex]_b[/tex],
where:
ρ[tex]_a[/tex] is the density of air = 1.23 kg/m³,
g is the acceleration due to gravity = 9.8 m/s²,
V[tex]_b[/tex] is the volume of the balloon = 0.084 m³.
Therefore, F[tex]_b[/tex] = 1.23 kg/m³ × 9.8 m/s² × 0.084 m³ = 1.012488 N.
Next, calculate the gravitational force (weight) of the balloon (W[tex]_b_a_l_o_o_n[/tex]):
W[tex]_b_a_l_o_o_n[/tex] = m[tex]_b_a_l_o_o_n[/tex] × g,
where, m[tex]_b_a_l_o_o_n[/tex] is the mass of the balloon given by the product of its volume and density:
m[tex]_b_a_l_o_o_n[/tex] = ρ[tex]_b[/tex] × V[tex]_b[/tex] = 0.27 kg/m³ × 0.084 m³ = 0.02268 kg.
Thus, W[tex]_b_a_l_o_o_n[/tex] = 0.02268 kg × 9.8 m/s² = 0.222264 N.
The net force (F[tex]_n_e_t[/tex]) acting on the balloon is the difference between the buoyant force and the weight of the balloon:
F[tex]_n_e_t[/tex] = F[tex]_b[/tex] - W[tex]_b_a_l_o_o_n[/tex] = 1.012488 N - 0.222264 N = 0.790224 N.
Finally, use Newton's second law to find the acceleration (a) of the balloon:
F[tex]_n_e_t[/tex] = m[tex]_b_a_l_o_o_n[/tex] × a
⇒ a = F[tex]_n_e_t[/tex] / m[tex]_b_a_l_o_o_n[/tex],
a = 0.790224 N / 0.02268 kg ≈ 34.84 m/s².
Therefore, the acceleration of the balloon is approximately 34.84 m/s²
An archer puts a 0.4 kg arrow to the bowstring. An average force of 190.4 N is exerted to draw the string back 1.47 m. The acceleration of gravity is 9.8 m/s². Assuming no frictional loss, with what speed does the arrow leave the bow? Answer in units of m/s. If the arrow is shot straight up, how high does it rise? Answer in units of m.
Answer:
v = 37.4 m/s , h = 71.39m
Explanation:
To find the velocity given:
m = 0.4 kg
F =190.4 N
d = 1.47 m
g = 9.8 m/s^2
So use the equation of work to solve the kinetic energy
W = F *d = 190.4 N * 1.47m
W = 279.88 J
Ke = 1 / 2 * m* v^2
v = √2*Ke / m =√ 2 *279.88 / 0.4 kg
v = 37.4 m/s
Now to find the high to rise can use the conserved law so:
Ke = Pe
279.88 = m*g*h
Solve to h'
h = 279.88 / 0.4kg * 9.8m/s^2
h =71.39 m
Final answer:
The arrow leaves the bow with a speed of approximately 37.49 m/s and, when shot straight up, rises to a maximum height of about 71.4 m.
Explanation:
To determine the speed with which the arrow leaves the bow, we apply the work-energy principle, which states that work done on the arrow is converted into its kinetic energy. The work done W by the bow can be calculated by multiplying the force F exerted by the distance d over which the force is applied: W = F × d. It is given that F is 190.4 N and d is 1.47 m; thus, W = 190.4 N × 1.47 m = 279.888 J.
The kinetic energy KE of the arrow can be given by KE = ½ mv², where m is the mass of the arrow and v is its velocity. Since work done equals the kinetic energy, we get 279.888 J = ½ × 0.4 kg × v². Solving for v gives us a velocity of approximately 37.49 m/s.
To find how high the arrow goes if it's shot straight up, we use the conservation of energy, where the initial kinetic energy is converted to gravitational potential energy at the highest point. The potential energy PE at maximum height can be given by PE = mgh, where g is the acceleration due to gravity (9.8 m/s²) and h is the height. Setting KE equal to PE, we have 279.888 J = 0.4 kg × 9.8 m/s² × h. Solving for h gives us a maximum height of approximately 71.4 m.
A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the stick a distance D from the middle of the stick. Before the collision, the puck was moving with speed v0 in a direction perpendicular to the stick, as indicated in the figure. The collision is completely inelastic, and the puck remains attached to the stick after the collision.
What is the angular momentum Lcm of the system before the collision, with respect to the center of mass of the final system?
Express Lcm in terms of the given variables.
The angular momentum of the hockey stick-puck system before the collision, with respect to the center of mass of the final system, is given by Lcm = mp * v0 * D, where mp is the puck's mass, v0 is its initial velocity, and D is the distance from the center of the stick.
Angular Momentum of Hockey Stick-Puck System
To find the angular momentum Lcm of the system before the collision with respect to the center of mass of the final system, we follow these steps:
Before the collision, the puck is moving with velocity v0. Since the collision is completely inelastic, the stick and puck move as a single system after the collision.The angular momentum L before the collision is given by: This expression simplifies to:Therefore, the angular momentum of the system before the collision, with respect to the center of mass of the final system, is Lcm = mp * v0 * D.
"A high-mass star near the end of its life undergoes successive cycles of energy generation within its core in which gravitational collapse increases the temperature to the point where a new nuclear fusion cycle generates sufficient energy to stop the collapse. This process does not work beyond the silicon-fusion cycle that produces iron. Why is this?A. Electrostatic forces between the highly charged iron nuclei are sufficient to overcome the collapse and stabilize the stellar core.B. Iron nuclei are so large that they occupy all remaining space and so the collapse cannot continue.C. Fusion of iron nuclei into heavier nuclei requires energy rather than producing excess energy and therefore will not produce the additional gas pressure to halt the collapse.D. The pressure from high-energy photons and neutrinos at the very high core temperatures reached at this stage of development is finally sufficient to halt the collapse.
Answer:
C
Explanation:
The correct answer C part.
The phenemonon mention the question above happens only because Fusion of iron nuclei into heavier nuclei requires energy rather than producing excess energy and therefore will not produce the additional gas pressure to halt the collapse, hence the process does not work beyond the silicon- fusion cycle that produces iron.
A box is moved a distance of 10 meters with a force of 5 newtons. The amount of work done is:
Answer:
50 J
Explanation:
Work = force × distance
W = Fd
W = (5 N) (10 m)
W = 50 J
The amount of work done is 50 J (Joule)
How to find the amount of work done?To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd. If the force is being exerted at an angle θ to the displacement, the work done is W = fd cos θ.
By applying the formula, we get:
Work = force × distance
Force = 5 newtons ...(given)
distance = 10 meters ...(given)
⇒ W = Fd
⇒ W = (5 N) (10 m)
⇒ W = 50 J
What is the work done?
Work is done whenever a force moves something over a distance. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.
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Flow around curved height contours requires the incorporation of the centrifugal force. What is the general term to describe the winds that flow along a curved trajectory above the level where friction plays a role?
Answer: Gradient Wind
Explanation:
Gradient wind, is the wind that accounts for air flow along a curved trajectory. It is an extension of the concept of geostrophic wind; for example the wind assumed to move along straight and parallel isobars (lines of equal pressure). The gradient wind represents the actual wind better than the geostrophic wind, especially when both wind speed and trajectory curvature are large, because they are in hurricanes and jet streams.
Planetary orbits around a star can be modeled with the following potentialU(r) =ar+br2(1)(a) Show that the equilibrium position for this potential is equal tore= 2b=a.(b) Use the Taylor expansion on the potential about the equilibrium position to show thatthe \spring" constant of small oscillations around this equilibrium position isa4=8b3
Answer:
a) r eq = -a/(2b)
b) k = a/r eq = -2b
Explanation:
since
U(r) = ar + br²
a) the equilibrium position dU/dr = 0
U(r) = a + 2br = 0 → r eq= -a/2b
b) the Taylor expansion around the equilibrium position is
U(r) = U(r eq) + ∑ Un(r eq) (r- r eq)^n / n!
,where Un(a) is the nth derivative of U respect with r , evaluated in a
Since the 3rd and higher order derivatives are =0 , we can expand until the second derivative
U(r) = U(r eq) + dU/dr(r eq) (r- r eq) + d²U/dr²(r eq) (r- r eq)² /2
since dU/dr(r eq)=0
U(r) = U(r eq) + d²U/dr²(r eq) (r- r eq)² /2
comparing with an energy balance of a spring around its equilibrium position
U(r) - U(r eq) = 1/2 k (r-r eq)² → U(r) = U(r eq) + 1/2 k (r-r eq)²
therefore we can conclude
k = d²U/dr²(r eq) = -2b , and since r eq = -a/2b → -2b=a/r eq
thus
k= a/r eq
A steel aircraft carrier is 370 m long when moving through the icy North Atlantic at a temperature of 2.0 °C. By how much does the carrier lengthen when it is traveling in the warm Mediterranean Sea at a temperature of 21 °C?
Answer:
The carrier lengthen is 0.08436 m.
Explanation:
Given that,
Length = 370 m
Initial temperature = 2.0°C
Final temperature = 21°C
We need to calculate the change temperature
Using formula of change of temperature
[tex]\Delta T=T_{f}-T_{i}[/tex]
[tex]\Delta T=21-2.0[/tex]
[tex]\Delta T=19^{\circ}C[/tex]
We need to calculate the carrier lengthen
Using formula of length
[tex]\Delta L=\alpha_{steel}\times L_{0}\times\Delta T[/tex]
Put the value into the formula
[tex]\Delta L=1.2\times10^{-5}\times370\times19[/tex]
[tex]\Delta L=0.08436\ m[/tex]
Hence, The carrier lengthen is 0.08436 m.
The purpose of a master production schedule (MPS) is to break down the aggregate planning decisions into such details as order sizes and schedules for individual subassemblies and resources by week and day. True or False?
Answer:
False.
Explanation:
Master production schedule (MPS) is nothing but plan for the individual commodities to be produced in a factory, during to a time period. MPS includes Planning, production, staffing , inventory, etc. It preferably used in places where it is know that when and how each product is demanded. It has nothing to deal with decision and breaking down of aggregate planning.
The definition of MPS given in question is wrong. There the given statement is false.
On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m and may be treated as uniform spherical objects.
Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.
Answer:
[tex]F = 1.489*10^{-7} N[/tex]
Explanation: Weight of space probes on earth is given by:[tex]W= m*g[/tex]
W= weight of the object( in N)
m= mass of the object (in kg)
g=acceleration due to gravity(9.81 [tex]\frac{m}{s^{2} }[/tex])
Therefore,
[tex]m_{1} = \frac{14500}{9.81}[/tex]
[tex]m_{1} = 1478.08 kg[/tex]
Similarly,
[tex]m_{2} = \frac{4800}{9.81}[/tex]
[tex]m_{2} = 489.29 kg[/tex]
Now, considering these two parts as uniform spherical objects
Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.
Force between these two objects is given by:
[tex]F = \frac{Gm_{1} m_{2}}{R^{2} }[/tex]
G= gravitational constant ([tex]6.67 * 10^{-11} m^{3} kg^{-1} s^{-2}[/tex])
[tex]m_{1} , m_{2}[/tex]= masses of the object
R= distance between their centres (in m)(18 m)
Substituiting all these values into the above formula
[tex]F = 1.489*10^{-7} N[/tex]
This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind is blowing. If the wind's force on the rope is negligible, what drag force in Newtons does the wind exert on the ball? B)A box is sliding down an incline tilted at a 15° angle above horizontal. The box is initially sliding down the incline at a speed of 1.4 m/s. The coefficient of kinetic friction between the box and the incline is 0.37. How far does the box slide down the incline before coming to rest?C)An object weighing 3.9 N falls from rest subject to a frictional drag force given by Fdrag = bv2, where v is the speed of the object and b = 2.5 N ∙ s2/m2. What terminal speed will this object approach?
Answer:
Part a)
[tex]F_v = 4.28 N[/tex]
Part B)
[tex]L = 1.02 m[/tex]
Part C)
[tex]v = 1.25 m/s[/tex]
Explanation:
Part A)
As we know that ball is hanging from the top and its angle with the vertical is 20 degree
so we will have
[tex]Tcos\theta = mg[/tex]
[tex]T sin\theta = F_v[/tex]
[tex]\frac{F_v}{mg} = tan\theta[/tex]
[tex]F_v = mg tan\theta[/tex]
[tex]F_v = 1.2\times 9.81 (tan20)[/tex]
[tex]F_v = 4.28 N[/tex]
Part B)
Here we can use energy theorem to find the distance that it will move
[tex]-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2[/tex]
[tex](-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2[/tex]
[tex](-3.5 + 2.54)L = - 0.98[/tex]
[tex]L = 1.02 m[/tex]
Part C)
At terminal speed condition we know that
[tex]F_v = mg[/tex]
[tex]bv^2 = mg[/tex]
[tex]2.5 v^2 = 3.9[/tex]
[tex]v = 1.25 m/s[/tex]
A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50.00 cm mark. (a) The period of oscillation is 2.50 s. Find d. (b) If you moved the pivot 5.00 cm closer to the 50.00 cm mark, what would the period of oscillation be
Answer:
(a). The value of d is 0.056 cm and 1.496 cm.
(b). The time period is 1.35 sec.
Explanation:
Given that,
Length = 50.00 cm
Time period = 2.50 s
Time period of pendulum is defined as the time for one complete cycle.
The period depends on the length of the pendulum.
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{I}{mgh}}[/tex]
Where, I = moment of inertia
We need to calculate the value of d
Using parallel theorem of moment of inertia
[tex]I=I_{cm}+md^2[/tex]
For a meter stick mass m , the rotational inertia about it's center of mass
[tex]I_{cm}-\dfrac{mL^2}{12}[/tex]
Where, L = 1 m
Put the value into the formula of time period
[tex]T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}[/tex]
[tex]T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}[/tex]
[tex]T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})[/tex]
Multiplying both sides by d
tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]
[tex](\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0[/tex]
Put the value of T, L and g into the formula
[tex]4.028d^2-6.25d+0.336=0[/tex]
[tex]d = 0.056\ m, 1.496\ m[/tex]
The value of d is 0.056 cm and 1.496 cm.
(b). Given that,
L = 50-5 = 45 cm
We need to calculate the time period
Using formula of period
[tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]
Put the value into the formula
[tex]T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}[/tex]
[tex]T=1.35\ sec[/tex]
Hence, (a). The value of d is 0.056 cm and 1.496 cm.
(b). The time period is 1.35 sec.
"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in groups dozens of millions of lightyears away. What method do astronomers use with the Hubble to find such distances
Answer:
finding Cepheid variable and measuring their periods.
Explanation:
This method is called finding Cepheid variable and measuring their periods.
Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.
A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.
Answer:
finding Cepheid variable and measuring their periods.
Explanation:
This method is called finding Cepheid variable and measuring their periods.
Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.
A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.
The velocity function (in meters per second) is given for a particle moving along a line. v(t) = 5t − 9, 0 ≤ t ≤ 3 (a) Find the displacement. -9/2 Correct: Your answer is correct. m (b) Find the distance traveled by the particle during the given time interval.
Answer:
a) The displacement is -4.5 m.
b) The traveled distance is 11.7 m.
Explanation:
Hi there!
a)The velocity of the particle is the derivative of the displacement function, x(t):
v(t) = dx/dt = 5t - 9
Separating varibles:
dx = (5t - 9)dt
Integrating both sides from x = x0 to x and from t = 0 to t.
x - x0 = 1/2 · 5t² - 9t
x = 1/2 · 5t² - 9t + x0
If we place the origin of the system of reference at x = x0, the displacement at t = 3 will be x(3):
x(3) = 1/2 · 5 · (3)² - 9(3) + 0
x(3) = -4.5
The displacement at t = 3 s is -4.5 m. It means that the particle is located 4.5 m to the left from the origin of the system of reference at t = 3 s.
b) When the velocity is negative, the particle moves to the left. Let´s find the time at which the velocity is less than zero:
v = 5t - 9
0 > 5t - 9
9/5 > t
1.8 s > t
Then until t = 1.8 s, the particle moves to the left from the origin of the reference system.
Let´s find the position of the particle at that time:
x = 1/2 · 5t² - 9t
x = 1/2 · 5(1.8 s)² - 9(1.8 s)
x = -8.1 m
From t = 0 to t = 1.8 s the traveled distance is 8.1 m. After 1.8 s, the particle has positive velocity. It means that the particle is moving to the right, towards the origin. If at t = 3 the position of the particle is -4.5 m, then the traveled distance from x = -8.1 m to x = -4.5 m is (8.1 m - 4.5 m) 3.6 m.
Then, the total traveled distance is (8.1 m + 3.6 m) 11.7 m.
The distance travelled by the particle during the given time interval is 45/2 meters.
Explanation:The distance travelled by the particle can be found by integrating the absolute value of the velocity function over the given time interval:
The distance is, therefore, the integral of |5t - 9| from 0 to 3:
D = ∫(5t - 9)dt = [5(t^2/2) - 9t] from 0 to 3 = (45/2) meters
So, the distance travelled by the particle during the given time interval is 45/2 meters.
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Let's begin by determining the equilibrium position of a seesaw pivot. You and a friend play on a seesaw.Your mass is____
90 kg, and your friend’s mass is 60 kg. The seesaw board is 3.0 m long and has negligible mass. Where should the pivot be placed so that the seesaw will balance when you sit on the left end and your friend sits on the right end?
To balance a seesaw with a 90 kg person on one end and a 60 kg person on the other end, the pivot should be placed 1.2 m from the 90 kg person. This is calculated using concepts of physics, specifically torque and equilibrium, assuming the force is applied at the person's center of mass.
Explanation:In physics, this problem can be solved using the concept of torque and the conditions for equilibrium. For the seesaw to be in balance or equilibrium, the total torque about the pivot point must be zero. Torque (τ) is defined as the product of the force (F) applied and the distance (d) from the pivot point where the force is applied, i.e., τ = Fd.
In this case, let's assume the pivot is placed x meters from your end of the seesaw. The weights of you and your friend can be represented as forces through multiplication by gravity (approx. 9.81 m/s^2). So, for you, the torque is (90 kg x 9.81 m/s^2) x and for your friend, it is (60 kg x 9.81 m/s^2) (3 m - x).
In equilibrium, these two torques should be equal, so we get the equation: (90 kg x 9.81 m/s^2) x = (60 kg x 9.81 m/s^2) (3 m - x). Solving this equation gives x = 1.2 m. So, the pivot should be placed 1.2 m from your end (90 kg person) for the seesaw to balance.
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Plane a flies at a constant speed from new york to los angeles along a route which is 2000 miles. Plane b flies in the opposite direction at a constant speed which is 100 mph faster than plane a. Plane b takes off one hour after plane a. They land at the same moment. How far are they from los angeles when they pass?
Answer:
from los angeles distance plan a = 1111.08 mi
from los angeles distance plan b = 888.92 mi
Explanation:
given data
new york to los angeles distance = 2000 miles
Plane b speed = 100 mph faster than plane a
Plane b takes off time = 1 hour after plane a
to find out
How far are they from los angeles when they pass
solution
we consider speed of plan a is = x mph
so speed of plan b will be = x + 100 mph
and we know plan b take here 1 hour less time than plan a so it mean time is distance divide speed i.e
[tex]\frac{2000}{x} - 1 =\frac{2000}{x+100}[/tex]
solve it we get x = 400 mph
it mean here
plan a speed is 400 mph
and plan b speed is 500 mph
and
now we consider they meet at time = t hour after a take off
then plan a travel = 400 t
and plan b travel = 500 ( t - 1 )
add both distance that is equal to 2000 mi
so 400 t + 500 ( t -1 ) = 2000
400 t + 500 ( t -1 ) = 2000
400 t + 500 ( t-1) = 2000
solve we get
t = 2.777
so total distance travel plan a = 400 × 2.777 = 1111.08 mi
total distance travel plan b = 2000 - 1111.08
total distance travel plan b = 888.92 mi
A 11.0 g rifle bullet is fired with a speed of 380 m/s into a ballistic pendulum with mass 10.0 kg, suspended from a cord 70.0 cm long.
a) Compute the vertical height through which the pendulum rises.(cm)
b) Compute the initial kinetic energy of the bullet;(j)
c) Compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum.(j)
Answer:
a) h = 0.0088 m
b) Kb = 794.2J
c) Kt = 0.88J
Explanation:
By conservation of the linear momentum:
[tex]m_b*V_b = (m_b+m_p)*Vt[/tex]
[tex]Vt = \frac{m_b*V_b}{m_b+m_p}[/tex]
[tex]Vt=0.42m/s[/tex]
By conservation of energy from the instant after the bullet is embedded until their maximum height:
[tex]1/2*(m_b+m_p)*Vt^2-(m_b+m_p)*g*h=0[/tex]
[tex]h =\frac{Vt^2}{2*g}[/tex]
h=0.0088m
The kinetic energy of the bullet is:
[tex]K_b=1/2*m_b*V_b^2[/tex]
[tex]K_b=794.2J[/tex]
The kinetic energy of the pendulum+bullet:
[tex]K_t=1/2*(m_b+m_p)*Vt^2[/tex]
[tex]K_t=0.88J[/tex]
a. The vertical height through which the pendulum rises is equal to 0.9 cm.
b. The initial kinetic energy of the bullet is equal to 794.2 Joules.
c. The kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum is equal to 0.883 Joules.
Given the following data:
Mass of bullet = 11.0 gSpeed = 380 m/sMass of pendulum = 10.0 kgLength of cord = 70.0 cma. To determine the vertical height through which the pendulum rises:
First of all, we would find the final velocity by applying the law of conservation of momentum:
Momentum of bullet is equal to the sum of the momentum of bullet and pendulum.
[tex]M_bV_b = (M_b + M_p)V[/tex]
Where:
[tex]M_b[/tex] is the mass of bullet.[tex]M_p[/tex] is the mass of pendulum.[tex]V_b[/tex] is the velocity of bullet.V is the final velocity.Substituting the given parameters into the formula, we have;
[tex]0.011\times 380 = (0.011+10)V\\\\4.18 = 10.011V\\\\V = \frac{4.18}{10.011}[/tex]
Final speed, V = 0.42 m/s
Now, we would find the height by using this formula:
[tex]Height = \frac{v^2}{2g} \\\\Height = \frac{0.42^2}{2\times 9.8} \\\\Height = \frac{0.1764}{19.6}[/tex]
Height = 0.009 meters.
In centimeters:
Height = [tex]0.009 \times 100 = 0.9 \;cm[/tex]
b. To compute the initial kinetic energy of the bullet:
[tex]K.E_i = \frac{1}{2} M_bV_b^2\\\\K.E_i = \frac{1}{2} \times 0.011 \times 380^2\\\\K.E_i = 0.0055\times 144400\\\\K.E_i = 794.2 \; J[/tex]
Initial kinetic energy = 794.2 Joules
c. To compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum:
[tex]K.E = \frac{1}{2} (M_b + M_p)V^2\\\\K.E = \frac{1}{2} \times(0.011 + 10) \times 0.42^2\\\\K.E = \frac{1}{2} \times 10.011 \times 0.1764\\\\K.E = 5.0055 \times 0.1764[/tex]
Kinetic energy = 0.883 Joules.
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Your grandfather clock's pendulum has a length of 0.9930 m. Part A If the clock runs slow and loses 19 s per day, how should you adjust the length of the pendulum? Note: due to the precise nature of this problem you must treat the constant g as unknown (that is, do not assume it is equal to exactly 9.80 m/s2).
To compensate for the clock's slowness, you should adjust the length of the pendulum to [tex]\( 0.9939 \)[/tex] meters.
To calculate the necessary adjustment to the length of the pendulum to compensate for the clock's slowness, we can start by determining the time period of the pendulum using the given length and the unknown acceleration due to gravity, denoted as [tex]\( g \).[/tex] The time period [tex]\( T \)[/tex] of a simple pendulum is given by the formula:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
where [tex]\( L \)[/tex] is the length of the pendulum.
Given that the clock loses 19 seconds per day, we convert this to seconds per period, as [tex]\( 1 \) day has \( 24 \) hours, \( 60 \)[/tex] minutes per hour, and [tex]\( 60 \)[/tex] seconds per minute:
[tex]\[ \text{Seconds lost per period} = \frac{19 \, \text{s}}{24 \times 60 \times 60 \, \text{s/day}} \][/tex]
Let's calculate the time period [tex]\( T \)[/tex] and the adjustment needed.
[tex]\[ T = 2\pi \sqrt{\frac{0.9930 \, \text{m}}{g}} \][/tex]
Now, we'll solve for [tex]\( g \):[/tex]
[tex]\[ T = 2\pi \sqrt{\frac{0.9930 \, \text{m}}{g}} \]\[ T^2 = 4\pi^2 \frac{0.9930 \, \text{m}}{g} \]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{T^2} \][/tex]
Substitute [tex]\( T \) with \( 24 \, \text{hours} \)[/tex] converted to seconds:
[tex]\[ T = 24 \times 60 \times 60 \, \text{s} = 86,400 \, \text{s} \]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{(86,400 \, \text{s})^2} \][/tex]
Calculate [tex]\( g \):[/tex]
[tex]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{7.48224 \times 10^9 \, \text{s}^2} \]\[ g = 0.0000016512 \, \text{m/s}^2 \][/tex]
Now that we have [tex]\( g \)[/tex], we can find the adjusted length of the pendulum. The new time period [tex]\( T' \)[/tex] can be calculated as follows:
[tex]\[ T' = 2\pi \sqrt{\frac{L'}{g}} \][/tex]
Where [tex]\( L' \)[/tex] is the adjusted length. Rearrange to solve for [tex]\( L' \):[/tex]
[tex]\[ L' = \frac{T'^2 \times g}{4\pi^2} \][/tex]
Given that the clock loses [tex]\( 19 \)[/tex] seconds per day and we want it to lose [tex]\( 0 \)[/tex] seconds, the new time period [tex]\( T' \) is \( 86,400 \)[/tex] seconds. Substitute into the formula:
[tex]\[ L' = \frac{(86,400 \, \text{s})^2 \times 0.0000016512 \, \text{m/s}^2}{4\pi^2} \][/tex]
[tex]\[ L' = 0.9939 \, \text{m} \][/tex]
So, to compensate for the clock's slowness, you should adjust the length of the pendulum to [tex]\( 0.9939 \)[/tex] meters.
Complete Question:
Your grandfather clock's pendulum has a length of 0.9930 m. Part A If the clock runs slow and loses 19 s per day, how should you adjust the length of the pendulum? Note: due to the precise nature of this problem you must treat the constant g as unknown (that is, do not assume it is equal to exactly 9.80 m/s2).
Which of the following systems acquisition methods requires staff to systematically go through every step in the development process and has a lower probability of missing important user requirements?
a.Systems development life cycle
b. Prototyping
c. End-user development
d. External acquisition
e. Object-oriented development
Answer:
a.Systems development life cycle
Explanation:
Of all the options the correct answer is a.Systems development life cycle.
Systems development life cycle: The life cycle phases of systems development include preparation, system assessment, system design, advancement, application, inclusion and testing, as well as maintenance and support.
So, we can see that the Systems development life cycle enables staff to systematically go through every step in the development process and has a lower probability of missing important user requirements.
Final answer:
The Systems Development Life Cycle (SDLC) is a methodical approach that encompasses a thorough step-by-step process to ensure all user requirements are captured, reducing the risk of overlooking important needs.
Explanation:
The system acquisition method that requires staff to systematically go through every step in the development process and has a lower probability of missing important user requirements is the Systems Development Life Cycle (SDLC). SDLC is a structured process that involves detailed planning, building, testing, and deployment, ensuring that all user requirements are met comprehensively.
Unlike prototyping, which may be quicker but less thorough, or end-user development which might miss broader system requirements, SDLC's methodical approach reduces the chance of overlooking user needs. Furthermore, external acquisition and object-oriented development are not as specifically focused on capturing all user requirements through a step-by-step process.
Two students on roller skates stand face-toface, then push each other away. One student has a mass of 93 kg and the second student 65 kg. Find the ratio of the magnitude of the first student’s velocity to the magnitude of the second student’s velocity.
Answer:
[tex]\frac{v_1}{v_2} = 0.698[/tex]
Explanation:
As we know that the two students are standing on skates
So there is no external force on the system of two students
So we can say that momentum is conserved
So here initially both students are at rest and hence initial momentum is zero
So we have
[tex]P_i = P_f[/tex]
[tex]m_1v_1 + m_2v_2 = 0[/tex]
[tex]\frac{v_1}{v_2} = \frac{m_2}{m_1}[/tex]
[tex]\frac{v_1}{v_2} = \frac{65}{93}[/tex]
[tex]\frac{v_1}{v_2} = 0.698[/tex]
The correct ratio of the magnitude of the first student's velocity to the magnitude of the second student's velocity is 65:93.
To find the ratio of the magnitudes of the velocities of the two students after they push each other away, we can use the principle of conservation of momentum. According to this principle, in the absence of external forces, the total momentum of a system remains constant.
Let's denote the velocities of the first and second students as[tex]\( v_1 \)[/tex]and [tex]\( v_2 \)[/tex] respectively. Since the students push each other in opposite directions, their momenta will be equal in magnitude but opposite in direction. We can write the conservation of momentum as:
[tex]\[ m_1 \cdot v_1 = m_2 \cdot v_2 \][/tex]
where [tex]\( m_1 = 93 \)[/tex] kg is the mass of the first student and[tex]\( m_2 = 65 \) kg[/tex] is the mass of the second student.
To find the ratio of the velocities, we divide both sides of the equation by[tex]\( m_2 \cdot v_2 \)[/tex]:
[tex]\[ \frac{m_1 \cdot v_1}{m_2 \cdot v_2} = 1 \][/tex]
[tex]\[ \frac{v_1}{v_2} = \frac{m_2}{m_1} \][/tex]
Substituting the given masses:
[tex]\[ \frac{v_1}{v_2} = \frac{65 \text{ kg}}{93 \text{ kg}} \][/tex]
Simplifying the ratio, we get:
[tex]\[ \frac{v_1}{v_2} = \frac{65}{93} \][/tex]
The acceleration due to gravity on the surface of a planet is five times as large as it is on the surface of Earth. The mass density of the planet is known to be four times that of Earth. What is the radius of this planet in terms of Earth's radius?
Answer:
1.25 R
Explanation:
Acceleration due to gravity on earth, ge = g
Acceleration due to gravity on planet, gP = 5 times the acceleration due to gravity on earth
gP = 5 g
Density of planet = 5 x density of earth
Let the radius of earth is R
Let the radius of planet is Rp.
Use the for acceleration due to gravity
[tex]g = \frac{4}{3}G\pi R\rho[/tex]
where, G s the universal gravitational constant and ρ be the density of planet.
For earth
[tex]g = \frac{4}{3}G\pi R\rho[/tex] .... (1)
For planet
[tex]g_{P} = \frac{4}{3}G\pi R_{P}\rho_{P}[/tex]
According to the question
gp = 5 g, ρP = 4 ρ
Substitute the values
[tex]5g = \frac{4}{3}G\pi R_{P}\4rho[/tex] .... (2)
Divide equation (2) by equation (1), we get
[tex]5=\frac{R_{p\times 4\rho }}{R\rho }[/tex]
Rp = 1.25 R
Thus, the radius of planet 1.25 R.
At a rock concert, a dB meter registered 128 dB when placed 2.9 m in front of a loudspeaker on the stage. The intensity of the reference level required to determine the sound level is 1.0×10−12W/m2. Determine the following:
A. What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?
B. How far away would the sound level be 82 dB?
Answer:
A)
665.5 W
B)
575.5 m
Explanation:
A)
[tex]S[/tex] = Sound level registered = 128 dB
[tex]I_{o}[/tex] = Intensity at reference level = [tex]1\times10^{-12}[/tex] Wm⁻²
[tex]I[/tex] = Intensity at the location of meter
Sound level is given as
[tex]S = 10 log_{10}\left ( \frac{I}{I_{o}} \right )[/tex]
[tex]128 = 10 log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\12.8 = log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\10^{12.8} = \frac{I}{1\times10^{-12}} \right \\I = 6.3[/tex]
[tex]P[/tex] = power output of the speaker
[tex]r[/tex] = distance from the speaker = 2.9 m
Power output of the speaker is given as
[tex]P = I(4\pi r^{2} )\\P = (6.3) (4) (3.14) 2.9^{2}\\P = 665.5 W[/tex]
B)
[tex]S[/tex] = Sound level = 82 dB
[tex]I_{o}[/tex] = Intensity at reference level = [tex]1\times10^{-12}[/tex] Wm⁻²
[tex]I[/tex] = Intensity at the location of meter
Sound level is given as
[tex]S = 10 log_{10}\left ( \frac{I}{I_{o}} \right )[/tex]
[tex]82 = 10 log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\8.2 = log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\10^{8.2} = \frac{I}{1\times10^{-12}} \right \\I = 1.6\times10^{-4}Wm^{-2}[/tex]
Power of the speaker is given as
[tex]P = I(4\pi r^{2} )\\665.5 = (1.6\times10^{-4}) (4) (3.14) r^{2}\\r = 575.5 m[/tex]
The power output of the speaker at the rock concert was approximately 137 Watts. The sound level would be 82 dB at a distance of about 103 meters from the speaker.
Explanation:This problem involves the use of the formula for calculating sound levels in decibels (dB). The formula is L = 10 log(I/I0) where L is the sound level in dB, I is the intensity of the sound wave, and I0 is the reference intensity. Now, let's solve both parts.
A. At the dB meter location, reverse calculate the intensity I: I = I0 * 10^(L/10) = 1.0 10^−12 * 10^(128/10) = 1.26 W/m^2. Since the sound spreads spherically, the power P is the intensity times the area of the sphere: P = I * 4πr^2 = 1.26 * 4π*(2.9)^2 = 137 Watts.
B. To find the distance where the sound would register at 82 dB, we first find the intensity I at that level: I = I0*10^(82/10) = 1.0x10^−12 * 10^8.2 = 6.31x10^-5 W/m^2. With I, we can solve for r: r = sqrt(P/(4πI)) = sqrt(137/(4π*(6.31x10^-5))) = 103 meters.
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Sometimes, when the wind blows across a long wire, a low-frequency "moaning" sound is produced. The sound arises because a standing wave is set up on the wire, like a standing wave on a guitar string. Assume that a wire (linear density = 0.0180 kg / m ) sustains a tension of 350 N because the wire is stretched between two poles that are 17.43 m apart. The lowest frequency that an average, healthy human ear can detect is 20.0 Hz. What is the lowest harmonic number n that could be responsible for the "moaning" sound?
Answer:
N = 5 harmonics
Explanation:
As we know that frequency of the sound is given as
[tex]f = \frac{N}{2L}\sqrt{\frac{T}{\mu}}[/tex]
now we have
[tex]T = 350 N[/tex]
[tex]\mu = 0.0180 kg/m[/tex]
L = 17.43 m
now we have
[tex]f = \frac{N}{2(17.43)}\sqrt{\frac{350}{0.0180}}[/tex]
[tex]f = 4 N[/tex]
if the lowest audible frequency is f = 20 Hz
so number of harmonics is given as
[tex]20 = 4 N[/tex]
N = 5 harmonics
A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of the U-shaped tube until the vertical height of the water column is hW = 19.0 cm .
The gauge pressure at the water/mercury interface is simply the "head" due to the water:
p = ρgh = 1000kg/m³ * 9.8m/s² * 0.19m = 1862 Pa
If we take the specific gravity of mercury to be 13.6, then the difference in height between the water and mercury columns is
h' = h(1 - 1/s.g.) = 19cm * (1 - 1/13.6) = 17.6 cm