Answer: the at which the bar conducts now is 5 Js⁻¹
Explanation:
Given the data in the question;
we know that; Heat transfer by conduction is given by;
Q ∝ [tex]A / l[/tex]
such that,
[tex]Q_{1}/Q_{2}[/tex] = [tex]A_{1}l_{2} / A_{2}l_{1}[/tex]
[tex]Q_{2} = Q_{1} A_{2}l_{1}/A_{1}l_{2}[/tex]
so
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r}{2}[/tex])² × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r^{2} }{4}[/tex]) × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × πr² × 1/4 × [tex]l[/tex] ) / (πr² × 1/2 × [tex]l[/tex] )
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × 1/4) / ( 1/2)
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × 1/4) / ( 1/2)
[tex]Q_{2} =[/tex] 5 Js⁻¹
Therefore, the at which the bar conducts now is 5 Js⁻¹
what is a asteroid traveling rapidly called
Answer:
meteor
Explanation:
A asteroid stays still and a meteor goes fast
Answer:
meteor
Explanation:
or some people call it a shooting star
Energy cannot be created nor destroyed, only changed from one form to another. How does listening to music on a radio obey the law of conservation of energy?
Group of answer choices
Sound energy is changed into potential energy.
Electrical energy is converted into other forms of energy, such as sound.
Electrical energy remains unchanged.
Electrical energy is gradually destroyed as the radio plays.
Answer: Electrical energy is converted into other forms of energy, such as sound.
Explanation: Using the law of conservation of energy we know that energy can never be destryped it can be transferred or be transformed into from one form to another.
Electrical energy is converted into other forms of energy, such as sound.
The law of conservation of energy or matter, states that energy can neither be created nor destroyed but can be converted from one form to another.
When listening to music on radio, the electric energy supplied to the radio will be converted to mechanical energy of the moving parts of the radio which is then converted to sound energy.
Thus, we can conclude that electrical energy is converted into other forms of energy, such as sound.
Learn more here:https://brainly.com/question/2828402
A woman on a bridge 84.5 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 6.00 m more to travel before passing under the bridge. The stone hits the water 4.00 m in front of the raft. Find the speed of the raft.
Answer:
V = 0.48 m/s
Explanation:
In this case, we need to analyze the given data by parts.
At first, we know that the woman is on a height of 84.5 m of a river. She drops a stone thinking that she may hit the raft that is traveling with a constant speed. When the raft is 6 m near the bridge, the woman drops the stone, and the stone hits the water when the raft is still 4 m far of the bridge.
With this given data, we can calculate the distance covered by the raft, because is traveling at a constant speed:
X = 6 - 4 = 2 m
And as it's traveling at constant speed then:
X = V.t
We have the distance of the raft, but not the time it took to cover that distance. This time will be the same time that the stone took to hit the water, therefore, if we can determine the time of the rock, well be determining the time of the raft to cover the distance, and then, we can determine it speed.
To determine the time of the rock, as the stone is going on a free fall, with an innitial speed of 0, the flight time of the rock will be:
y = gt²/2 ---> solving for t
2y/g = t²
t = √2y/g
If g = 9.8 m/s, and replacing the data we have that the flight time of the rock is:
t = √2*84.5 / 9.8
t = 4.15 s
So the rock took 4.15 s to hit the water, and it's also the time that the raft took to cover the distance of 2 m, then, it's speed:
V = X/t
V = 2 / 4.15
V = 0.48 m/sHope this helps
Microbes such as bacteria have small positive charges when in solution. Public health agencies are exploring a new way to measure the presence of small numbers of microbes in drinking water by using electric forces to concentrate the microbes. Water is sent between the two oppositely charged electrodes of a parallel-plate capacitor. Any microbes in the water will collect on one of the electrodes.
Required:
a. On which electrode will the microbes collect?
b. How could the microbes be easily removed from the electrodes for analysis?
Answer:
The answer is below
Explanation:
a) Coulomb's law of electric force for charges at rest states that Like charges repel each other while unlike charges attract one another.
Therefore since the microbes has small positive charges, the microbe would be repelled by the positively charged electrodes and attracted by the negative charged electrodes.
Hence, the microbes would collect on the negatively charged electrodes.
b) The microbes can easily removed from the negative electrode for analysis by discharging the electrode from the source. Thereby making the electrode to be incapable of attracting the microbe.
This problem has been solved! See the answer A 6.0 kg object, initially at rest in free space, "explodes" into three segments of equal mass. Two of these segments are observed to be moving with equal speeds of 20 m/s with an angle of 60 degrees between their directions of motion. How much kinetic energy is released in this explosion?
Answer:
Explanation:
mass of each part = 6 / 3 = 2 kg .
momentum of each of given part = 2 x 20 = 40 kg m/s
Two momentum of 40 kg m/s , acting at angle 60 degree .
Resultant momentum = 2 x 40 cos 30 = 69.28 kg m/s
The third mass will have equal and opposite momentum to this momentum , following law of conservation of momentum .
If its velocity be v .
2 x v = 69.28
v = 34.64 m /s
Third mass will have velocity of 34.64 m /s
Total kinetic energy of all three mass
KE = 1/2 x 2 ( 20² + 20² + 34.64² )
= 400 + 400 + 1199.92
= 1999.93 J .
When the moving sidewalk at the airport is broken, as it often seems to be, it takes you 41 ss to walk from your gate to baggage claim. When it is working and you stand on the moving sidewalk the entire way, without walking, it takes 80 ss to travel the same distance. How long will it take you to travel from the gate to baggage claim if you walk while riding on the moving sidewalk?
Answer:
41
Explanation:
An airplane used to drop water on brushfires is flying horizontally in a straight line at 180 mi/h at an altitude of 450 ft. Determine the distance d at which the pilot should release the water so that it will hit the fire at B. The distance at which the pilot should release the water so that it will hit the fire at B is ft.
Answer:
1399.2 ft
Explanation:
The initial velocity = 180 mph = [(180 * 5280) / (1 * 3600)] ft/s = 264 ft/s
[tex]In\ the \ horizontal\ direction(x)\\\\Initial\ velocity = v_{ox}=264\ ft/s\\\\distance\ travelled\ in\ x \ direction(x) =v_{ox}t\\\\\\For\ the\ vertical\ direction:\\\\initial\ velocity(y_{oy})=0\\\\vertical\ distance(y)=y_{oy}t+0.5gt^2\\\\but\ g\ =-32\ ft/s^2. Hence:\\\\y=0t+0.5(-32)t^2\\\\y=-16t^2\\\\At\ point\ B, y=-450, therefore:\\\\-450=-16t^2\\\\t^2=28.125\\\\t=5.3\ s\\\\The\ distance\ at\ which\ the\ pilot\ should\ release\ the\ water=x=v_{ox}t=264*5.3\\\\x=1399.2\ ft[/tex]
why would the bulb not light?
Im not sure this is correct but, i think thats because the wire connected at the negative side. The current go from positive to negative and thats why the bulb not light, also the bulb location is not at the center of the battery which explain why there's no current at the upper side.
Hope you understand. Sorry if my english sucks YEET
The light bulb isn't connected to the positive thing sticking out of the battery, both ends must connect back to the battery
A Typical operating voltage of an electron microscope is 50 kV. A Typical experimental operating voltage range of a Scanning electron microscope is 1kV to 30kV. Higher voltages can penetrate and causes deformation on the sample. Lets assume it operates at 10kV. (i)What is the smallest distance that it could possibly resolve
Answer:
y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]
Explanation:
Let's solve this exercise in parts. Let's start by finding the wavelength of the electrons accelerated to v = 10 103 V, let's use the DeBroglie relation
λ= [tex]\frac{h}{p} = \frac{h}{mv}[/tex]
Let's use conservation of energy for speed
starting point
Em₀ = U = e V
final point
Em_f = K = ½ m v²
Em₀ = Em_f
eV = ½ m v²
v =[tex]\sqrt{\frac{2eV}{m} }[/tex]
we substitute
λ= [tex]\sqrt{ \frac{h^2 m}{2eV}}[/tex]
the diffraction phenomenon determines the minimum resolution, for this we find the first zero of the spectrum
a sin θ = m λ
first zero occurs at m = 1, also these experiments are performed at very small angles
sin θ = θ
θ = λ / a
This expression is valid for linear slits, in the microscope the slits are circular, when solving the polar coordinates we obtain
θ = 1.22 λ / D
where D is the diameter of the opening
we substitute
θ = [tex]\frac{1.22}{D}[/tex] \sqrt{ \frac{h^2 m}{2eV}}
this is the minimum angle that can be seen, if the distance is desired suppose that the distance of the microscope is L, as the angles are measured in radians
θ = y / L
when substituting
where y is the minimum distance that can be resolved for this acceleration voltage
y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]
If you blow harder into a trumpet you will create a sound wave with a greater
A. pitch
B. amplitude
C wavelength
Answer is C
Explanation:
Answer:
C. wavelength
good luck, i hope this helps :)
SERIOUSLY HELP NOWWW LIKE NOWW I REALLY NEED THIS ANSWERED
Which of the following is the best way to measure the age of the Earth?
Examine rocks collected in my backyard.
Analyze maps of all known fossil sites in the United States.
Test the rate of decay of specific elements in rock samples.
Explore leaves embedded in ash from recent volcanic eruptions.
Answer:
the second one Test the rate of decay of specific elements in rock samples.
A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.76 rad/sec. The moment of inertia of the student plus the stool is 5 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis.
Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.
Answer:
a) the new angular speed of the student is 0.9642 rad/s
b)
the kinetic energy of the student before the objects are pulled in is 1.9119 J
the kinetic energy of the student after the objects are pulled in is 2.4252 J
Explanation:
Given that;
mass of each object m = 1 kg
distance of objects from axis of rotation r = 0.9 m
Moment of inertia of each object initially [tex]I_{oi}[/tex]
[tex]I_{oi}[/tex] = mr² = 1kg ×(0.9m)² = 1 kg × 0.81 m² = 0.81 kg.m²
moment of inertia of each object finally [tex]I_{of}[/tex]
[tex]I_{of}[/tex] = mr² = 1kg × (0.33 m)² = 0.1089 kg.m²
Now
moment of inertia of student plus stool [tex]I_{}[/tex] = 5 kg.m²
initial angular speed ω₀ = 0.76 rad/sec
final angular speed ω = ?
Now using conservation of angular momentum;
([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀ = ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω
so we substitute
(5 + 2 (0.81) )0.76 = (5 + 2 (0.1089) )ω
5.0312 = 5.2178 ω
ω = 5.0312 / 5.2178
ω = 0.9642 rad/s
Therefore, the new angular speed of the student is 0.9642 rad/s
b)
K.E of student before = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀²
= (0.5) (5 + 2 (0.81) )(0.76)²
= 0.5 × 6.62 × 0.5776
= 1.9119 J
Therefore, the kinetic energy of the student before the objects are pulled in is 1.9119 J
KE of student finally = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω²
= (0.5) (5 + 2 (0.1089) ) (0.9642)²
= 0.5 × 5.2178 × 0.9296
= 2.4252 J
Therefore, the kinetic energy of the student after the objects are pulled in is 2.4252 J
Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees rest on the flowers. The charged bees both generate an electric field, and while the charged bees are resting on the flowers, the net electric field at some distance between them is zero.
(a) Do the bees have the same or opposite signs of charge?
Same � the electric fields point in opposite directions and therefore cancel at some midpoint.
Same � the electric fields multiply together to equal zero.
Opposite � the electric fields point in the same direction summing to zero.
Opposite � the net electric field due to the two bees points in a direction perpendicular to the direction from one bee to the other.
(b) Suppose the net electric field is zero at a distance that is closer to bee 1. Does bee 1 have a magnitude of charge greater than or less than that of bee 2?
greater than
less than
Answer:
a. Same � the electric fields point in opposite directions and therefore cancel at some midpoint.
b. bee 1 has a magnitude of charge less than bee 2
Explanation:
a. Do the bees have the same or opposite signs of charge?
They have the same charge. This is because since same charges would produce electric fields in opposite directions, that is the only way they can cancel out at some point. So, the charges are the same and the electric fields point in opposite directions and therefore cancel at some midpoint.
b. Suppose the net electric field is zero at a distance that is closer to bee 1. Does bee 1 have a magnitude of charge greater than or less than that of bee 2?
Let q be the charge on bee 1 and r its distance from the neutral electric field point. So, it electric field E = kq/r².
Also, let q' be the charge on bee 2 and d its distance from the neutral electric field point. So, it electric field E' = kq'/d².
Since E = E' at the neutral point.
kq/r² = kq'/d²
q/q' = r²/d² = (r/d)²
Given that r < d, so r/d < 1 and (r/d)² < 1
So, q/q' < 1
q < q'
So, the charge on bee 1 is less than that on bee 2
Temperature and salinity difference in ocean water cause...
Answer:
high and low tied maybe? not sure
A block of concrete has a mass of 48kg a crane lifts the block to a height of 12m above the ground calculate the gravitational potential energy stored by the block (gravitational field strength g=10N/kg)
Answer:
5760 J
Explanation:
From the question given above, the following data were obtained:
Mass of block = 48 kg
Height (h) = 12 m
Gravitational field strength (g) = 10 N/Kg
Gravitational potential energy (PE) =?
The gravitational potential energy stored by the block can simply be obtained as follow:
PE = mgh
PE = 48 × 10 × 12
PE = 5760 J
Therefore, the gravitational potential energy stored by the block is 5760 J
A dumped harmonic oscillator of a mass of 500 g has a period of 0.5 second. The amplitude of the oscillation is decreasing 2.0 % during each oscillation. a./ The initial amplitude is 10 cm. What will be the amplitude after 25 oscillation
Answer:
The answer is below
Explanation:
The amplitude decreases by 2% during each oscillation. Hence the decrease in amplitude can be represented by an exponential decay in the form:
y = abˣ; where x ad y are variables, a is the initial value and b is the factor.
Let y represent the amplitude after x oscillations. Since the initial amplitude is 10 cm, hence:
a = 10 cm, b = 2% = 0.02.
Therefore:
y = 10(0.02)ˣ
The amplitude after 25 oscillations is gotten by substituting x = 25 into the equation. Hence:
y = 10(0.02)²⁵
y= 3.355 * 10⁻⁴² cm
The amplitude after 25 oscillations is 3.355 * 10⁻⁴² cm
When the skin produces hair, it is removing excess water and salts from the body.
True or false?
The cart is given an initial push up the ramp. After this push, as the car moves up the ramp, the direction of the acceleration of the cart is ________ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _______.
Answer:
Explanation:
The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .
Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .
When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .
After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.
where is a neutron located within an atom
Answer:
the neutron is located in the nucleus of an atom
Answer:
nucleus
Explanation:
Atoms are made up of protons and neutrons located within the nucleus, with electrons in orbitals surrounding the nucleus.
There are 5 planets visible to the naked eye in the sky.
True
False
In a crash test, a car with a mass of 1600 kg is initially moving at a speed of 20 m/s just before it collides with a barrier. The final speed of the car after the collision is zero. The original length of the car is 4.50 m , but after the collision, the smashed car is only 3.60 m long.
Required:
a. What is the average speed Of the car during the period from first contact with the barrier to the moment the car comes to a stop? You may assume the force that the barrier exerts on the car is constant during this period.
b. How much time elapses between the moment the car makes first contact with the barrier and the moment it comes to a stop?
c. Making the very rough approximation that the large force that the barrier exerts on the car is approximately constant during contact, determine the approximate magnitude of this force?
Answer:
The answer to the given points can be defined as follows:
Explanation:
In point 1:
[tex]\bold{v_f^2= v_i^2+2as}\\\\\to v_f=0\\\\\to v_i=20 \frac{m}{s}\\\\\to s= 4.50\ m -3.60 \ m \\\\[/tex]
[tex]=0.9 \ m \\[/tex]
put the value in the above formula:
[tex]\to 0= 20^2+2 \times a \times 0.9\\\\\to -1.8\ a=400\\\\\to -a= \frac{400}{1.8} \\\\ \to a= -222.22\ \ \frac{m}{s^2}[/tex]
[tex]\bold{v_f=v_i+at}\\\\\to 0=20+ (-222.22)t\\\\\to 222.22t=20\\\\\to t=\frac{20}{222.22}\\\\\to t= 0.0900 \ s\\\\\to v_{avg}=\frac{s}{t}=\frac{0.9}{t}= 10\ \frac{m}{s}[/tex]
for point 2:
[tex]t= 0.0900 \ s -\text{found above}[/tex]
for point 3:
[tex]\to |a| = 222.22 \frac{m}{s^2} \text{found above}\\\\\to \bold{|F| = m \cdot |a|}\\\\[/tex]
[tex]=1600 \ kg \times 222.22 \ \frac{m}{s^2} \\\\= 3.55\times 10^{5} \ N[/tex]
A pitching machine is programmed to pitch baseballs horizontally at a speed of 134 km/h. The machine is mounted on a truck and aimed forward. As the truck drives toward you at a speed of 85 km/h, the machine shoots a ball toward you. A pickup truck moves to the left at a constant velocity. A pitching machine sits in the bed of the pickup truck. The pitching machine launches a baseball to the right with a different constant velocity. A man with a baseball mitt stands at rest some distance to the right of the truck. For each of the object pairings listed, determine the correct relative speed. The speed of the pitching machine relative to the truck The speed of the pitched ball relative to the truck The speed of the pitching machine relative to you The speed of the pitched ball relative to you
Answer: 134 = 143 = 151 = 166 = 176
Hope this helps!!
Sorry if it's incorrect!!
:'(
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta leading away from the heart. Since blood within the heart is essentially stationary, this pressure difference can be inferred from a measurement of the speed of blood flow in the aorta. Take the speed of sound in stationary blood to be c.
a. Sound sent by a transmitter placed directly inline with the aorta will be reflected back to a receiver and show a frequency shift with each heartbeat. If the maximum speed of blood in the aorta is v, what frequency will the receiver detect? Note that you cannot simply use the textbook Doppler Shift formula because the detector is the same device as the source, receiving sound after reflection.
b. Show that in the limit of low blood velocity (v <
f= 2fo v/c
Answer:
a) f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex] , b) Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo[tex]\frac{c+v}{c}[/tex]
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’ [tex]\frac{c}{ c-v}[/tex]
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀ [tex]\frac{c}{c-v}[/tex]
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex]
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
[tex]( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}[/tex]
f ’’ = fo [tex]( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )[/tex]
f ’’ = fo [tex]( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )[/tex]
leave the linear term
f ’’ = f₀ + f₀ 2[tex]\frac{v}{c}[/tex]
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
How old do you need to be in order to qualify to be a U.S. Senator
Answer: 30 Years Old
Explanation: The constitution has around three qualifications for service in the U.S. Senate, Your age must be at least 30 years.
can an object have kinetic energy if there is no motion.
No. The object has to have motion for it to have kinetic energy.
The x-component and y-component of two vectors A & B are Ax = 9, Ay = 12,Bx =
15 & By = 20. Find:
/A+B/
Answer:
40Explanation:
Given the following
Ax = 9,
Ay = 12,
Bx = 15
By = 20
Get A and B
A = √9²+12²
A= √81+144
A = √225
A = 15
Get B;
B = √15²+20²
B = √225+400
B = √625
B = 25
get /A+B/
A+B = 15+25
/A+B/ = /40/
Hence the value of /A+B/ is 40
Students want to investigate the inverse relationship between the pressure and temperature of an ideal gas as predicted by the ideal gas law. Their plan is to use a gas filled cylinder with a movable piston on one end and a heater inside that can be turned on and off. The students will the measure the pressure and temperature of the gas. Which of the following refinements to this procedure will allow the students to observe the predicted relationship between pressure and temperature? Select two answers
A. Start with enough gas to have a pressure near atmospheric pressure, and repeat the experiment, removing gas from the cylinder each time.
B. Fix the piston in place so the volume of the pas remains constant.
C. Ensure the piston and cylinder walls are insulated to the gas can reach equilibrium for each set of measurements
D. Conduct the investigation under conditions of very high pressure to ensure ideal gas behavior
Answer:
Option B, Fix the piston in place so the volume of the pas remains constant
Explanation:
As we know
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
The effect on variable due to another variable can be studied by keeping the third variable constant.
Hence, in order the study the variation of temperature with pressure or vice versa, the volume needs to fixed at a certain value.
Hence, option B is correct
It is common to see birds of prey rising upward on thermals. The paths they take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius 6.00 m every 5.00 s and rises vertically at a rate of 3.00 m/s. Part APart complete Find the speed of the bird relative to the ground. Express your answer using three significant figures. v
Answer:
8.11 m/s
Explanation:
The bird has two velocity, one velocity is in the upward direction (+y direction) and the other velocity is in the x direction (that is along the radius of circle of the bird).
the vertical speed [tex]v_y=3\ m/s[/tex]
The horizontal speed ([tex]v_x[/tex]):
[tex]v_x=\frac{2\pi R}{T} =\frac{2\pi *6\ m}{5\ s} =7.54\ m/s[/tex]
The bird's speed relative to the ground (v) is given by:
v = [tex]\sqrt{v_x^2+v_y^2}[/tex]
[tex]v=\sqrt{7.54^2+3^2} \\\\v=8.11\ m/s[/tex]
Lolliguncula brevis squid use a form of jet propulsion to swim—they eject water out of jets that can point in different directions, allowing them to change direction quickly. When swimming at a speed of 0.15m/s0.15m/s or greater, they can accelerate at 1.2m/s21.2m/s 2 .
(a) Determine the time interval needed for a squid to increase its speed from 0.15m/s0.15m/s to 0.45m/s0.45m/s.
(b) What other questions can you answer using the data?
Answer:
a) t = 0.25 s, b) x = 0.075 m
Explanation:
a) For this exercise we will use kinematic relationships in one dimension
v = v₀ + a t
in the problem they indicate the initial velocity v₀ = 0.15 m / s, the final velocity v = 0.45 m / s and the acceleration of the squid a = 1.2 m / s²
t = [tex]\frac{v -v_o}{a}[/tex]
we calculate
t = [tex]\frac{0.45 - 0.15}{1.2}[/tex]
t = 0.25 s
b) We can also find the distance traveled during this acceleration
v² = v₀² + 2a x
x = [tex]\frac{v^2 -v_o^2 }{2a}[/tex]
let's calculate
x = [tex]\frac{0.45^2 - 0.15^2 }{2 \ 1.2}[/tex]
x = 0.075 m
Which would you choose to keep rods or cones? And why?