Answer:
F= 15.6 N, a= 33.2 m/s^2
Explanation:
mass= m = 0.47 kg
spring constant= k = 130N/m
spring compression = x = 0.12 m
a).
force on the mass= F = k*x
F = 130 * 0.12 N
F= 15.6 N
b).
Acceleration of mass= a=?
F= ma
a=F / m
a= 15.6/ 0.47 m/s^2
a= 33.2 m/s^2
Final answer:
The force on the mass when the spring is released is 15.6 N, and the acceleration of the mass at that instant is approximately 33.19 m/s^2, calculated using Hooke's Law and Newton's second law, respectively.
Explanation:
Understanding Spring Force and Acceleration
To find the force on the mass at the instant the spring is released (in part A), we use Hooke's Law, which states that the force exerted by a spring (F) is equal to the negative spring constant (k) times the displacement from equilibrium (x), so F = -kx. Here, k = 130 N/m and x = 0.12 m, so the force is F = 130 N/m * 0.12 m = 15.6 N. The negative sign indicates that the force is in the opposite direction of the displacement.
To find the acceleration of the mass at the instant the spring is released (in part B), we apply Newton's second law, which relates force (F), mass (m), and acceleration (a) as F = ma. Rearranging for acceleration, we get a = F/m. Substituting the values, we have a = 15.6 N / 0.47 kg = approximately 33.19 m/s2.
How do the source regions of polar and tropical air masses differ
Answer:
Maritime polar and tropical which are cool/humid and warm/humid respectively. Continental polar and tropical which are dry/cold and dry/hot respectively. The maritime is associated with water while the continental is associated with land.
Explanation:
Maritime/polar air masses are humid and cool in nature. They move toward the top at the continental tropical air masses which are also humid but warm at the bottom. In addition, the maritime is associated with water while the continental is associated with land. We can have maritime polar and tropical which are cool/humid and warm/humid respectively. Furthermore, we can have continental polar and tropical which are dry/cold and dry/hot respectively.
Final answer:
Polar air masses form in high-latitude, cold regions and can be dry (continental) or moist (maritime), while tropical air masses form near the equator in warm regions and are generally moist (maritime) or dry (continental).
Explanation:
The source regions of polar and tropical air masses differ mainly in terms of their geographical locations and their temperature and humidity characteristics, which are influenced by whether they form over land or water. Polar air masses originate in high-latitude regions around 60° north or south, and are typically cold with continental polar air masses being dry while maritime polar air masses are somewhat moist. In contrast, tropical air masses form closer to the equator, between 15° and 35° north and south latitude, and are warm, with maritime tropical air masses being moist and continental tropical air masses being dry due to the deserts they often originate from, such as the Sahara and Australian deserts.
Compare the Lagrangian method with the Newton-Euler method in detail.
The comparison of the Lagrangian and Newton methods are explained. Explanation:
The Newton-Euler Method is derived by Newton's Second Law of Motion, that describes the dynamic systems in terms of force and momentum.
It deals with concentration of particles to calculate the overall diffusion and convection of a number of particles.
The Lagrangian Method the dynamic behavior is described in terms of work and energy.
It deals with individual particles to calculate the trajectory of each particle separately.
The Lagrangian method and the Newton-Euler method are two approaches used to solve problems in classical mechanics involving Newton's laws of motion. The Newton-Euler method relies directly on Newton's second law and free-body diagrams, while the Lagrangian method is based on the principle of least action and the Euler-Lagrange equations. The choice between the two methods often depends on the complexity and nature of the physical system in question.
Explanation:When comparing the Lagrangian method with the Newton-Euler method, it is important to understand that both approaches are used to solve complex problems in classical mechanics, which often involve the application of Newton's laws of motion. The Newton-Euler method is more traditional and is founded upon direct application of Newton's second law, F=ma (force equals mass times acceleration), and the related concepts for rotary motion. This method involves creating a free-body diagram, identifying all the forces acting on the object, and applying Newton's second law to find the accelerations and subsequently the positions and velocities of the object in question.
Contrastingly, the Lagrangian method is a more modern approach, which is grounded in the principle of least action. Instead of focusing on forces, it involves the calculation of the Lagrangian, which is the difference between an object's kinetic and potential energies, and applying the Euler-Lagrange equations to find the equations of motion. This method is particularly powerful in systems where the forces are conservative and can be derived from a potential energy; the Lagrangian method is also more convenient when dealing with complex constraints and coordinate systems that are not Cartesian.
While the Newton-Euler method is practical and straightforward, especially in simple scenarios with few forces or when numerical solutions are required, the Lagrangian approach offers more flexibility and can simplify calculations in systems with symmetries or non-standard geometries. Understanding the application of these methods is an integral part of a problem-solving procedure when using Newton's laws of motion, and both methods reinforce concepts useful across various areas of physics.
A drone traveling horizontally at 110 m/s over flat ground at an elevation of 3000 meters must drop an emergency package on a target on the ground. The trajectory of the package is given by x = 110 t , y = − 4.9 t 2 + 3000 , t ≥ 0 where the origin is the point on the ground directly beneath the drone at the moment of release. How many horizontal meters before the target should the package be released in order to hit the target? Round to the nearest meter.
Answer:
The horizontal distance of the target should be 2721,4 meters.
Explanation:
First of all we need to find the time that the emergency package hits the ground after the moment of release:
y=0 (because when it hits the ground it is on the level of 0m);
[tex]0=-4,9*t^2+3000\\t=24,74[/tex]
The emergency package hits the ground after 24,74 seconds from release.
Lets assume that package preserves his 110 m/s horizontal speed during the free fall. The targets horizontal distance is:
[tex]110*24,74=2721,4[/tex]
2721,4 meters
A 1100 kg automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a constant acceleration of 5.0 m/s2. At the same instant a 2000 kg truck, traveling at a constant speed of 7.0 m/s, overtakes and passes the automobile. A)How far is the com of the automobile-truck system from the traffic light at t=3.0 s? B)What is the speed of the com then?
Answer:
a ) 21 m b) 7 m/s
Explanation:
For this case, we can consider for our reference system, that point zero (this is, at the rest of the traffic signal) is when X₀ = 0, and direction of movement is from left to right (positive sign)
For the car:
a= acceleration is constant = 5 m/s² m1 , mass = 1100 Kg , V = not constant
This is an uniformly accelerated rectilinear movement, and applicable formulas are:
V = V₀ + at , X = X₀ + V₀t + 1/2at²
For the truck:
V = speed is constant = 7 m/s , mass = 2000 Kg, a = 0
This is an uniform rectilinear movement, and the applicable formula is:
X = X₀ + Vt
a) At t = 3.0 sec, we can use the formula for the track, considering that X₀ = 0 (when it pass the rest of the traffic signal)
X = 0 + (7 m/s)x(3 s) = 21 m
So, at 3 sec, truck will be at 21 m from the rest of traffic light and, as truck has a constant speed; at that exact second; truck and car will be together as a com, so both will be 21 m away from point zero
b) At the exact time (3 sec, or in distance words, 21 m) car and truck will be together as a com, so they will both have the exact speed, hence, speed of car at that point will be 7 m/s
The position of a particle moving along a straight line at any time t is given by s(t)=t^3+9t^2-27.
Find the velocity of the particle at the time when the acceleration is zero.
Answer:
v= - 27 m/s
Explanation:
Given that
s= t³- 9 t²-27 ( Correct from sources)
As we know that velocity given as
[tex]v=\dfrac{ds}{dt}[/tex]
v=3 t ² - 18 t ------------1
As we know that acceleration given as
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]v=\dfrac{d^2s}{dt^2}[/tex]
v=3 t ² - 18 t
a=6 t -18
Given that acceleration is zero (a= 0 )
0 = 6 t - 18
t= 3 sec
Now by putting the values in the equation 1
v=3 t ² - 18 t m/s
v=3 x 3 ² - 18 x 3 m/s
v= 27 - 54 m/s
v= - 27 m/s
Given:
[tex]s(t)=t^3+9t^2-27[/tex]As we know,
The velocity:
→ [tex]v = \frac{ds}{dt}[/tex]
[tex]v = 3t^2 -18 t[/tex] ...(eqn 1)
and,
The acceleration:
→ [tex]a = \frac{dv}{dt}[/tex]
[tex]a = 6t -18[/tex]
When, a = 0
→ [tex]0 = 6t -18[/tex]
[tex]6t = 18[/tex]
[tex]t = \frac{18}{6}[/tex]
[tex]= 3 \ sec[/tex]
By putting the values in "eqn 1", we get
→ [tex]v = 3t^2-18t[/tex]
[tex]= 3\times 3^2-18\times 3[/tex]
[tex]= 27-54[/tex]
[tex]= -27 \ m/s[/tex]
Thus the above approach is appropriate.
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TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 72.0-kg physicist placed himself and 400 kg of equipment at the top of a 610-m-high antenna to perform gravity experiments. By how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.150 m in radius?
Answer:
1.894 × 10^-4 m
Explanation:
mass of the man = 72 kg, mass of the equipment = 400 kg
total mass on top of the antenna = 72 + 400 = 472 kg
total weight of the man and equipment = total mass × acceleration due to gravity (g, 9.81 m/s²) = 472 × 9.81 = 4630.32 N
to calculate the ΔL ( how much the antenna was compressed), we use the formula below
E = stress/ strain = (F/A) / (ΔL/L) = FL / ΔLA where F is the force in Newton, A is the surface area of the circular face of the antenna in m²
E = 2.1 × 10^11 N/m² which is the young modulus of steel
A = πr² = 3.142 × (0.150²) = 0.071m²
make ΔL the subject of the formula
ΔL = FL / AE
substitute the values into the equation
ΔL = (4630.32 × 610) / ( 0.071 × 2.1 × 10^11) = 1.894 × 10^-4 m
A box is 25 kg with a kinetic force of .35 is pushed horizontally. What force must you continue continue to exert if you want to accelerate the box at a rate of 2.5 m/s squared
The force required is equal to 148.33 N.
Why?
To answer your question, I'll assume that the value of 0.35 is referring to the coefficient of kinetic friction. So, we can solve the problem using the following equations:
[tex]F-F_{f}=m*a\\\\FrictionalForce=\mu *m*g\\\\F_{f}=0.35*25kg*9.81m\frac{m}{s^{2}}=85.83N[/tex]
Now, substituting, we have:
[tex]F-F_{f}=m*a\\\\F=25kg*2.5\frac{m}{s^{2}}+85.83N\\\\F=62.5N+85.83N=148.33N[/tex]
Hence, we have that the force required to accelerate the box at a rate of 2.5 m/s2 is 148.33N.
Have a nice day!
An 880 kg cannon at rest fires a 12.4 kg cannonball forward at 540 m/s. What is the recoil velocity of the cannon? (Unit m/s)
Answer:
7.61 m/s backwards
Explanation:
Initial momentum = final momentum
0 = (880 kg) v + (12.4 kg) (540 m/s)
v = -7.61 m/s
The cannon's recoil is 7.61 m/s backwards.
Answer:
The recoil velocity vector of the cannon is [tex](7.609,0,0)\frac{m}{s}[/tex]
Explanation:
We can solve this problem by applying the Momentum Conservation Principle.
The principle of conservation of momentum states that when you have an isolated system with no external forces, we can use the following equation to calculate the final velocity of one object.
[tex]m1.v1=m2.v2[/tex] (I)
Where ''[tex]m1[/tex]'' and ''[tex]v1[/tex]'' are the mass and velocity of the first object.
And where ''[tex]m2[/tex]'' and ''[tex]v2[/tex]'' are the mass and velocity of the second object.
The momentum is a vectorial magnitude.
If we use the equation (I) with the data given :
[tex](880kg).v1=(12.4kg).(540\frac{m}{s})[/tex]
[tex]v1=7.609\frac{m}{s}[/tex]
If we considered as negative the sense of the velocity vector from the cannonball, the cannon's velocity vector will have the same direction but opposite sense that the cannonball's velocity vector (It will be positive).
We can give it a vectorial character like this :
[tex]v1=(7.609,0,0)\frac{m}{s}[/tex]
The velocity vector will be entirely in the x-axis.
An object falls freely from rest on a planet where the acceleration due to gravity is it is on Earth. In the first 5 seconds it falls a distance of twice as much as
A) 250 m.
B) 500 m.
C) 150 m.
D) 100 m. E) none of these
Answer:
A. The object falls a distance of 250 m
Explanation:
Hi there!
In the question, you have forgotten the acceleration due to gravity. However, looking on the web I´ve found a very similar problem in which the acceleration due to gravity was as twice as much as it is on Earth.
The equation of height of a falling object is the following:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height of the object after a time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (on Earth: ≅ -10 m/s² considering the upward direction as positive).
Let´s place the origin of the system of reference at the point where the object is released so that y0 = 0. Since the object falls from rest, v0 = 0.
Then, the height of the object after 5 s will be :
y = 1/2 · 2 · g · t² (notice that the acceleration due to gravity is 2 · g)
y = g · t²
y = -10 m/s² · (5 s)²
y = -250 m
The object falls a distance of 250 m.
The object falling on this other planet with the same gravity as Earth falls a distance closest to 250m in 5 seconds, according to the standard formula and doubling the Earth result.
Explanation:Given the planet where the object is falling has the same acceleration due to gravity as Earth, we can use the formula for the distance fallen, d (in meters), over a certain time, t (in seconds), assuming the object falls from rest. This formula is d = 0.5 * g * t^2, where g is the acceleration due to gravity. For Earth, g is approximately 9.8 m/s^2.
When t=5 seconds on Earth, using the given formula, d = 0.5 * 9.8 * (5^2) = 122.5 meters. Since the object on the other planet falls twice the distance in the same time, the object falls 2 * 122.5 = 245 meters. Thus, Choice A) 250 m is the closest to the calculated value.
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You attach a block to the bottom end of a spring hanging vertically. You slowly let the block move down and find that it hangs at rest with the spring stretched by 18.0 cm. Next, you lift the block back up and release it from rest with the spring unstretched. What maximum distance does it move down
Answer:36 cm
Explanation:
Given
At Equilibrium extension a is 18 cm
suppose m is the mass of block
thus
[tex]ka=mg[/tex]
[tex]k=\frac{mg}{a}[/tex]
[tex]k=\frac{mg}{0.18}[/tex]
Now if the block is released from un-stretched position
conserving Energy and supposing block moves x units down
[tex]\frac{1}{2}kx^2=mgx[/tex]
[tex]x=\frac{2mg}{k}[/tex]
[tex]x=2mg\times \frac{0.18}{mg}[/tex]
[tex]x=0.36 \approx 36 cm[/tex]
What are the potentially negative impacts of movement corridors?
Answer:
These corridors can decrease the inbreeding in declining populations and enhance the spread of different diseases.
Explanation:
Movement corridors generally will enhance a process such as dispersal which is the spread or distribution of things over a considerably large area. These corridors can decrease the inbreeding in declining populations. They are very vital to species that usually migrate seasonally. On the other day, the movement corridor can be highly harmful in nature due to its ability of enhancing the spread of different diseases. These corridors facilitate the movement of species and allow the spread of harmful diseases.
Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.
a) Through how many radians must the spool be turned to wrap one even layer of wire?
b) What is the length of this wound wire?
Answer:
a) # lap = 301.59 rad , b) L = 90.48 m
Explanation:
a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm
# turns = 1 turn (24 cm / 0.5 cm)
# laps = 48 laps
Let's reduce to radians
# laps = 48 laps (2 round / 1 round)
# lap = 301.59 rad
b) Each lap gives a length equal to the length of the circle
L₀ = 2π R
L = # turns L₀
L = # turns 2π R
L = 48 2π 30
L = 9047.79 cm
L = 90.48 m
The spool needs to be turned through 2π radians to wrap one even layer of wire. The length of the wound wire can be calculated using the circumference formula.
Part a: To wrap one even layer of wire around the spool, the spool must be turned through an angle of 2π radians because one complete revolution is equal to 2π radians.
Part b: To calculate the length of the wire wound on the spool, we find the circumference of the spool using the formula: circumference = 2πr, where r is the radius of the spool (30 cm). The length of the wound wire is the product of this circumference and the height of the spool (24 cm).
When AC Electric is producing 120 volts with 20 amps, What controls the the 120 volts at 60cycles a second?
Answer:
Cycles per second is dependent on the construction of the alternator and the 120 volts is dependent upon the current and resistance in the circuit according to the ohms law.
Explanation:
We are given with AC of 120 volts, 20 amperes and 60 hertz frequency.
According to the Ohm's law, we find its resistance:
[tex]R=\frac{V}{I}[/tex]
[tex]R=\frac{120}{20}[/tex]
[tex]R=6\ \Omega[/tex]
So, this 6 ohm resistance controls the current controls the magnitude of the AC current, while the frequency of the current remains constant and depends upon the construction and rotational speed of the armature of the alternator producing the current.
Here the value of frequency is the number of times the current changes its direction or the polarity in one second.
The control of the AC power frequency is managed by power plant generators, which use turbine controls to ensure electricity alternates at a specific frequency, such as the standard 60 Hz in North America. Voltage and current in AC power cross zero 120 times a second due to this oscillation.
Alternating Current (AC) electric power is characterized by a voltage that alternates in polarity from positive to negative, resulting in a sinusoidal waveform. The frequency of this oscillation in North America is 60 times per second (60 Hz). This oscillation means that the voltage swings from a peak value to the opposite peak value, crossing zero in the process, twice every cycle. Therefore, for a 60 Hz AC supply, voltage and current cross zero 120 times a second.
The control of the frequency at which this oscillation occurs is managed by the generator at the power plant. In the case of a hydro-electric dam, turbine controls adjust the rotational frequency to produce the desired AC frequency. For example, in North America, these turbines are regulated to rotate in such a way that the electricity they generate alternates at 60 Hz. This regulated rotation is what ensures the consistency of the 120 volts at 60 cycles per second.
The power consumption of an electrical device such as a light bulb in an AC circuit also swings. Peak power is calculated by multiplying peak current by peak voltage. Given that the root mean square (RMS) voltage is 120 volts for a standard AC outlet, the peak voltage is actually higher, around 170 volts. Thus, for a 60-watt light bulb, the power consumption will pulse from zero up to its peak value 120 times per second, averaging out to the rated 60 watts.
What is the period of a simple pendulum 47 cm long (a) on the Earth, and ( b) when it is in a freely falling elevator?
Answer:
a)1.37 s
b)∞ ( Infinite)
Explanation:
Given that
L= 47 cm ( 1 m =100 cm)
L= 0.47 m
a)
On the earth :
Acceleration due to gravity = g
We know that time period of the simple pendulum given as
[tex]T=2\pi\sqrt{ \dfrac{L}{g_{{eff}}}[/tex]
Here
[tex]g_{eff}= g[/tex]
Now by putting the values
[tex]T=2\pi \times\sqrt{ \dfrac{0.47}{9.81}}[/tex]
T=1.37 s
b)
Free falling elevator :
When elevator is falling freely then
[tex]g_{eff}= 0[/tex] ( This is case of weightless motion)
Therefore
[tex]T=2\pi\sqrt{ \dfrac{L}{0}[/tex]
T=∞ (Infinite)
(a) The period of a simple pendulum 47 cm long when on the earth = 1.38 seconds
(b) The period of a simple pendulum when it is in a freely falling elevator = infinity (∞)
Period: This can be defined as the time taken for an object to complete one oscillation. The s.i unit is seconds (s)
The formula for the period of a simple pendulum is
T = 2π√(L/g).................... Equation 1
Where T = period of the simple pendulum, L = length of the simple pendulum, g = acceleration due to gravity.
(a) From the question,
Given: L = 47 cm = 0.47 m,
Constant: g = 9.8 m/s², π = 22/7 ≈ 3.14
Substitute these values into equation 1
T = 2(3.14)√(0.47/9.8)
T = 6.284√(0.048)
T = 6.284(0.219)
T = 1.38 seconds
(b) When it is in a free-falling elevator,
Then g = 0 m/s²
T = 2(3.142)√(0.47/0)
T = Infinity (∞)
Therefore, The period of the simple pendulum is (a) 1.38 seconds when it is on the earth and (b) infinity (∞) when it is in a freely falling elevator.
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Read the false statement. In an atom, protons and electrons are in the nucleus, which is surrounded by neutrons. Which option rewords the false statement so it is true?
1. In an atom, protons and neutrons are in the nucleus, which is surrounded by electrons.
2. In an atom, electrons and neutrons are in the nucleus, which is surrounded by protons.
3. In an atom, electrons and neutrons are in the nucleus, which is surrounded by photons.
4. In an atom, electrons and photons are in the nucleus, which is surrounded by neutrons.
Answer: A.
In an atom, protons and neutrons are in the nucleus, which is surrounded by electrons.
Explanation:
It must first be noted that an atom consists of protons, neutrons and electrons.
The proton and neutron is contained in the atom while the electron is found in the outer most shell of the atoms. It can be concluded then that in an atom, protons and neutrons are in the nucleus, which is surrounded by electrons.
Answer:
A
Explanation:
Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 36.0 m. During the collision at the bottom of the elevator shaft, a 85.0 kg passenger is stopped in 5.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision
Answer:
a) I = -2257.6 Kg*m/s
b) F = -451,520N
Explanation:
part a.
we know that:
I = [tex]P_f-P_i[/tex]
where I is the impulse, [tex]P_f[/tex] the final momentum and [tex]P_i[/tex] the initial momentum.
so:
I = [tex]MV_f-MV_i[/tex]
where M is the mass, [tex]V_f[/tex] the final velocity and [tex]V_i[/tex] the initial velocity.
Therefore, we have to find the initial velocity or the velocity of the passenger just before the collition.
now, we will use the law of the conservation of energy:
[tex]E_i=E_f[/tex]
so:
mgh = [tex]\frac{1}{2}MV_i^2[/tex]
where g is the gravity and h the altitude. So, replacing values, we get:
(85kg)(9.8m/s^2)(36m)= [tex]\frac{1}{2}(85kg)V_i^2[/tex]
solving for [tex]V_i[/tex]:
[tex]V_i = 26.56m/s[/tex]
Then, replacing in the initial equation:
I = [tex]MV_f-MV_i[/tex]
I = [tex](85kg)(0m/s)-(85kg)(26.56m/s)[/tex]
I = -2257.6 Kg*m/s
Then, the impulse is -2257.6 Kg*m/s, it is negative because it is upwards.
part b.
we know that:
Ft = I
where F is the average force, t is the time and I is the impulse. So, replacing values, we get:
F(0,005s) = -2257.6 Kg*m/s
solving for F:
F = -451520N
Finally, the force is -451,520N, it is negative because it is upwards.
A compressed spring in a toy is used to launch a 5.00-gram ball.
If the ball leaves the toy with an initial horizontal speed of 5.00 meters per second, the minimum amount of potential energy stored in the compressed spring was:
a) 0.0125 J
b) 0.0250 J
c) 0.0625 J
d) 0.125 J
Answer:
c) 0.0625 J
Explanation:
How the mechanical energy is conserved, then ball’s kinetic energy is equal to stored energy in compressed spring.
Then:
[tex]K = U_{e}[/tex]
Where K: kinetic energy
[tex]U_{e}[/tex]: elastic potential energy
[tex]K = \frac{mv^{2} }{2}[/tex]
[tex]K = \frac{(0.005)(5)^{2} }{2}[/tex]
K = 0.0625 J
and
[tex]U_{e}[/tex] = 0.0625
What are the 3 energy-matter components and their percentage of the energy pie, and how do we come to know about each?
Answer:
The 3 main energy matter components are Coal, natural gas and hydro
Explanation:
The 3 most important energy components that are frequently being used for the generation of the energy i.e. coal which constitutes 41% it has the maximum contribution or can say mostly utilized after that comes natural gas which again constitutes 22% and followed by Hydro which is just 16%. with the rise of civilization people get aware of these energy components by coming across it in various phase of life.
Because of its center-surround organization, a neuron that has its entire receptive field exposed to bright light will:
a. fire slowly until the light turns off, then begin firing rapidly.
b. maintain the same rate of firing as if there was no light presented.
c. stop firing entirely.
d. fire rapidly.
Answer:
b. maintain the same rate of firing as if there was no light presented.
Explanation:
The neuron will maintain the same rate of firing as if there was no light presented.
Receptive filed of neuron is place on its sensory surface generally on the back of eye that an stimulus must reach to activate the neuron. The bright light cannot change the rate of firing of the neuron.
Neurons are messengers of information. A neuron that has its full receptive field exposed to strong light will maintain the same rate of firing as if there was no light present because of its center-surround arrangement.
What are neurons?Neurons are messengers of information. They transfer information between different parts of the brain and between the brain and the rest of the nervous system through electrical impulses and chemical signals.
A neuron that has its full receptive field exposed to strong light Hence it will maintain the same rate of firing as if there was no light present because of its center-surround arrangement.
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Two physics students, Will N. Andable and Ben Pumpiniron, are in the weightlifting room. Will lifts the 100-pound barbell over his head 10 times in one minute; Ben lifts the 100-pound barbell over his head 10 times in 10 seconds. Which student does the most work? ______________ Which student delivers the most power? ______________ Explain your answers.
Answer:
"Same"
"Ben Pumpiniron"
Explanation:
Given that
Will lifts 100 pound ,10 times in 1 minute
Ben lifts 100 pound ,10 times in 10 sec
As we know that work done given as
W= m g h
The height and mass are same for both therefore there will do the same work.
But we know that rate of work is known as power.Therefore Ben is more power .
Power
[tex]P=\dfrac{W}{t}[/tex]
t is less for Ben that is why power is more in Ben.
The answer will be "Same" and "Ben Pumpiniron".
Explanation:
Both Ben and Will are doing same amount of work. that is lifting 100 pounds of barbell 10 times.
They apply the same force to lift the same barbell up to same distance hence, they do equal work.
Yet Ben is more powerful than Will because rate of work done (Power) by Ben is higher than the that of Will.
active and passive secruity measures are employed to identify, detect, classify and analyze possible threats inside of which zone?
Answer:
Assessment zone
Explanation:
It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.
2 litre of water at is poured into a plastic bucket containing 10 litre of water at 20 degree Celsius.What is the final temperature of water?(Density of water=1kg/litre)
Answer:[tex]30^{\circ}[/tex]
Explanation:
Given
[tex]v_1=2 liter[/tex]
volume of water in bucket [tex]v_2=10 liter[/tex]
density of water [tex]\rho =1 kg/liter[/tex]
thus [tex]m_1=\rho \cdot v_1=2 kg[/tex]
[tex]m_2=\rho \cdot v_2=10 kg[/tex]
[tex]T_2=20^{\circ}C[/tex]
suppose [tex]T_1=80^{\circ}C[/tex]
Conserving heat energy i.e. heat lost by water is gained by water in bucket
[tex]m_1cT_1+m_2cT_2=(m_1+m_2)T[/tex]
where T=final Temperature
[tex]T=\frac{m_1T_1+m_2T_2}{m_1+m_2}[/tex]
[tex]T=\frac{160+200}{12}[/tex]
[tex]T=30^{\circ}C[/tex]
A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle of elevation changing when the balloon is at an altitude of 2500 feet?
Answer:
0.0031792338 rad/s
Explanation:
[tex]\theta[/tex] = Angle of elevation
y = Height of balloon
Using trigonometry
[tex]tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta[/tex]
Differentiating with respect to t we get
[tex]\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta[/tex]
Now, with the base at 200 ft and height at 2500 ft
The hypotenuse is
[tex]h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft[/tex]
Now y = 2500 ft
[tex]cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974[/tex]
[tex]\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s[/tex]
The angle is changing at 0.0031792338 rad/s
The rate of change of the angle of elevation in this problem would be calculated using the principles of related rates problems in calculus. This involves setting up an equation relating the quantities of interest (here, the altitude of the balloon and the angle of elevation seen by the observer), and differentiating this equation with respect to time. The precise calculation could not be completed as the question lacked certain specific data.
Explanation:Rate of change of the angle of elevation is being calculated in this question. This question falls into the category of related rates problems in calculus, where the rate of change of one quantity (here, the altitude or the height to which the balloon has ascended) affects the rate of change of another quantity (the angle of elevation seen by the observer on the ground). However, the provided reference information does not appear to give specific details needed to answer this question.
Generally, you would use the properties of a right angled triangle and the concept of derivatives to set up an equation relating the quantities and find the value. For example, if we consider a right triangle where the observer and the balloon at a particular moment form the ends of a line representing the hypotenuse, and the angle of elevation to the balloon as θ, then tan(θ) = (altitude of the balloon) / (distance of observer from the launch point). And you'd differentiate this equation with respect to time to find the rate of change of the angle θ. However, again, the specific calculation may require more detailed information not provided here.
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Galileo found that a ball rolling down one inclined plane would roll how far up another inclined plane?
A) the ball would not roll up the other plane at all
B) to nearly its original height
C) to about one quarter its original height
D) to nearly twice its original height
E) to nearly half its original height
Answer: B.) To nearly its original height
Explanation:
Because if you have a ball than the inclined plane is heading up you would get almost the same height as it started.
Galileo found that a ball rolling down one inclined plane then, the ball will reach about the original height. Hence, option B is correct.
What is momentum?The momentum is the result of a particle's velocity and its mass. Force and motion, meaning it has both magnitude and the direction.
According to Isaac Newton's second equation of motion, the force acting on a particle equals the time rate of increase of momentum.
The impulse, which is the product of the force and the intervals (the impulse), is equal to the difference in momentum, according to Newton's 2nd law, if a steady force operates on a particle for a specific amount of time.
On the other hand, a particle's momentum is the amount of time needed for a consistent action to fight it to rest.
Because if you were to throw a ball while the inclined plane was moving upward, you would nearly reach its starting height.
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A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by a current flowing southward at 2.0 meter a second if he ended up exactly across the stream from where he began at what angel to the shore must he swim upstream
Answer:
Angle: [tex]48.19^o[/tex]
Explanation:
Two-Dimension Motion
When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.
Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.
To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is
[tex]v_b\ cos\alpha[/tex]
where [tex]v_b[/tex] is the speed of the boy in still water and [tex]\alpha[/tex] is the angle respect to the shoreline. If the river flows at speed [tex]v_s[/tex], we now set
[tex]v_b\ cos\alpha=v_s[/tex]
[tex]\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}[/tex]
[tex]\alpha=48.19^o[/tex]
Viscosity: Blood is about 4 to 5 times ____________ viscous than water. Viscosity of blood depends upon the amount of dissolved substances in the blood relative to the amount of fluid. Viscosity is ____________ if the amount of substances increases, the amount of fluid decreases, or both.
Answer:more, increased
Explanation:
Blood is four to five times more viscous than water.
Viscosity depends upon dissolved substances in blood and it increases as the amount is increases or if the fluid in the blood decreases.
Viscosity is the resistance offered by liquid to the flow and it also depends upon temperature and it decreases with increase in temperature.
Two men decide to use their cars to pull a truck stuck in the mud. They attach ropes and one pulls with a force of 615 N at an angle of 31◦with respect to the direction in which the truck is headed, while the other car pulls with a force of 961 N at an angle of 25◦with respect to the same direction. What is the net forward force exerted on the truck in the direction it is headed? Answer in units of N.
Answer:
1398.12 N
Explanation:
We define the x-axis in the direction parallel to the movement of the truck on and the y-axis in the direction perpendicular to it.
x-components of the ropes forces
T₁x = 615N*cos31°=527.1579 N :Tension in direction x of the rope of the car 1
T₂x= 961 N*cos25°=870.96 N :Tension in direction x of the rope of the car 2
Net forward force exerted on the truck in the direction it is headed (Fnx)
Fnx = T₁x + T₂x
Fnx = 527.1579 N + 870.96 N
Fnx = 1398.12 N
Economic growth can be illustrated by: a. an inward shift of the production possibilities curve. b. a movement along the production possibilities curve. c. a movement from a point on the production possibilities curve to a point inside the production possibilities curve. d. an outward shift of the production possibilities curve.
Answer:
Economic growth can be illustrated by:
d. an outward shift of the production possibilities curve.
Explanation:
Economic growth is the process of increasing the economy's ability to produce goods and services. It is achieved by increasing the quantity or quality of resources.
Production Possibilities refers to the ability of a country to produce goods or services given the limited resources and technology. It is therefore possible to increase production of both goods at the same time as long as resources allow it.
The Production Possibilities Curve, also known as the production possibilities frontier, is a graph that shows the maximum number of possible units a company can produce if it only produces two products using all of its resources efficiently. Firstly, and most commonly, growth is defined as an increase in the output that an economy produces over a period of time, the minimum being two consecutive quarters. An increase in an economy's productive potential can be shown by an outward shift in the economy's production possibility frontier (PPF).
Each point on the curve shows how much of each good will be produced when resources shift from making more of one good and less of the other. The curve measures the trade-off between producing one good versus another.PPC or production possibility curve is a curve whose basic purpose is to show the different possible combinations of two goods that can be produced within the given available resource.
The two main characteristics of PPC are: slopes downwards to the right: PPC slopes downwards from left to right. It is because in a situation of fuller utilization of the given resources, production of both the goods cannot be increased simultaneously.
Economic growth is represented by an outward shift of the production possibilities curve, indicating that the economy can now produce more goods and services than before.
Explanation:Economic growth can be illustrated by an outward shift of the production possibilities curve. An increase in the quality or quantity of factors of production, such as labor, capital, and technology, can enhance an economy's ability to produce goods and services, which is represented graphically by the production curve moving outward. This shift indicates that the economy can now produce more than it could before, making previously unattainable levels of production possible.
After sliding down a snow-covered hill on an inner tube, Ashley is coasting across a level snowfield at a constant velocity of +3.0 m/s. Miranda runs after her at a velocity of +4.2 m/s and hops on the inner tube.
How fast do the two of them slide across the snow together on the inner tube? Ashley's mass is 69 kg and Miranda's is 59 kg. Ignore the mass of the inner tube and any friction between the inner tube and the snow.
Answer:
Their combined velocity on the inner tube is +3.55 m/s
Explanation:
This question deals with the conservation of linear momentum. The total linear momentum in a closed system is conserved.
Therefore,
P_i = P_f
m₁ v₁ + m₂ v₂ = (m₁ + m₂) v₁₂
where
m₁ is the mass of Ashleym₂ is the mass of Mirandav₁ is Ashley's velocityv₂ is Miranda's velocityv₁₂ is their combined velocityTherefore,
v₁₂ = (m₁ v₁ + m₂ v₂) / (m₁ + m₂)
v₁₂ = ( (69 kg)(3 m/s) + (59 kg)(4.2 m/s) ) / (69 kg + 59 kg)
v₁₂ = +3.55 m/s
Therefore, Ashley and Miranda's combined velocity on the inner tube is +3.55 m/s.
The positive sign shows that the velocity is in the positive direction.
A 1.00-kg sample of steam at 100.0 °C condenses to water at 100.0 °C. What is the entropy change of the sample if the latent heat of vaporization of water is 2.26 x 10⁶ J/kg?
Answer:
The entropy change of the sample of water = 6.059 x 10³ J/K.mol
Explanation:
Entropy: Entropy can be defined as the measure of the degree of disorder or randomness of a substance. The S.I unit of Entropy is J/K.mol
Mathematically, entropy is expressed as
ΔS = ΔH/T....................... Equation 1
Where ΔH = heat absorbed or evolved, T = absolute temperature.
Given: If 1 mole of water = 0.0018 kg,
ΔH = latent heat × mass = 2.26 x 10⁶ × 1 = 2.26x 10⁶ J.
T = 100 °C = (100+273) K = 373 K.
Substituting these values into equation 1,
ΔS =2.26x 10⁶/373
ΔS = 6.059 x 10³ J/K.mol
Therefore the entropy change of the sample of water = 6.059 x 10³ J/K.mol
The entropy change of the sample of steam is equal to 6,058.98 J/Kmol.
Given the following data:
Mass = 1.00 kgTemperature = 100°C to Kelvin = [tex]273+100=373\;K[/tex]Latent heat of vaporization of water = [tex]2.26 \times 10^6\;J/kg[/tex]To determine the entropy change of the sample of steam:
Mathematically, entropy change is given by the formula:
[tex]\Delta S = \frac{\Delta H}{T}[/tex]
Where:
[tex]\Delta H[/tex] is the enthalpy change.T is the absolute temperature.Substituting the given parameters into the formula, we have;
[tex]\Delta S = \frac{2.26 \times 10^6\times 1}{373}[/tex]
Entropy change = 6,058.98 J/Kmol.
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