Galactic nuclei that are still active as they get older tend to become less active. Imagine that over the course of a few days,
we see an X-ray source whose brightness swings noticeably. The diameter of the X-ray source is only a few light-days.
Previously, the cosmos was considerably denser and its galaxies were far closer to one another, increasing the likelihood of collisions. 5. We see that elliptical galaxies were more prevalent in the past because they are created when two spiral galaxies collide.
The reason a galaxy's core appears brightest is that there are more stars there. There are millions more stars per cubic parsec in the Galactic Bulge than in any other region.
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the heat of fusion of water is 6.008×103 j mol−1 at its normal melting point of 273.15 k.
1.86 Kg K/mol is the freezing point depression constant.
Given data
Heat of fusion of ice is
Δ[tex]H_{fusion}[/tex]= 6.008 × [tex]10^{3}[/tex]J/ mol
Melting point of water is Tm=273.15K
Molar mass of water is Mw= 18.0153 ×[tex]10^{-3}[/tex]Kg/mol
Gas constant is 8.3145 J/Kmol
The freezing point depression constant can be calculated by the formula given below:
[tex]K_{f}[/tex]= [tex]\frac{Mw *R*(Tm)^{2} }{H_{fusion}}[/tex]
Where,
[tex]K_{f}[/tex] is the freezing point depression constant.
Mw the molar mass of water.
Ris the universal gas constant.
By changing the variables in the aforementioned equation, we obtain
[tex]K_{f}[/tex]= [tex]\frac{Mw *R*(Tm)^{2} }{H_{fusion}}[/tex]
= [tex]\frac{18.0153*8.3145*10^{-3}* (273.15)^{2} }{6.008*10^{3} }[/tex]
=[tex]\frac{18.0153*8.3145*10^{-3}* 75610.9225 }{6.008*10^{3} }[/tex]
= [tex]\frac{11175164.6*10^{-3} }{6.008*10^{3} }[/tex]
On simplifying the above equation, we get
[tex]K_{f}[/tex]=1860.047×[tex]10^{-3}[/tex]KgK/mol
=1.86KgK/mol
freezing level depression a constant 1.86 Kg K/mol.
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Complete question:
The heat of fusion of ice is 6.008×[tex]10^{3}[/tex] J/mol at its normal melting point of 273.15 K. Calculate the freezing point depression constant, [tex]K_{f}[/tex].
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Complete question:
The heat of fusion of ice is 6.008×[tex]10^{3}[/tex] J/mol at its normal melting point of 273.15 K. Calculate the freezing point depression constant, [tex]K_{f}[/tex].
In the figure, determine the character of the collision. The masses of the blocks, and the velocities before and after, are shown. The collision is A) completely inelastic. B) perfectly elastic. C) characterized by an increase in kinetic energy. D) not possible because momentum is not conserved.
The collision shown in the figure is A) completely inelastic.
What is velocity?Velocity is a physical quantity that expresses the rate and direction of a moving object. It is the speed of an object in a particular direction. It is a vector quantity that is measured in meters per second (m/s). Velocity can also be described as the rate of change of an object's position with respect to time. It is an important concept in physics and is used to calculate the motion of objects.
This is because the kinetic energy of the system is not conserved, as the velocities of the blocks after the collision are lower than before. The masses of the blocks are not relevant, as the collision is completely inelastic regardless of their mass. Therefore, the collision is characterized by a decrease in kinetic energy and is not possible if momentum is to be conserved.
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suppose the collision between the packages is perfectly elastic. to what height does the package of mass mm rebound?
If there is no net loss of kinetic energy in the system as a result of the collision. This is known as an elastic collision. In elastic collisions, kinetic energy and momentum are both conserved.
In an elastic collision, does velocity conserve?There must be a change in velocities since each body's momentum is changing while its mass remains constant. Because of this, in elastic and inelastic collisions, velocity is not conserved.
Is it feasible to have an elastic collision?Only collisions between subatomic particles are truly elastic. There are no real-world instances of collisions that are totally elastic because friction always results in some kinetic energy being lost as heat transfer.
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a 500.0 lb man is crossing the street when the brakes on 454.5 kg car fail and she hits the gorilla head on at 50.0 mph. if the collision is perfectly inelastic, at what speed will the car and the gorilla travel after the collision?
If the collision is perfectly inelastic, the speed of the car and the gorilla travel after the collision is 14.9m/s.
Perfectly inelastic collision, also known as completely inelastic collision is one in which the maximum amount of kinetic energy has been lost during a collision, making it the most extreme case of an inelastic collision.
From the question given
mass of gorilla man, m = 500lb - 500 * 0.4536 = 226.8kg
Initial speed of the car, u = 50mph = 50 * 0.447 = 22.35 m/s
mass of the car, M = 454.5kg
applying conservation of momentum formula :
(m + M) * V = M * u
(226.8 + 454.5) * V = 454.5 * 22.35
681.3V = 10,158.075
V = 14.9m/s
Therefore, the speed of the car and the gorilla travel after the collision is 14.9 m/s.
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true or false: a compass is placed close to a stationary charge. if the charge remains stationary, the compass will not deflect, but if the charge moves, the compass will deflect.
True, if the charge remains stationary, the compass will not deflect, but if the charge moves, the compass will deflect.
Charge is the bodily assets of count number that reasons charged be counted to experience a force whilst located in an electromagnetic area. Electric powered rate can be fantastic or bad. Like expenses repel every different and not like expenses entice each other.
Charge is the physical belongings of remember that reasons it to revel in a pressure whilst positioned in an electromagnetic area. High-quality and negative electric powered prices are the two varieties of costs generally carried through rate carriers, protons and electrons. Electricity is created by way of the motion of costs.
Charges are of sorts: tremendous and terrible, typically carried by fee vendors protons and electrons. Examples of the sorts of costs are subatomic debris or the debris of depend: Protons are undoubtedly charged. Electrons are negatively charged.
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a pickup truck (2268 kg) and a compact car (1100 kg) have the same momentum. (a) what is the ratio of their kinetic energies? (b) if the same horizontal net force were exerted on both vehicles, pushing them from rest over the sa
A pickup truck (2268 kg) and a compact car (1100 kg) have the same momentum. The ratio of their kinetic energies (truck to car) is 0.485
The formulas for the momentum and the kinetic energy are:
p = m . v
Ek = 1/2 . mv²
Where:
p = momentum
m = mass
Ek = kinetic energy
v = speed or velocity
Ek = 1/2 . mv² = 1/2 . mv . v
= 1/2 . p . v
Since p or momentum of both vehicles are the same, hence, the ratio of their kinetic energies is
Ek_truck : Ek_car = v_truck : v_car
In the given problem:
m_truck = 2268 kg
m_car = 1100 kg
Both have the same momentum, hence,
m_truck x v_truck = m_car x v_car
v_truck / v_car = m_car / m_truck
= 1100 / 2268 = 0.485
Therefore,
Ek_truck / Ek_car = v_truck / v_car = 0.485
Your question is incomplete, but most probably your question was:
A pickup truck (2268 kg) and a compact car (1100 kg) have the same momentum. If the same horizontal net force were exerted on both vehicles, pushing them from rest over the same distance, what is the ratio of their final kinetic energies? (ratio: truck to car)
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the floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.33 with the floor. if the train is initially moving at a speed of 40 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?
The shortest braking distance at which the train should be stopped will be equal to 19.10 m.
In order to solve this problem we use the concept of Newton's Second Law of Motion according to which the force applied on any body is equal to the product of mass and acceleration. We divide the force into two components that is on horizontal axis and vertical axis.
The forces on vertical axis can be written as
N - W = 0
N = W = mg
where m is mass of the object and g is acceleration due to gravity, W is weight of object, N is external force.
The forces on horizontal axis can be written as
-Fr = ma
-μN = ma
-μ(mg) = ma
a = -μg
where a is acceleration, μ is coefficient of friction, Fr is the frictional force.
a = (-0.33)×(9.8) m/s²
a = -3.23 m/s² (it is negative because the train is stopping due to this)
We calculate the distance using the kinematics equation:
V² = U² + 2aS
where V is final velocity, U is initial velocity, a is acceleration and S is displacement.
S = (V² - U²)/2a
When the train stops the speed is zero that is final velocity is zero.
U = 40km/h = 40×(5/18) m/s = 11.11 m/s
S = ( 0 - 11.11²) / 2 (-3.23)
S = 19.10 m
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d047: a force of 100 lbs. is applied to a selection of 5 faces, 2 edges, and 2 vertices. what is the total magnitude of the force (in lbs.) solidworks simulation will apply?
total magnitude of the force (in lbs) is 900 lbs
On 5 faces total load = 500lbs
On 2 edges load = 200 lbs
On 2 vertices load= 200 lbs
Total load applied = 500+200+200 = 900 lbs
In science, the word "force" has a precise meaning. At this level it is appropriate to describe the force as pushing or pulling. A force does not contain or "have within" an object. A force is applied from one object to another. The idea of force is not limited to animate and inanimate objects.
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at strut intersection headed eastbound, atc instructs you to hold west on the 10 dme fix west of lch on v306, standard turns. what entry procedure is recommended? a. teardrop. b. direct. c. parallel.
ATC directs you to hold west on the 10 DME fix west of ICH on v306 at the strut intersection while travelling eastward. The standard turns direct entrance technique is advised.
Timing is initiated for the hold entry over/crossing the fix outbound in the entry. Here's when "over" enters the picture—on the first entry outbound. Timing begins when behind the fix on the outbound leg once the hold has been established. A STAR is an ATC-coded IFR arrival route that was created for use by IFR aircraft arriving at specific airports. The pilot-in-command must immediately contact ATC and obtain a revised clearance if the emergency authority of 14 CFR Section 91.3(b) is used to vary from the terms of an ATC clearance. You should decline the clearance and ask for a modified clearance if you receive ATC instructions that you believe could endanger safety or cause you to break a FAR. Take any quick, logical action necessary to maintain the safety of your aircraft.
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You throw a rock downward into water with a speed of 3mg/k, where k is the coefficient in Eq. f=k vf=kv. Assume that the relationship between fluid resistance and speed is as given in Eq. f=k vf=kv, and calculate the speed of the rock as a function of time.
The speed of the rock as a function of time is that is v(0) as the speed of the rock as a function of time.
{Newton's Second Law}}According to Newton's Second Law, the net force acting on a body is - mg-kv −mg+kv=−ma⇒mDV =mg−kv
Integrate (1) to get -k1 ∫mg-kv−kdv =m 1 ∫ger
-k 1 [ln(mg−kv)]=meters
t+c.
ln(mg−kv)= meters−kt−kC.
mg−kv=exp(−c/k)exp(−kt/m)
kv=mg−C exp(−kt/m)
v=k mg-C1exp(−kt/m)
Use the initial conditions to get \dfrac{3mg}{k}=\dfrac{mg-C}=-2mg}k3mgmg-C⇒C. = -2mg
The final form of (6) is
v=\dfrac{mg+2mg\exp(-kt/m)}{k}
v=kmg+2mgexp(−kt/m).
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In a free-fall experiment in the lab, trainee found out that the average time of fall of the teel ball i 0. 550. What i fall height?
In a free-fall experiment, the average time of fall of the ball is 0.550 seconds. Its height is 1.48 meters.
The equation of motion is given by
s = ut + 1/2 . at²
Where:
s = distance
a = acceleration
u = initial velocity
t = time
In the free fall motion, the initial velocity is zero and the acceleration is equal to g or acceleration due to gravity. The distance is equal to height. Hence,
h = 1/2 . gt²
In the given problem:
t = 0.550 s
Hence, the height of the object is:
h = 1/2 x 9,8 x 0.55²
= 1.48 meters
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when a coin is placed 16.9 cm away from the center of a concave mirror, its image is located 42.9 cm behind the mirror. what is the focal length of the mirror?
The focal length of the concave mirror is 26.2 cm.
object distance , u = 16.7 cm
image distance , v = -46.1 cm
Now, let the focal length of the mirror is f
The mirror formula is given by,
1/f = 1/u + 1/v
Putting the values of the given parameters to evaluate the focal length of the concave mirror.
1/f = 1/16.7 - 1/46.1
We solve the above equation and thus the value of the focal length of the concave mirror comes out to be,
f = 26.2 cm
The focal length of the mirror is 26.2 cm
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Two 1000 kg cars drive east; the first mov-
ing at 45 m/s, the second at 15 m/s.
What is the magnitude of the total momen-tum of the svstem?
Answer in units of kg • m/s.
Answer:
30000kg.m/s
Explanation:
two cars mass 1000
first moving with 45m/s
second moving with 15m/s
find the total magnitude of momentum = ?
momentum formula=mv
so,
mass is common
mass(v1 -v2)
putting the value
1000(45-15)
30000kg.m/s
why is the incoming wavelength different from the underwater wavelength of the light ray? responses most of the light ray is reflected by the water. most of the light ray is reflected by the water. the light ray passes through two different materials. the light ray passes through two different materials. the light ray gains energy the further it moves from its source. the light ray gains energy the further it moves from its source. the light ray gradually loses energy as it moves further from its source.
As the wavelength of light shrinks, more refraction occurs. Light with shorter wavelengths, like violet and blue, is slowed more and, as a result, bends more than light with longer wavelengths.
How do you calculate wavelength?The distance between any two identical locations on adjacent waves may always be used to calculate the wavelength. When dealing with a longitudinal wave, the distance between one compression and the next or between one rarefaction and the next is used to calculate the wavelength.
The wavelength, what is it?A waveform signal's wavelength is defined as the separation between two identical points (adjacent crests) in adjacent cycles as the signal travels through space or along a wire. This length is often stated in meters (m) and centimeters in wireless systems (cm).
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Uppoe that a phone i dropped from a light pole 70. 0 meter high. How far will the phone have fallen after 3. 00 econd
The formula for horizontal distance is x = Vx * t, where t is the time. The formula for vertical distance from the ground is y = h + Vy * t - g * t2 / 2, where g is the acceleration of gravity.
What is the free fall equation?
Where g is the acceleration of gravity, vf is equal to g * t. On Earth, the value of g is 9.8 m/s/s. The velocity of the object after being dropped from rest for any given period of time can be calculated using the aforementioned equation.
How is height determined?
At the spot where the ruler, book, or other flat object meets your head, lightly mark the wall with a pencil. Measure the distance from the floor to the mark on the wall using a tape measure, preferably one made of metal that will stay straight.
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The soccer player is running at a speed of 4m/s when he slides on the ground, coming to a stop distance of 2m. What is the acceleration?
The acceleration is -1.5m/s.
What is acceleration?
The speed at which velocity changes over time. Due to its magnitude and direction, acceleration is a vector quantity. The first derivative of velocity with respect to time or the second derivative of position with respect to time are further examples.
What is speed ?
Speed is defined as the ratio of distance to the amount of time it took to cover that distance.
speed= distance/ time
s= 4.6m/s
d=60m
4.6= 60/t
4.6t= 60
t= 13.04 seconds
acceleration= (the change in velocity/speed)/(time) this is the same as a=Δv/t
a= unkown
Δv= (final speed - initial speed) = 3.1m/s - 4.6m/s= -1.5m/s
Therefore, the acceleration is -1.5m/s.
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How does the shape of an object affect how it interacts with light?
The shape of an object can affect how it interacts with light in several ways.
First, the shape of an object can determine how light is reflected off of its surface. A smooth, curved surface will reflect light in a more diffuse manner, producing a soft, even reflection. A rough, uneven surface will scatter light in many different directions, producing a more irregular reflection.
Second, the shape of an object can also affect how light is absorbed by the object. An object with a highly absorbent surface will absorb more light than an object with a less absorbent surface. This is why black objects tend to absorb more light and heat than white objects, which tend to reflect more light.
Finally, the shape of an object can affect how light passes through it. An object with a transparent or translucent surface will allow light to pass through it, while an object with an opaque surface will block the passage of light. The shape of the object can also affect how light is refracted, or bent, as it passes through the object. An object with a curved surface, such as a lens, will refract light in a specific manner, while a flat surface will not.
determine the force (in n) exerted by only the fluid on the window of an instrument probe at this depth if the window is circular and has a diameter of 3.30 cm.
The result of the calculation is 385500 Pa and 460.5 N. They held that the Force was far too nuanced to be simply divided between light and dark.
They believed that Force-users were not purely good or evil, but rather that they had a role in a wider spectrum.
3.9 cm = 0.039 m, 101.3 kPa = 101300 Pa
The air pressure at the surface, the water density, the gravitational acceleration, and the height difference are all taken into account when determining the pressure at a depth of d = 29 m.
The instrument probe's pressure field is located where:
Therefore, the force applied to the area under pressure would be 385500 Pa.
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a mass of sunken lead is resting against the bottom in a glass of water. you take this lead, put it in a small boat of negligible mass, and float the boat in the water.
A. The sunken lead displaces a volume of water equal to the lead's own volume.
B. The floating lead displaces a volume of water equal to the lead's own volume.
C. The sunken lead displaces a volume of water whose weight equals the lead's weight.
D. The floating lead displaces a volume of water whose weight equals the lead's weight.
When a mass of sunken lead is resting against the bottom in a glass of water, and put it in a small boat of negligible mass, then float the boat in the water; what will happen is: The floating lead displaces a volume of water whose weight equals the lead's weight, and the sunken lead displaces a volume of water equal the lead's own volume. So, the correct answers are option A and D.
This is happened because according to Archimede's Principle which states that a body immersed in a fluid experiences an upthrust equal to the weight of the fluid displaced, and this is fundamental to the equilibrium of a body floating in still water.
What is mass?Mass is the amount of matter which an object contains. The more matter something has, the more it will weigh. For example, a horse has more matter in it than a cat thus, its mass is heavier. The size of something does not determine how much matter an object has.
Mass (symbolized m) is a dimensionless quantity representing the amount of matter in a particle or object. The standard unit of mass in the International System (SI) is the kilogram (kg).
In physics, mass is a quantitative measure of inertia, a fundamental property of all matter. It is, in effect, the resistance which a body of matter offers to a change in its speed or position upon the application of a force.
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A 2150 kg car, moving east at 10.0 m/s, collides and joins with a 3250 kg car. The cars move east together at 5.22 m/s.
The momentum during the collision is conserved. Hence, the total initial momentum is equal to the total final momentum. Therefore, the initial velocity of the car with a mass of 3250 Kg was, 2.05 m/s.
What is momentum?Momentum of a body is the product of its mass and velocity. Momentum is a vector quantity having both magnitude and direction.
The momentum during a collision is conserved. Thus,the sum of initial momentum of the colliding bodies is equal to the final momentum of the combined mass.
Given that, mass of car 1 = 2150 Kg
velocity = 10 m/s
Momentum = mass × velocity.
= 10 m/s × 2150 Kg = 21500 kg m/s.
Mass of car 2 = 3250 kg
the combined mass = 3250 + 2150 = 5400 Kg
final velocity = 5.22 m/s
final velocity = 5400 Kg × 5.22 = 28188 kg m/s .
momentum of car 1 + momentum of car 2 = 281888 kg m/s
21500 kg m/s + (3400 Kg × V) = 281888 kg m/s
v = 2.05 m/s.
Therefore, the initial velocity of the car with a mass of 3250 Kg is 2.05 kg m/s.
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Your question is incomplete. But your complete question probably was:
A 2150 kg car, moving east at 10.0 m/s, collides and joins with a 3250 kg car. The cars move east together at 5.22 m/s. What was the initial speed of car with 3250 kg mass.
planet a has twice the mass of planet b. from this information, what can we conclude about the escape speed for planet a compared to that of planet b?
Answer:
Assume that the two planets have the same radius. The escape velocity of planet a(with twice the mass) will be [tex]\sqrt{2}[/tex] times the escape velocity of planet b.
Explanation:
Let [tex]G[/tex] denote the gravitational constant.
Consider a spherical planet of mass [tex]M[/tex] and radius [tex]r[/tex]. If an object of mass [tex]m[/tex] is on the surface of the planet, the gravitational potential energy [tex]\text{GPE}[/tex] of that object will be:
[tex]\displaystyle \text{GPE} = \left(-\frac{G\, M\, m}{r}\right)[/tex].
If this object is moving at a speed of [tex]v[/tex], the kinetic energy [tex]\text{KE}[/tex] of this object will be [tex]\text{KE} = (1/2)\, m\, v^{2}[/tex].
If this object is moving at the escape velocity [tex]v_{e}[/tex] of the planet, the [tex]\text{KE}[/tex] of this object will be equal to the opposite [tex]\text{GPE}[/tex]. In other words:
[tex]\begin{aligned}\frac{1}{2}\, m\, {v_e}^{2} &= \text{KE} \\ &= (-\text{GPE}) \\ &= \frac{G\, M\, m}{r}\end{aligned}[/tex].
Rearrange this equation to find escape velocity [tex]v_{e}[/tex]:
[tex]\begin{aligned}{v_{e}}^{2} &= \frac{2\, G\, M}{r}\end{aligned}[/tex].
[tex]\begin{aligned}{v_{e}} &= \sqrt{\frac{2\, G\, M}{r}}\end{aligned}[/tex].
Assume that the radius [tex]r[/tex] of the planet is constant. Based on this equation, escape velocity [tex]v_{e}[/tex] will be portional to [tex]\sqrt{M}[/tex], the square root of the mass of the planet.
Hence, if the radius of planet a and planet b are equal, the escape velocity of planet a will be [tex]\sqrt{2}[/tex] times the escape velocity of planet b.
what is the maximum speed with which a 1200- kg car can round a turn of radius 90.0 m on a flat road if the coefficient of static friction between tires and road is 0.70?
The the speed of the car is 24m/s.
The radius of curve is, 90
The mass of the car is, 1200 kg
The coefficient of friction between the tires and the road is, 0.70
The acceleration due to gravity is, .9.81
Here, mg is the weight of the car Ftr , is the frictional force, is the normal force Fn m is the mass of the car.
Fn = mg
V² = μrg
V = √90 × 0.70 × 9.81 = 24m/s
The rate at which an object moves in relation to a reference point is measured as its speed. It is regarded as a magnitude or scalar quantity and has no direction. The formula: can be used to calculate speed.
Speed equals Distance/Time.
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the heat of vaporization of water at 100°c is 40.66 kj/mol. calculate the quantity of heat that is absorbed/released when 7.00 g of steam condenses to liquid water at 100°c
The quantity of heat released is -11.30 kJ
First, we have to calculate the number of moles of water.
Molar mass of water = 18 g/mole
Moles of water = 0.278 mole
ΔH= -q/n
heat of vaporization = 40.66 kJ/mol
we have to find the heat absorbed/released
n = number of moles of water
40.66 = q/ 0.278
q = -11.30 KJ
Heat is taken during the vaporization process, but it is released during the condensation step.
What is heat of vaporization?
The heat of vaporisation (Hvap), also known as enthalpy of vaporisation, is determined by the quantity of enthalpy (heat energy) necessary to change a liquid substance into a gas or vapour. Joules per mole (J/mol) or, in some cases, calories are used to measure it (C). Because enthalpy must constantly be added to a system in order to evaporate a liquid, the heat of vaporisation always has a positive value. The likelihood that the molecules will separate from the liquid and turn into a gas increases as their kinetic energy increases.
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Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.
(c) displacement may or may not be equal to distance
ab and ccall
The statement that is true about displacement is that displacement may or may not be equal to the distance travelled.
Displacement can be defined as the shortest distance between the initial and the final position of the body.
Displacement is a vector quantity it has a magnitude as well as direction.
Displacement can be zero, can be negative and it also can be positive.
Displacement may or may not be equal to the distance travelled. Because the object travels in the shortest distance possible then the distance will be equal to the displacement of the object. But if the object take any other path to travel from the initial to final position then that is placement may not be equal to the distance travelled.
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A student is trying to calculate the coefficent of friction between a box and a surface. She measures that the 80kg box will slide if the student pushes with a force greater than 408N.
Answer each of the following questions for the case in which the student is pushing as hard as possible without causing the box to slide.
How many forces are acting on the box?
What is the magnitude of the friction?
What is the magnitude of the normal force from the floor?
Which coefficient of friction can be determined in this case?
What is the value of this coefficient friction?
in a mechanical wave, the restoring force is the force that actually causes the oscillation.a. opposite the direction of the propagationb. opposite the direction of the dicplacement from equilibriumc. in the same direction as the dicplacement from equilibriumd. in the same direction as the propagaion
Option B; In a mechanical wave, the restoring force is the force that actually causes the oscillation opposite the direction of the displacement from equilibrium.
Oscillation is the repeating or periodic oscillation of a quantity, often in time, around a central value (often an equilibrium point) or between two or more states. A swinging pendulum and alternating current are two common examples of oscillation. Physics can employ oscillations to simulate intricate interactions, as those between atoms.
The beating of the human heart (for circulation), business cycles in economics, predator-prey population cycles in ecology, geothermal geysers in geology, the vibration of the strings in guitars and other string instruments, the periodic firing of nerve cells in the brain, and the periodic swelling of Cepheid variable stars in astronomy are just a few examples of the dynamic systems that exhibit oscillations in nearly every branch of science.
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true or false: a leaf floating up and down in a pond is an example of a longitudinal wave. a. true b. false
A leaf floating up and down in a pond is an example of a longitudinal wave that is False.
Particle motion is perpendicular to energy motion. Shear waves exhibit wave polarization that longitudinal waves do not. There is a distinct peak-to-valley movement. The node remains stationary. Here you can touch standing waves with a rope without disturbing the waves.
The position on the standing wave of maximum amplitude is known as the antinode. Antinodes occur halfway between nodes. Standing waves are undesirable. Some of the energy is reflected off the ends of the line and actually travels up the line.
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suppose that the color and behavior of a star identify it as a type that we know has absolute magnitude 4.8. if the star's apparent magnitude is found to be 14.8, how far away is it? vcu
If the star's apparent magnitude is found to be 14.8, how far away is 1000 parsecs.
calculation:-
subtract absolute from apparent to get DM:
14.8 - 4.8 = 10
divide DM by 5:
10/5 = 2
raise 10 to that power:
10^2 = 100
multiple products by 10 parsecs:
10 x 100 parsecs = 1000 parsecs
The apparent magnitude of a celestial object such as B. A star or galaxy is the brightness measured by an observer at a given distance from the object. The shorter the distance between the observer and the object, the greater the apparent brightness.Two stars of the same visual magnitude A and B. The apparent magnitude depends on the position of the observer. Different observers give different measurements.
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at what two points between object and screen may a converging lens with a 3.00- cm focal length be placed to obtain an image on the screen?
The focal length be placed to obtain an image on the screen f>10cm. When an object is placed between F and a concave mirror, the image will appear behind the mirror, enlarged, and upright.
The focal length of your lens is the distance (in millimeters) between its point of convergence and the sensor or film used to capture the image. How much of the scene your camera can record depends on the focal length of your film or digital camera lens. The focal length of a narrow lens in air is the separation between the lens's primary plane and its center. The focal length of a converging lens is positive and determines how far a collimated light beam must travel to focus on a single point.
If a screen image is to be created, v=u=40
Additionally, v 1 u 1 = f 1
Using the equations above, one can write f 1 = 40+u 1 u 1.
So, u 2 + 40 u + 40 f = 0.
As a result, the determinant of the equation 0, there is no such u.
Consequently, 40 2 4(1)(40f)0
hence, f>10cm
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how many times does the light beam shown in the figure reflect from the top mirror? (figure 1)
5 times the light beam shown in the figure reflect from the top mirror.
The assumption we must make in this situation is that the angles of incidence and reflection will be equal. We can observe from the illustration that the angle of incidence is [tex]$15^{\circ}$[/tex].
Provided that there are 68 cm in the vertical direction. This allows us to calculate the horizontal distance traveled by light before it reflects off the bottom mirror at a half-length.
[tex]x & =68 \mathrm{~cm} \times \tan 15^{\circ}[/tex]
=18.36 cm
The number of x lengths that may fit within a 168 cm length is,
[tex]k & =\frac{168 \mathrm{~cm}}{18.36 \mathrm{~cm}}[/tex]
=9.15
[tex]& \approx 9[/tex]
There will be reflections from the top mirror nine times out of ten, and the bottom mirror nine times out of ten.
Hence, number of reflections the top mirror is 5
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Note:- The question could be as bellow,
How many times does the light beam shown in the figure reflect from the top mirror? (figure 1)