Answer:
18.5 kg
Explanation:
The mass M2 is approximately 18.5 kg.
To determine the mass M2, we need to use the given information to solve for the tensions T1 and T3, and then use these tensions and the horizontal tension of 39 N to calculate the mass of the weight on the right.
We can use the angles and the horizontal tension to calculate the vertical components of T1 and T3. The vertical component of T1 is given by T1 * sin 55°, and the vertical component of T3 is given by T3 * sin 44°.
We can use the principle of equilibrium to set the sum of the forces in the vertical direction equal to zero. In this case, the forces are the weight of the left weight W1, the vertical component of T1, the vertical component of T3, and the weight of the right weight M2 * g, where g is the acceleration of gravity (9.8 m/s^2).
We can write this as an equation: W1 + T1 * sin 55° + T3 * sin 44° - M2 * g = 0.
We know that the weight of the left weight is W1 = M1 * g, where M1 is the mass of the left weight. We also know that the horizontal tension is 39 N, and we can calculate the vertical components of T1 and T3 using the given angles and the horizontal tension. When we substitute these values into the equation above and solve for M2, we get:
M2 = (W1 + T1 * sin 55° + T3 * sin 44°) / g
= (M1 * g + 39 N * sin 55° + 39 N * sin 44°) / g
≈ 18.5 kg
Therefore, the mass M2 is approximately 18.5 kg.
One object has twice as much mass as another object. The first object also has also twice as much...
a. velocity
b. gracitational acceleration
c. inertia
d. all of these
Answer: (c) inertia
Explanation: Mass is directly proportional to inertia. Mass is not related to velocity, and the gravitational acceleration of all objects a given distance from the same massive body is the same. This is tricky, as the force of gravity (weight) is larger on heavier objects, but one must remember that weight is calculated by multiplying an object's mass by g. Option b is only considering g.
could the young two-slit interference experiment be performed with gamma rays? if not, why not? if so, discuss differences in the experimental design compared to the experiment with visible light.
While there would be interference, the phase difference between the two sources would fluctuate and charge too quickly for you to see the pattern.
Young's double-slit experiment makes use of two coherent light sources that are spaced closely apart. Only a few orders of magnitude above the wavelength of light are typically employed. Young's double-slit experiment, which is illustrated by a figure, contributed to our knowledge of the wave theory of light. Since the pattern is made up of parallel straight-line-shaped fringes rather than the fringes of overlapping circles, Young's interference experiment with slits results in a crisper fringe pattern than with pinholes. Compared to the wider overlap of parallel straight lines, circles have relatively fewer portions of overlap.
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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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a simple motor is powered by a 5.00 v source and consists of a coil rotating in a uniform magnetic field. it is observed that the current in the motor decreases from 165 ma to 45.0 ma as the rotation rate goes from zero to a steady 125 rpm. determine the back emf as the motor turns at 125 rpm.
The process of building a straightforward direct current (d.c.) electric motor.
What is Electric motor?It is made out of a rectangular coil of wire positioned on a spindle so that it can move back and forth between the curved poles of a permanent magnet in the shape of a U.
The coil's two ends are soldered to a copper split ring, also known as a commutator, and its two halves, respectively. When two carbon brushes are placed in circuit with a battery and rheostat, they are made to mildly press against the commutator by means of springs, which causes the coil to revolve.
Assume that when the current is turned on for the first time, the coil is in the horizontal position. According to Fleming's left-hand rule, current will travel through the coil in the direction depicted.
Therefore, The process of building a straightforward direct current (d.c.) electric motor.
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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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What is the independent variable on this graph?
Oonly temperature (*C)
Oboth flow time (s) and temperature (C)
only flow time (s)
suppose we observe a source of x-rays that varies substantially in brightness over a period of a few days. what can we conclude?
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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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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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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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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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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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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What is the magnitude of the force needed to keep a 60.-newton rubber block moving across level, dry asphalt in a straight line at a constant speed of 2.0 meters per second?
The magnitude of the force needed to keep the block moving at a constant speed is 60 N.
his is the same as the weight of the block. This is because the only force acting on the block is gravity. As the block is moving at a constant speed, the force of friction is equal to the force of gravity. This means that the magnitude of the force needed to keep the block moving is equal to the weight of the block, which is 60 N. To calculate this, we used the equation F = ma, where F is the force, m is the mass of the block, and a is the acceleration, which is 0 in this case. This gave us the equation F = m x 0, which is simplified to F = 0, meaning the magnitude of the force needed is 0, which is equal to the weight of the block, which is 60 N.
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one coulomb is defined as the amount of charge that passes a point in a wire in 1 second when a current of 1 ampere is flowing through the wire group of answer choices true false
True, One coulomb is defined as the amount of charge that passes a point in a wire in 1 second when a current of 1 ampere is flowing through the wire.
An electric current is a circulation of charged particles, including electrons or ions, shifting through an electrical conductor or space. Its miles are measured because of the net fee of drift of electric fee through a floor or into a managed extent.
The current is one of the most important and essential elements within electric and electronic technology. The contemporary flowing in a circuit may be utilized in an expansion of ways from generating warmness to inflicting circuits to replace, or information to be saved in an included circuit.
Current is decided by the number of electrons passing through a cross-section of a conductor.
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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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The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed. (a) Consider a pipe of length L open at both ends. If the speed of sound in air is v, what is the fundamental frequency f1 of the sound wave produced when you blow into the pipe (in terms of L and v)? (b) If you drill a hole at a position half the length of the pipe, what is the new fundamental frequency f1' of the sound that can be produced in the pipe (in terms of the original length L and v)? (c) For which values of n does the nth harmonic of the original pipe of length L from part (a) also exist as a harmonic of the pipe in part (b). (d) What length of open-closed pipe would you need to achieve the same fundamental frequency f1'' as the original open-open pipe of length L from part (a)? (e) What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe from part (d) (in terms of L and v)?
The frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe 428.75 Hz.
Calculation:-
A. L = 80 cm
= 0.8 m
V = 343 m/s ( sound speed in the air )
V1 = n V / 2 L
= 1 X 343 / 2 X 0.8
V1 = 214.375 Hz
B. using the above equation
Vf = V / 2 L
[tex]Vf\alpha 1 / L[/tex]
so the new frequency is high than the previous or before.
C. Now L = L / 2 = 0.8 / 2
L = 0.4 m
Vn = n V / 2 L
= 1 X 343 / 2 X 0.4
Vn = 428.75 Hz
so the new frequency is 428.75 Hz
D. Vn = n V / 2 L
L = L /2
Vn = 2 n V / 2 L
Vn = 2 n Vn
n = 2n
Vn = n' Vn
even multiples of the fundamental frequency
E. for open-closed pipe
fn = n V / 4 L
and for an open- open pipe
Vn = n V / 2 L
n V / 2 L = n V / 2 L
2 L = L
L = L / 2
So half the length of open-open Pipe.
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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.
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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Please help: "A runner moves 2.38 m/s north. She accelerates at 0.310 m/s² at a -62.0° angle. How many seconds later is she running directly east?" Thanks!
According to equations of motion, we get after 8.70 seconds she running directly east.
Which are the 3 equations of motion?Three equations of motion, commonly referred to as the laws of constant acceleration, exist for uniform acceleration. In order to derive the components such as displacement(s), velocity(initial and final), time(t), and acceleration(a), these equations are used.These are the three equations:
v = v₀ + ats = v₀ t + ½at. v= v₀ + 2as.Given :
v₀ = 2.38 m/s
v = 0 m/s
a = 0.310 sin (-62.0°) m/s²
to find time t :
v = at + v₀
0 m/s = (0.310 sin (-62.0°) m/s²) t + (2.38 m/s)
t = 8.70 s
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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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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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If a ball i thrown horizontally with an initial velocity of 20. 0 Meter per econd from a 60. 0 meter high tower what i the horizontal velocity of the ball before it reache the ground
The horizontal velocity of the ball before it reaches the ground, where the initial velocity of 20 m/s from a 60m high tower, is 20m/s
Velocity is the directional speed of an object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.
From the question given
Initial horizontal verical velocity (ux) = 20 m/s
height of tower (h) = 60m
Basically, after the ball is thrown, it becomes a projectile. It means that the acceleration along the horizontal direction is ax = 0 m/s² over the whole motion.
Now, let's use the formula
Vx = Ux + ax * t
= 20 + (0 m/s²) * t
Vx = 20 m/s
Therefore, the horizontal velocity of the ball before it reaches the ground is 20 m/s.
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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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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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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)
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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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 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].
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 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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