a The kinetic energy of an object always has a positive value.
Explanation:
The kinetic energy of an object is the energy possessed by an object due to its motion, and it can be calculated as
[tex]K=\frac{1}{2}mv^2[/tex]
where
m is the mass of the object
v is its speed
Let's now analyze each statement:
a The kinetic energy of an object always has a positive value. --> TRUE. In fact, the mass of the object is always positive, and the term [tex]v^2[/tex] is always positive as well, so the kinetic energy is always positive.
b The kinetic energy of an object is directly proportional to its speed. --> FALSE. Looking at the formula, we see that the kinetic energy is proportional to the square of the speed, [tex]K\propto v^2[/tex].
c The kinetic energy of an object is expressed in watts. --> FALSE. Watts is the units for measuring power, while the kinetic energy is measured in Joules, the units for the energy.
d The kinetic energy of an object is a quantitative measure of its inertia. --> FALSE. The inertia of an object depends only on its mass, not on its speed.
e The kinetic energy of an object is always equal to the object’s potential energy. --> FALSE. The potential energy depends on the altitude from the ground, not from the speed, so the two energies can be different.
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Kinetic energy, a concept in physics, refers to the energy that an object possesses due to its motion and is always positive. It is proportional to the square of speed, expressed in Joules, and distinct from inertia and potential energy.
Explanation:In physics, kinetic energy pertains to the energy possessed by an object due to its motion. The kinetic energy (K) of a moving object can be calculated using the formula K = (1/2)mv², where 'm' stands for mass and 'v' for velocity.
Concerning the statements included in the question, option a) 'The kinetic energy of an object always has a positive value.' is correct. This is due to the fact that kinetic energy is product of the mass and the square of the speed, which are always positive or zero, thereby making kinetic energy always positive.
Regarding option b) 'The kinetic energy of an object is directly proportional to its speed', it's essential to note that kinetic energy is proportional not to the speed, but the square of the speed, hence this statement is partially correct.
The kinetic energy is not expressed in watts (option c). It's actually measured in Joules, which is equal to kg · m²/s² (kilogram meter squared per second squared).
Kinetic energy is not a measure of an object’s inertia (option d). It measures an object's energy due to motion, whereas inertia is an object’s resistance to change in motion.
Lastly, the kinetic energy of an object is not always equal to the object’s potential energy (option e). These are two different forms of energy that can convert into each other under certain circumstances, but they are not always equal.
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Assume that the total cholesterol levels for adults are normally distributed with mean cholesterol level of 51.6 mg/dL and standard deviation 14.3 mg/dL. Find the probability that an individual will have a cholesterol level greater than 60 mg/dL.
Answer:
therefore, the probability that an individual will have a cholesterol level greater than 60 mg/dL.= 0.27
Explanation:
given data
Normal distribution
mean cholesterol level μ= 51.6 mg/dL
Standard deviation σ= 14.3 mg/dL
x= 60 mg/dL
We have to find out P(x>60)
We Know that [tex]P(x>a) =P(Z>(a-\mu)/\sigma)[/tex]
therefore, [tex]P(x>60) =P(Z>(60-51.3)/14.3)[/tex]
= P(Z>0.61)
= 1 - P(Z<0.61)
= 1 - 0.7291
= 0.27
therefore, the probability that an individual will have a cholesterol level greater than 60 mg/dL.= 0.27
To find the probability that an individual will have a cholesterol level greater than 60 mg/dL, we calculate the z-score and find the corresponding probability using a Z-table or calculator.
Explanation:To find the probability that an individual will have a cholesterol level greater than 60 mg/dL, we need to use the z-score formula. The z-score is calculated by subtracting the mean from the value of interest and dividing it by the standard deviation. In this case, the z-score is (60 - 51.6) / 14.3 = 0.589. Using a Z-table or a calculator, we can find that the probability of a z-score greater than 0.589 is approximately 0.279.
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Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to a halt after traveling 1.50 m along the incline. (a) If the initial speed of the crate was 1.77 m/s and the angle of inclination is 30.0°, how much energy was dissipated by friction? (b) What is the coefficient of sliding friction?
Answer:(a)891.64 N
(b)0.7
Explanation:
Mass of crate [tex]m=100 kg[/tex]
Crate slows down in [tex]s=1.5 m[/tex]
initial speed [tex]u=1.77 m/s[/tex]
inclination [tex]\theta =30^{\circ}[/tex]
From Work-Energy Principle
Work done by all the Forces is equal to change in Kinetic Energy
[tex]W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2[/tex]
[tex]W_{gravity}=mg(0-h)=mgs\sin \theta [/tex]
[tex]W_{gravity}=-mgs\sin \theta[/tex]
[tex]W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N[/tex]
change in kinetic energy[tex]=\frac{1}{2}\times 100\times 1.77^2=156.64 J[/tex]
[tex]W_{friction}=156.64+735=891.645[/tex]
(b)Coefficient of sliding friction
[tex]f_r\cdot s=W_{friciton}[/tex]
[tex]891.645=f_r\times 1.5[/tex]
[tex]f_r=594.43 N[/tex]
and [tex]f_r=\mu mg\cos \theta [/tex]
[tex]\mu 100\times 9.8\times \cos 30=594.43[/tex]
[tex]\mu =0.7[/tex]
Final answer:
To answer part (a), calculate the work done by friction using the equation work = force × distance. Calculate the change in kinetic energy of the crate using the equation change in kinetic energy = 0.5 × mass × (final velocity)^2 - 0.5 × mass × (initial velocity)^2 and subtract it from the work done by friction. To answer part (b), use the equation frictional force = coefficient of friction × normal force and calculate the normal force using the equation weight = mass × acceleration due to gravity.
Explanation:
To answer part (a), we need to calculate the work done by friction. The work done by friction can be calculated using the equation work = force × distance. In this case, the force is the frictional force and the distance is the distance traveled by the crate. The work done by friction is equal to the change in kinetic energy of the crate, which can be calculated using the equation change in kinetic energy = 0.5 × mass × (final velocity)^2 - 0.5 × mass × (initial velocity)^2. Subtracting the change in kinetic energy from the work done by friction will give us the energy dissipated by friction.
To answer part (b), we can use the equation frictional force = coefficient of friction × normal force. The normal force is equal to the weight of the crate, which can be calculated using the equation weight = mass × acceleration due to gravity. By substituting the known values into these equations, we can find the coefficient of sliding friction.
A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 9.1 m/s. After the collision, the 0.20-kg puck has a speed of 5.5 m/s at an angle of θ = 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision, (b) Find the fraction of kinetic energy lost in the collision.
The problem presents a collision scenario where principles of momentum conservation and kinetic energy are applied to determine the final velocity of a striking puck and the fraction of kinetic energy lost during the collision.
Explanation:The question describes a collision scenario in physics involving two pucks, where conservation of momentum and kinetic energy principles are applied. The question can be resolved by dividing it into two sections: (a) Determination of the velocity of the 0.30-kg puck after the collision, and (b) Determination of the fraction of kinetic energy lost in the collision.
For part (a), we use the law of conservation of momentum, which states that the total linear momentum of an isolated system remains constant, regardless of whether the objects within the system are at rest or in motion. To find the final velocity of the 0.30-kg puck, we subtract the momentum vector of the 0.20-kg puck after the collision from the momentum vector before the collision. For part (b), we first find the initial and final total kinetic energy of the system. The fraction of kinetic energy lost is given by the difference between the initial and final kinetic energies, divided by the initial kinetic energy.
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The principle of conservation of momentum and kinetic energy principles are employed to figure out the final velocity of the 0.30-kg puck and the fraction of kinetic energy lost in the collision. Initial and final velocities of the pucks, as well as their masses, are used in the process.
Explanation:To solve this problem, we need to apply the principle of conservation of momentum, which states that the total momentum of a system before a collision is equal to the total momentum after the collision. To find out the velocity of the 0.30-kg puck after the collision (a), the momentum equation (m₁v₁_initial + m₂v₂_initial = m₁v₁_final + m₂v₂_final) is applied. It is known that initial velocities of 0.30-kg and 0.20-kg pucks are 0 and 9.1 m/s respectively; the final velocity of the 0.20-kg puck is 5.5 m/s at 53° to the x-axis.
To calculate the fractional kinetic energy lost in the collision (b), firstly, the kinetic energy before and after the collision is calculated using the equation KE = 1/2mv₂. The fraction of kinetic energy lost is then (KE_initial - KE_final)/KE_initial.
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A block is attached to a spring, with spring constant k, which is attached to a wall. It is initially moved to the left a distance d (at point A) and then released from rest, where the block undergoes harmonic motion. The floor is frictionless.The points labelled A and C are the turning points for the block, and point B is the equilibrium point.1)Which of these quantities are conserved for the spring and block system?mechanical energyx-direction linear momentumneither energy nor momentum
Answer:
Explanation:
When a spring attached with a block is stretched and released , the block starts moving under SHM. The elastic energy stored in it initially at the time of initial stretch is repeatedly converted into kinetic energy and vice - versa while the body keeps moving under SHM. So we can tell that the mechanical energy of the system remains constant.
In the whole process the velocity of the body keeps changing in terms of both magnitude and direction . It happens because a variable force acts on the body constantly towards the equilibrium point so its momentum also keeps changing all the time
At an accident scene on a level road, investigators measure a car's skid mark to be 98 m long. It was a rainy day and the coefficient of friction was estimated to be 0.40. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.
Answer:
The speed of the car was 28 m/s.
Explanation:
Hi there!
The initial kinetic energy of the car, KE, is equal to the negative work, W, done by friction to bring the car to stop. Let´s write the work-energy theorem:
W = ΔKE = final kinetic energy - initial kinetic energy
In this case, the final kinetic energy is zero, then:
W = - initial kinetic energy
Since the work done by friction is negative (the work is done in opposite direction to the movement of the car), then:
Wfr = initial kinetic energy
The work done by friction is calculated as follows:
Wfr = Fr · d
Where:
Fr = friction force.
d = distance.
The friction force is calculated as follows:
Fr = N · μ
Where:
N = normal force.
μ = coefficient of friction
Since the only vertical forces acting on the car are the weight of the car and the normal force, and the car is not being accelerated in the vertical direction, the normal force has to be equal to the weight of the car (with opposite sign).
Then the friction force can be written as follows:
Fr = m · g · μ
Where:
m = mass of the car.
g = acceleration due to gravity (9.8 m/s²)
The work done by friction will be:
W = m · g · μ · d
The equation of kinetic energy is the following:
KE = 1/2 · m · v²
Where
m = mass of the car.
v = speed.
Then:
W = KE
m · g · μ · d = 1/2 · m · v²
2 · g · μ · d = v²
2 · 9.8 m/s² · 0.40 · 98 m = v²
v = 28 m/s
The speed of the car was 28 m/s.
A 93 kg man holding a 0.653 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 11.3 m/s (relative to the ground) and then catches the ball after it rebounds from the wall. How fast is he moving after he catches the ball? Ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall. Answer in units of m/s.
The 93 kg man will be moving at a speed of 0.161 m/s after catching the ball that he threw at a wall and rebounded back to him, based on the principle of conservation of momentum.
Explanation:This question involves the principle of conservation of momentum. According to the law of conservation of momentum, the total momentum of the system (man and the ball) is conserved before and after the collision with the wall. The total initial momentum, before the ball is thrown, is zero as everything is at rest.
When the man throws the ball, the ball gains momentum in one direction, and to keep the total momentum zero, the man must gain an equal amount of momentum in the opposite direction. On returning, the ball has changed its direction, so it has lost the initial momentum and gained the same amount in the opposite direction.
The man catches the ball, meaning he gains the momentum that the ball brought back, causing him to move. To determine how fast he is moving, we set the final momentum of the system (which should still be zero) equal to the initial momentum.
To calculate the speed (v) of the man we'll use the equation: v = 2 × ball_mass × ball_speed / man_mass = 2 × 0.653 kg × 11.3 m/s / 93 kg = 0.161 m/s
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The man will be moving at approximately 0.078 m/s after he catches the ball. This is found using the conservation of momentum, considering the ball's momentum before and after it collides with the wall and bounces back to the man.
Explanation:The student's question involves the conservation of momentum, a fundamental concept in physics. When the man throws the ball and catches it after it bounces off the wall, both the ball and the man are involved in an isolated system where external forces are negligible. As momentum is conserved, the velocity of the man after catching the ball can be calculated using the conservation of momentum principle.
To solve this problem, we first consider the momentum before and after the collision with the wall. The man throws the ball at 11.3 m/s. The direction of the ball's momentum changes after it bounces back, but the speed remains the same due to the 'elastic' nature of the collision as mentioned in the problem statement (no energy loss).
The initial momentum of the system (man and ball) is zero because they're both at rest initially. When the man throws the ball, the ball gains momentum in the forward direction, and the man gains momentum in the opposite direction. This momentum for the man (in the opposite direction) is what will be calculated.
The initial momentum of the ball and man is:
Initial Ball: 0 kg·m/s (at rest)Initial Man: 0 kg·m/s (at rest)The final momentum after throwing the ball is:
Final Ball: -0.653 kg * 11.3 m/s (the negative sign indicates the direction change after bouncing)Final Man: 93 kg * v (where v is the velocity we want to calculate)By conservation of momentum:
Initial momentum = Final momentum
0 = -0.653 kg * 11.3 m/s + 93 kg * v
Solving for v gives:
v = (0.653 kg * 11.3 m/s) / 93 kg
v ≈ 0.078 m/s
This is the speed of the man after catching the ball, moving opposite to the direction in which he threw the ball.
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A block of mass 11.8 kg slides from rest down a frictionless 35.6° incline and is stopped by a strong spring with k = 3.11 x 10^4 N/m. The block slides 2.17 m from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?
Answer:
x = 6.58 m
Explanation:
For this exercise let's use the concept of conservation of mechanical energy. Let's look for energy at two points the highest and when the spring is compressed
Initial
Em₀ = U = m g and
Final
[tex]E_{mf}[/tex] = ke = ½ k x2
As there is no friction the mechanical energy is conserved
Em₀ = [tex]E_{mf}[/tex]
m g y = ½ k x²
Let's use trigonometry to face height
sin θ = y / L
y = L sin θ
x = √ 2mg (L synth) / k
x = √ (2 11.8 9.8 2.17 sin35.6 / 3.11 104)
x = 6.58 m
Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e., protons with one orbiting electron) are fed into a chamber where there is a strong magnetic field. Electrons in this chamber are trapped in tight orbits, which greatly increases the chance that they will collide with a hydrogen atom and ionizes it. One such source uses a magnetic field of 70 mT, and the electrons' kinetic energy is 1.2 eV. If the electrons travel in a plane perpendicular to the field, what is the radius of their orbits?
Answer:
[tex]r=5.278\times 10^{-4}\ m[/tex]
Explanation:
Given that:
magnetic field intensity, [tex]B=0.07\ T[/tex]kinetic energy of electron, [tex]KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J[/tex]we have mass of electron, [tex]m=9.1\times 10^{-31}\ kg[/tex]Now, form the mathematical expression of Kinetic Energy:
[tex]KE= \frac{1}{2} m.v^2[/tex]
[tex]1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2[/tex]
[tex]v^2=4.2198\times 10^{11}[/tex]
[tex]v=6.496\times 10^6\ m.s^{-1}[/tex]
from the relation of magnetic and centripetal forces we have the radius as:
[tex]r=\frac{m.v}{q.B}[/tex]
[tex]r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}[/tex]
[tex]r=5.278\times 10^{-4}\ m[/tex]
The radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.
Force of the electron
The magnetic force and centripetal force of the electron is calculated as follows;
qvB = mv²/r
qB = mv/r
r = mv/qB
where;
m is mass of electronv is speed of the electronq is the magnitude of the chargeB is magnetic field strengthKinetic energy of the electronThe kinetic energy of the electron is given as;
K.E = ¹/₂mv²
mv² = 2K.E
v² = 2K.E/m
v² = (2 x 1.2 x 1.6 x 10⁻¹⁹)/(9.11 x 10⁻³¹)
v² = 4.215 x 10¹¹
v = √4.215 x 10¹¹
v = 6.5 x 10⁵ m/s
Radius of their orbitThe radius of their orbit is calculated as follows;
r = mv/qB
r = (9.11 x 10⁻³¹ x 6.5 x 10⁵) / (1.6 x 10⁻¹⁹ x 70 x 10⁻³)
r = 5.287 x 10⁻⁵ m
Thus, the radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.
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During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a velocity of 4.1 m/s, grabs and holds onto him so that they move off together with a velocity of 2.2 m/s. If the mass of the tackler is 100 kg, determine the mass of the receiver in kilograms. Assume momentum is conserved.
Answer:m=86.36 kg
Explanation:
Given
mass of tackler [tex]m_T=100 kg[/tex]
initial velocity of Tackler [tex]u_t=4.1 m/s[/tex]
Final velocity of combined system [tex]v=2.2 m/s[/tex]
Let m be the mass of receiver
conserving momentum
[tex]m\times 0+m_t\times u_t=(m+m_t)v[/tex]
[tex]0+100\times 4.1=(100+m)\cdot 2.2[/tex]
[tex]410=(100+m)\cdot 2.2[/tex]
[tex]100+m=186.36[/tex]
[tex]m=86.36 kg[/tex]
A DC current of 60 mA can cause paralysis of the body's respiratory muscles and hence interfere with breathing, but only 15 mA (rms) of AC current will do the same thing. Suppose a person is working with electrical power lines on a warm humid day and therefore has a low body resistance of 1000 OHMS. What DC and AC (amplitude and rms) potentials would it take to cause respiratory paralysis?
Answer:
DC: 60V
AC: 15Vrms (21.2V amplitude)
Explanation:
Since we know the current values for each case, we can calculate voltage in both scenarios.
For DC voltage:
Vdc = Idc * R = 60mA * 1000 Ohms = 60Vdc
For AC voltage:
Vac = Iac * R = 15mA * 1000 Ohms = 15 Vac (rms) the amplitude of this signal would be Vmax = 15Vac * 1.414 = 21.2V
A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-glass prism perpendicular to one face and then refracts through the hypotenuse back into the air. The angle between these two faces is 37 ∘.
Answer:
The angle between the red and blue light is 1.7°.
Explanation:
Given that,
Wavelength of red = 656 nm
Wavelength of blue = 486 nm
Angle = 37°
Suppose we need to find the angle between the red and blue light as it leaves the prism
[tex]n_{r}=1.572[/tex]
[tex]n_{b}=1.587[/tex]
We need to calculate the angle for red wavelength
Using Snell's law,
[tex]n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}[/tex]
Put the value into the formula
[tex]1.572\sin37=1\times\sin\theta_{r}[/tex]
[tex]\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})[/tex]
[tex]\theta_{r}=71.0^{\circ}[/tex]
We need to calculate the angle for blue wavelength
Using Snell's law,
[tex]n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}[/tex]
Put the value into the formula
[tex]1.587\sin37=1\times\sin\theta_{b}[/tex]
[tex]\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})[/tex]
[tex]\theta_{b}=72.7^{\circ}[/tex]
We need to calculate the angle between the red and blue light
Using formula of angle
[tex]\Delta \theta=\theta_{b}-\theta_{r}[/tex]
Put the value into the formula
[tex]\Delta \theta=72.7-71.0[/tex]
[tex]\Delta \theta=1.7^{\circ}[/tex]
Hence, The angle between the red and blue light is 1.7°.
The Hydrogen discharge lamp emits light of specific wavelengths which pass through a flint-glass prism, refract (or bend), and display a color spectrum due to dispersion. The red and blue lights have different refraction angles because of their different wavelengths. The distinct color bands indicate discrete energy transitions in the hydrogen gas.
Explanation:Light emission and its properties are central to this question. The light emitted by a hydrogen discharge lamp, which contains hydrogen gas at low pressure, indicates that there are discrete energies emitted, thus producing light of specific wavelengths. These distinct wavelengths are a result of H2 molecules being broken apart into separate H atoms when an electric discharge passes through the tube.
When this light enters a prism, it refracts or bends, leading to a phenomenon called dispersion. This happens because not all colors or wavelengths of light are bent the same amount: the degree of bending depends on the wavelength of the light, as well as the properties of the material (in this case, flint-glass prism). For instance, red and blue light have different refraction angles due to their different wavelengths of 656 nm and 486 nm respectively.
So, in the context of a hydrogen discharge lamp, the two prominent wavelengths, 656 nm (red) and 486 nm (blue), will refract differently when passing through the flint-glass prism. Further, the separation (or dispersion) of light into these distinct colored bands as it exits the prism gives us a line spectrum - visual evidence of the discrete energy transitions happening at the atomic level in the hydrogen gas of the lamp.
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When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 45.5 N and it is acting with an effective lever arm of 2.45 cm, what is the torque that the muscle produces on the wrist?
Answer:
Torque on the rocket will be 1.11475 N -m
Explanation:
We have given that muscles generate a force of 45.5 N
So force F = 45.5 N
This force acts on the is acting on the effective lever arm of 2.45 cm
So length of the lever arm d = 2.45 cm = 0.0245 m
We have to find torque
We know that torque is given by [tex]\tau =F\times d=45.5\times 0.0245=1.11475N-m[/tex]
So torque on the rocket will be 1.11475 N -m
A bullet, m = 0.500 kg, traveling with some initial velocity strikes and embeds itself in the bob of a ballistic pendulum, M = 9.50 kg. a) The combined masses rise to a height h = 1.28 m. The speed Vf of the combined masses immediately following impact is ____
Answer:
5 m/s
Explanation:
mass of bullet, m = 0.5 kg
initial velocity of bullet is u.
mass of pendulum, M = 9.50 kg
height raised, h = 1.28 m
The kinetic energy of the bullet and pendulum system is equal to the potential energy of the system
Let vf be the final velocity after the impact
KE = PE
1/2 (M + m) Vf^2 = (M + m) x g x h
vf^2 = 2 x 9.8 x 1.28 = 25.088
vf = 5 m/s
Thus, the final speed of the combined mass system after the impact s 5 m/s.
Final answer:
The final speed of the combined masses after the collision can be found using the conservation of momentum and mechanical energy principles, resulting in a final speed of approximately 1.20 m/s.
Explanation:
The final speed Vf of the combined masses after the collision can be calculated using the principle of conservation of momentum and conservation of mechanical energy.
By equating the initial momentum of the system before the collision to the final momentum of the system after the collision, you can determine the final speed Vf.
In this case, the final speed Vf is approximately 1.20 m/s.
Police lieutenants, examining the scene of an accident involving two cars, measure the skid marks of one of the cars, which nearly came to a stop before colliding, to be 78 m long. The coefficient of kinetic friction between rubber and the pavement is about 0.90. Estimate the initial speed of that car assuming a level road.
Answer:
37.11231 m/s
Explanation:
u = Initial velocity
v = Final velocity = 0
s = Displacement = 78 m
g = Acceleration due to gravity = 9.81 m/s²
[tex]\mu[/tex] = Coefficient of kinetic friction = 0.9
Acceleration is given by
[tex]a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g[/tex]
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=-2\mu gs\\\Rightarrow -u^2=2(-\mu g)s-v^2\\\Rightarrow u=\sqrt{v^2-2(-\mu g)s}\\\Rightarrow u=\sqrt{0^2-2\times (-0.9\times 9.81)\times 78}\\\Rightarrow u=37.11231\ m/s[/tex]
The initial speed of that car is 37.11231 m/s
Unpolarized light of intensity 800 W/m2 is incident on two ideal polarizing sheets that are placed with their transmission axes perpendicular to each other. An additional polarizing sheet is then placed between the two, with its transmission axis oriented at 30∘ to that of the first. What is the intensity of the light passing through the stack of polarizing sheets? (Express your answer to two significant figures.)
Answer:
75 W/m²
Explanation:
[tex]I_0[/tex] = Unpolarized light intensity = 800 W/m²
[tex]\theta[/tex] = Angle between filters = 30°
Intensity of light after passing through first polarizer
[tex]I=\frac{I_0}{2}\\\Rightarrow I=\frac{800}{2}\\\Rightarrow I=400\ W/m^2[/tex]
Intensity of light after passing through second polarizer
[tex]I_1=Icos^2\theta\\\Rightarrow I_1=400\times cos^2(30)\\\Rightarrow I_1=300\ W/m^2[/tex]
Intensity of light after passing through third polarizer
[tex]I_2=I_1cos^2\theta\\\Rightarrow I_2=300\times cos^2(90-30)\\\Rightarrow I_1=75\ W/m^2[/tex]
The intensity of the light passing through the stack of polarizing sheets is 75 W/m²
The intensity of the light passing through three polarizing sheets, with the first and the third being perpendicular and the second making an angle of 30 degrees with the first, is zero, due to the rules of light polarization.
Explanation:Unpolarized light is composed of many rays having random polarization directions. When it passes through a polarizing filter, it decreases its intensity by a factor of 2, thus the intensity after the first polarizer is 800 W/m^2 /2 = 400 W/m^2. According to Malus's Law, the intensity of polarized light after passing through a second polarizing filter is given as I = Io cos² θ, where Io is the incident intensity (from the first polarizer) and θ is the angle between the direction of polarization and the axis of the filter.
The second polarizer is making an angle of 30 degrees with the first polarizer. Therefore, the intensity after the second polarizer is I = 400 W/m^2 * cos²(30) ≈ 346 W/m^2. The third polarizer is perpendicular to the first polarizer (or, equivalently, it makes an angle of 90 degrees with the second polarizer), so the final intensity is I = 346 W/m^2 * cos²(90), which is 0 because the cosine of 90 degrees is zero. So, no light (intensity=0) passes through the stack of these polarizing sheets.
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A transformer connected to a 120-v(rms) at line is to supply 13,000V(rms) for a neon sign.
To reduce shock hazard, a fuse is to beinserted in the primary circuit;the fuse is to blow when the rmscurrent in the secondary circuit exceeds 8.50 mA.
(a) What is theratio of secondary to primary turns of the transformer?
(b) Whatpower must be supplied to the transformer when the rms secondarycurrent is 8.50 mA?
(c) What current rating should the fuse in theprimary circuit have?
Answer:
(a) 108
(b) 110.500 kW
(c) 920.84 A
Solution:
As per the question:
Voltage at primary, [tex]V_{p} = 120\ V[/tex] (rms voltage)
Voltage at secondary, [tex]V_{s} = 13000\ V[/tex] (rms voltage)
Current in the secondary, [tex]I_{s} = 8.50\ mA[/tex]
Now,
(a) The ratio of secondary to primary turns is given by the relation:
[tex]\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}[/tex]
where
[tex]N_{p}[/tex] = No. of turns in primary
[tex]N_{s}[/tex] = No. of turns in secondary
[tex]\frac{N_{s}}{N_{p}} = \frac{13000}{120}[/tex] ≈ 108
(b) The power supplied to the line is given by:
Power, P = [tex]V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW[/tex]
(c) The current rating that the fuse should have is given by:
[tex]\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}[/tex]
[tex]\frac{13000}{120} = \frac{I_{p}}{8.50}[/tex]
[tex]I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A[/tex]
A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/s. A second piece has a mass of 300 g and flies off along the y-axis with a speed of 45.0 m/s. What are the magnitude and direction of the total momentum of these two pieces?A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along the -axis with a speed of 82.0 m/s. A second piece has a mass of 300 g and flies off along the -axis with a speed of 45.0 m/s. What are the magnitude and direction of the total momentum of these two pieces?361 kg·m/s at 0.983° from the x-axis361 kg·m/s at 56.3° from the x-axis93.5 kg·m/s at 28.8° from the x-axis21.2 kg·m/s at 39.5° from the x-axis21.2 kg·m/s at 56.3° from the x-axis
Answer:
The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.
Explanation:
Hi there!
The total momentum is calculated as the sum of the momenta of the pieces.
The momentum of each piece is calculated as follows:
p = m · v
Where:
p = momentum.
m = mass.
v = velocity.
The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:
p = (m · v, 0)
p = (0.200 kg · 82.0 m/s, 0)
p = (16.4 kg m/s, 0)
The 300 g-piece flies along the y-axis. Its momentum vector will be:
p =(0, m · v)
p = (0, 0.300 kg · 45.0 m/s)
p = (0, 13.5 kg m/s)
The total momentum is the sum of each momentum:
Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)
Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)
Total momentum = (16.4 kg m/s, 13.5 kg m/s)
The magnitude of the total momentum is calculated as follows:
[tex]|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s[/tex]
The direction of the momentum vector is calculated using trigonometry:
cos θ = px/p
Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.
cos θ = 16.4 kg m/s / 21.2 kg m/s
θ = 39.3 (39.5° if we do not round the magnitude of the total momentum)
Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.
An antitank weapon fires a 3.00-kg rocket which acquires a speed of 50.0 m/s after traveling 90.0 cm down a launching tube. Assuming the rocket was accelerated uniformly, what is the average force that acted on it?
Answer:
Force in the rocket will be 4166.64 N
Explanation:
We have given mass of the rocket m = 3 kg
Rocket acquires a speed of 50 m sec so final speed v = 50 m/sec
Initial speed u = 0 m/sec
Distance traveled s = 90 cm = 0.9 m
From third equation of motion we know that [tex]v^2=u^2+2as[/tex]
[tex]50^2=0^2+2\times a\times 0.9[/tex]
[tex]a=1388.88m/sec^2[/tex]
From newton's law we F = ma
So force will be [tex]F=1388.88\times 3=4166.64N[/tex]
So force in the rocket will be 4166.64 N
The average force that acted on the rocket can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 3.00 kg and the acceleration can be determined by dividing the change in velocity by the distance traveled. The average force that acted on the rocket is 166.8 Newtons.
Explanation:The average force that acted on the rocket can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 3.00 kg and the acceleration can be determined by dividing the change in velocity by the distance traveled. The change in velocity is 50.0 m/s and the distance traveled is 90.0 cm, which is equivalent to 0.90 m.
So, the acceleration of the rocket is 50.0 m/s divided by 0.90 m, which gives us 55.6 m/s². Now, we can use Newton's second law to calculate the force: F = m * a, where F is the force, m is the mass, and a is the acceleration.
Plugging in the values, we get F = 3.00 kg * 55.6 m/s² = 166.8 N. Therefore, the average force that acted on the rocket is 166.8 Newtons.
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Sound waves are a longitudinal wave that have a speed of about 340 m/s when traveling through room temperature air. What is the wavelength of a sound wave that has a frequency of 6,191 Hz?
The wavelength of the wave is 0.055 m
Explanation:
The relationship between speed, frequency and wavelength of a wave is given by the wave equation:
[tex]v=f\lambda[/tex]
where
v is the speed
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the sound wave in this problem we have
v = 340 m/s is the speed
f = 6,191 Hz is the frequency
Solving for [tex]\lambda[/tex], we find the wavelength:
[tex]\lambda=\frac{v}{f}=\frac{340}{6191}=0.055 m[/tex]
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A garden hose is attached to a water faucet on one end and a spray nozzle on the other end. The water faucet is turned on, but the nozzle is turned off so that no water flows through the hose. The hose lies horizontally on the ground, and a stream of water sprays vertically out of a small leak to a height of 0.78 m.
What is the pressure inside the hose?
To solve this problem it is necessary to use the concepts related to pressure depending on the depth (or height) in which the object is on that fluid. By definition this expression is given as
[tex]P = P_{atm} +\rho gh[/tex]
Where,
[tex]P_{atm} =[/tex] Atmospheric Pressure
[tex]\rho =[/tex]Density, water at this case
g = Gravity
h = Height
The equation basically tells us that under a reference pressure, which is terrestrial, as one of the three variables (gravity, density or height) increases the pressure exerted on the body. In this case density and gravity are constant variables. The only variable that changes in the frame of reference is the height.
Our values are given as
[tex]P_{atm} = 1.013*10^5Pa[/tex]
[tex]\rho = 1000Kg/m^3[/tex]
[tex]g = 9.8m/s^2[/tex]
[tex]h = 0.78m[/tex]
Replacing at the equation we have,
[tex]P = P_{atm} +\rho gh[/tex]
[tex]P = 1.013*10^5 +(1000)(9.8)(0.78)[/tex]
[tex]P = 108944Pa[/tex]
[tex]P = 0.1089Mpa[/tex]
Therefore the pressure inside the hose is 0.1089Mpa
When light of wavelength 350 nm is incident on a metal surface, the stopping potential of the photoelectrons is measured to be 0.500 V. What is the work function of this metal? (c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s, 1 eV = 1.60 × 10-19 J) 3.05 eV 4.12 eV 3.54 eV 0.500 eV
The work function of the metal can be calculated using Einstein's photoelectric equation and the given information about the wavelength of light, stopping potential, Planck's constant, and speed of light. The initial step is to calculate the energy of the photon; thereafter, calculate the kinetic energy of the photoelectron with the stopping potential, and finally find the work function by subtracting the kinetic energy from the photon energy.
Explanation:To calculate the work function of the metal, we need to first figure out the energy of the photon hitting the metal surface. This can be found using Einstein's photoelectric equation: E = hc/λ. Here, h is Planck's constant (6.626 x 10-34 J · s), c is the speed of light (3.00 x 108 m/s), and λ is the wavelength of the light (350 nm or 350 x 10-9 m).
After calculating the energy of the photon, we can determine the energy required to eject the photoelectrons, known as the stopping potential, which in this case is 0.500 V. Note this is the extra energy required to stop the photoelectron and is equal to the kinetic energy of the photoelectron. Hence, 0.500V = 0.500 eV because 1V = 1eV for a single electron with charge e. To convert this to joules, we use the conversion 1eV = 1.60 x 10-19J.
Finally, in order to find the work function, we use Einstein's photoelectric equation again which states that the energy of the incident photon is equal to the work function (φ) plus the kinetic energy of the photoelectron. That is, E = φ + KE. From this, we can rearrange the equation to solve for φ, the work function of the metal where: φ = E - KE.
By substituting the appropriate values into this equation, the work function of the metal can be found out. But remember, work function is a property of the metal only and is independent of the incident radiation. Also, it determines the threshold frequency or cutoff wavelength of the metal beyond which photoelectric effect will not occur.
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(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. The electrical conductivity of and density of gold are 4.3 X 107 (O-m)-1 and 19.32 g/cm3, respectively.
(b) Now compute the electron mobility for Au.
Answer:
Part A:
[tex]n=8.85*10^{28}m^{-3}[/tex]
Part B:
[tex]Electron Mobility=3.03*10^{-3} m^2/V[/tex]
Explanation:
Part A:
To calculate the number of free electrons n we use the following formula::
n=1.5N-Au
Where N-Au is number of gold atoms per cubic meter
[tex]N-Au=\frac{Density*Avogadro Number}{atomic weight}[/tex]
[tex]Density = 19.32g/cm^3[/tex]
[tex]Avogadro Number=6.02*10^{23} atoms/mol[/tex]
[tex]Atomic weight=196.97g/mol[/tex]
So:
[tex]n=1.5*\frac{Density*Avogadro Number}{atomic weight}[/tex]
[tex]n=1.5*\frac{19.32*6.02*10^{23}}{196.97}[/tex]
[tex]n=8.85*10^{28}m^{-3}[/tex]
Part B:
[tex]Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}[/tex]
n is calculated above which is 8.85*10^{28}m^{-3}
Charge on electron=1.602*10^{-19}
Elec- Conductivity= 4.3*10^{7}
[tex]Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}[/tex]
[tex]Electron Mobility=3.03*10^{-3} m^2/V[/tex]
You have a glass ball with a radius of 2.00 mm and a density of 2500 kg/m3. You hold the ball so it is fully submerged, just below the surface, in a tall cylinder full of glycerin, and then release the ball from rest. Take the viscosity of glycerin to be 1.5 Pa s and the density of glycerin to be 1250 kg/m3. Use g = 10 N/kg = 10 m/s2. Also, note that the drag force on a ball moving through a fluid is: Fdrag = 6πηrv . (a) Note that initially the ball is at rest. Sketch (to scale) the free-body diagram of the ball just after it is released, while its velocity is negligible. (b) Calculate the magnitude of the ball’s initial acceleration. (c) Eventually, the ball reaches a terminal (constant) velocity. Sketch (to scale) the free-body diagram of the ball when it is moving at its terminal velocity. (d) Calculate the magnitude of the terminal velocity. (e) What is the magnitude of the ball’s acceleration, when the ball reaches terminal velocity? (f) Let’s say that the force of gravity acting on the ball is 4F, directed down. We can then express all the forces in terms of F.
Answer:
(a) check attachment
(b)5 m/s²
Explanation:
Given: radius = 2.00mm: density = 2500kg/m³: viscosity of glycerin = 1.5pa: decity of glycerin = 1250kg/m³: g = 10N/kg = 10m/s²: Fdrag = 6πnrv
(a) for answer check attachment.
(b) For the magnitude of the balls initial acceleration:
Initial net force(f) = mg - upthrust
= [tex]mg - (\frac{m}{p} )pg.g[/tex]
acceleration (a) = [tex]Acceleration(a)=\frac{f}{m}\\=g - (\frac{pg}{p})g\\=g(1-\frac{pg}{p} )\\=10(1-\frac{1250}{2500} )\\a=10(1-0.5)\\a=5 m/s²[/tex]
c.) fromthe force diagram in the attachment; when the ball attains terminal velocity the net force will be zero(0)
[tex]mg=6πnrv + upthrust[/tex]
d.) For the magnitude of terminal velocity:
[tex]mg=6πnrv + (\frac{m}{p})pg.g\\\\(\frac{4}{3}πr^{3} p)g=6πnrv +\frac{4}{3}πr^3pg.g\\\\V = \frac{2}{9}.\frac{(2*10^{-3})^{2}*(2500-1250)*10}{1.5}\\\\=0.79cm/s[/tex]
e.) when the ball reaches terminal velocity, the acceleration is zero (0)
A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg. The child grabs and clings to a bar that is 1.55 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 35.0 rpm to 19.0 rpm. What is the moment of inertia of the merry‑go‑round with respect to its central axis?
Answer:
72.75 kg m^2
Explanation:
initial angular velocity, ω = 35 rpm
final angular velocity, ω' = 19 rpm
mass of child, m = 15.5 kg
distance from the centre, d = 1.55 m
Let the moment of inertia of the merry go round is I.
Use the concept of conservation of angular momentum
I ω = I' ω'
where I' be the moment of inertia of merry go round and child
I x 35 = ( I + md^2) ω'
I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19
35 I = 19 I + 1164
16 I = 1164
I = 72.75 kg m^2
Thus, the moment of inertia of the merry go round is 72.75 kg m^2.
For the following reaction, 24.8 grams of diphosphorus pentoxide are allowed to react with 13.2 grams of water . diphosphorus pentoxide(s) + water(l) phosphoric acid(aq) What is the maximum mass of phosphoric acid that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams
The maximum mass of phosphoric acid that can be formed is 34.2 g.
The formula for the limiting reagent is [tex]\bold{P_4O_1_0}[/tex]
The excess reagent that remains after the reaction is complete is 3.76 g.
What is phosphoric acid?Phosphoric acid (H3PO4), also known as orthophosphoric acid, is the most important oxygen acid of phosphorus.
It is a crystalline solid and is used to generate fertilizer phosphate salts.
Given,
The balanced chemical equation is
[tex]P_4O_1_0 + 6 H_2O =4 H_3PO_4[/tex]
The molar mass of the following
[tex]\bold{P_4O_1_0}[/tex] = 284 g
[tex]\bold{H_2O}[/tex] = 16 g
[tex]\bold{H_3PO_4}[/tex] = 98 g
Calculating the number of moles
[tex]\dfrac{\bold{P_4O_1_0}}{284 g} = 0.0873\; mol/P_4O_1_0[/tex]
[tex]\dfrac{\bold{13.2\;g\;H_2O}}{18 g} = 0.733\; mol/P_4O_1_0[/tex]
From equation, 1 mol of [tex]\bold{P_4O_1_0}[/tex] reacts with 6 mol of water, then for 0.0873 mol of [tex]\bold{P_4O_1_0}[/tex] will react
[tex]0.0873\; mol\; P_4O_1_0 \times 6 mol H_2O/ 1 molP_4O_1_0 = 0.524\; mol\; H_2O[/tex]
The number of moles of water remaining are:
0.733 mol - 0.524 mol = 0.209 mol
Convert the amount of water into mass unit
0.209 mol · 18 g / mol = 3.76 g
The number of moles of water remaining is 3.76 g
Assuming that the reaction has a yield of 100%
Since 1 mol of [tex]\bold{P_4O_1_0}[/tex] produces 4 mol of phosphoric acid, 0.0873 mol [tex]\bold{P_4O_1_0}[/tex] will produce:
[tex]0.0873\; mol\; P_4O_1_0 \times 4 mol H_3PO_4/ molP_4O_1_0 = 0.349\; mol\; of\;phosphoric\;acid[/tex]
Then, the maximum mass of phosphoric acid that can be formed is
[tex]0.349\; mol\; H_3PO_4 \times 98\;g/ mol = 34.2\; g[/tex]
Thus, The maximum mass of phosphoric acid is 34.2 g, The formula for the limiting reagent is [tex]\bold{P_4O_1_0}[/tex], and the excess reagent that remains after the reaction is complete is 3.76 g.
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Final answer:
The maximum mass of phosphoric acid that can be formed is 35.6 grams, with diphosphorus pentoxide being the limiting reagent and 8.77 grams of water remaining as excess.
Explanation:
The maximum mass of phosphoric acid that can be formed is 35.6 grams. The formula for the limiting reagent is diphosphorus pentoxide (P4O6). After the reaction is complete, 8.77 grams of water will remain as the excess reagent.
Usually, to find the maximum mass of the product formed (phosphoric acid), and the mass of the excess reagent that remains after the reaction, it involves determining the limiting reagent by calculating the moles of reactants and using stoichiometry. The limiting reagent is the reactant that will be completely used up first during the chemical reaction, limiting the amount of product formed.
A large aquarium of height 6 m is filled with fresh water to a depth of D = 1.50 m. One wall of the aquarium consists of thick plastic with horizontal length w = 8.40 m. By how much does the total force on that wall increase if the aquarium is next filled to a depth of D = 4.30 m? (Note: use g = 9.81 m/s2 and rho = 998 kg/m3.)
Answer
given,
height of aquarium = 6 m
Depth of fresh water = D = 1.50 m
horizontal length of the aquarium(w) = 8.40 m
total force increased when liquid is filled to depth = 4.30 m
g = 9.81 m/s²
ρ = 998 Kg/m³
force in the aquarium.
dF = PdA
[tex]F = \int PdA[/tex]
[tex]F = \int \rho\ g\ y\ (wdy)[/tex]
[tex]F = \rho\ g\ w \int ydy[/tex]
[tex]F = \rho\ g\ w\dfrac{y^2}{2}[/tex]
[tex]F = \dfrac{\rho\ g\ w\ y^2}{2}[/tex]
At D = 1.5 m
[tex]F = \dfrac{980\times 9.8\times 8.4\times 1.5^2}{2}[/tex]
F = 9.08 x 10⁴ N
At D = 4.30 m
[tex]F = \dfrac{980\times 9.8\times 8.4\times 4.3^2}{2}[/tex]
F = 7.46 x 10⁵ N
Total force on the wall increased by
ΔF = 74.6 x 10⁴ - 9.08 x 10⁴
ΔF = 65.52 x 10⁴ N
A graduated cylinder is half full of mercury and half full of water. Assume the height of the cylinder is 0.25 m and don't forget about atmospheric pressure. 1) Calculate the pressure at the bottom of a graduated cylinder. (Express your answer to two significant figures.)
To develop this problem it is necessary to apply the concepts related to the definition of absolute, manometric and atmospheric pressure.
The pressure would be defined as
[tex]P = P_0 + P_w +P_m[/tex]
Where,
[tex]P_0 =[/tex] Standard atmosphere pressure
[tex]P_W =[/tex]Pressure of Water
[tex]P_m =[/tex]Pressure of mercury
Therefore we have that
[tex]P = P_0 \rho_w g \frac{h}{2} + \rho_m g\frac{h}{2}[/tex]
Here g is the gravity, \rho the density at normal conditions and h the height of the cylinder.
Replacing with our values we have that,
[tex]P = 1.013*10^5+10^3*9.81*\frac{0.25}{2}+13.6*10^3*9.81*\frac{0.25}{2}[/tex]
[tex]P = 1.21*10^5Pa \approx 0.121Mpa[/tex]
Therefore the pressure at the Bottom of a graduated cylinder is 0.121Mpa
The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.
A graduated cylinder is half full of mercury and half full of water. If the height of the cylinder is 0.25 m, the height of each column of liquid is:
[tex]\frac{0.25m}{2} = 0.13 m[/tex]
We can calculate the pressure at the bottom of the cylinder (P) using the following expression.
[tex]P = P_{atm} + P_{Hg} + P_w[/tex]
where,
[tex]P_{atm}[/tex]: atmospheric pressure (101,325 Pa)[tex]P_{Hg}[/tex]: pressure exerted by Hg[tex]P_w[/tex]: pressure exerted by the waterWe can calculate the pressure exerted by each liquid (P) as:
[tex]P = \rho \times g \times h[/tex]
where,
ρ: density of the liquidg: gravity (9.81 m/s²)h: height of the liquid column (0.13 m)The pressure exerted by the column of Hg is:
[tex]P_{Hg} = \frac{13.6 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 17 \times 10^{3} Pa[/tex]
The pressure exerted by the column of water is:
[tex]P_{w} = \frac{1.00 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 1.3 \times 10^{3} Pa[/tex]
The pressure at the bottom of the cylinder is:
[tex]P = P_{atm} + P_{Hg} + P_w = 101325 Pa + 17 \times 10^{3} Pa + 1.3 \times 10^{3} Pa = 1.2 \times 10^{5} Pa[/tex]
The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.
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The universe is filled with photons left over from the Big Bang that today have an average energy of about 4.9 ✕10-4 (corresponding to a temperature of 2.7 K).
What is the number of available energy states per unit volume for these photons in an interval of 4 ✕10-5eV?
Answer:
The number of available energy is [tex]4.820\times10^{45}[/tex]
Explanation:
Given that,
Energy [tex]E=4.9\times10^{-4}\ J[/tex]
Temperature = 2.7 K
Energy states per unit volume [tex]dE=4\times10^{-5}\ eV[/tex]
We need to calculate the number of available energy
Using formula of energy
[tex]N=g(E)dE[/tex]
[tex]N=\dfrac{8\pi\times E^2 dE}{(hc)^3}[/tex]
Where, h = Planck constant
c = speed of light
E = energy
Put the value into the formula
[tex]N=\dfrac{8\pi\times(4.9\times10^{-4})^2\times4\times10^{-5}\times1.6\times10^{-19}}{(6.67\times10^{-34}\times3\times10^{8})^3}[/tex]
[tex]N=4.820\times10^{45}[/tex]
Hence, The number of available energy is [tex]4.820\times10^{45}[/tex]
A spring has a 12 cm length. When a 200-g mass is hung from the spring, it extends to 27 cm. The hanging mass were pulled downward a further 5 cm. What is the time-dependent function of the position, in centimeters, of the mass, assuming that the phase angle LaTeX: \phi=0ϕ=0?
LaTeX: x(t)=15\cos{(8.08\textrm{ }t)}x(t)=15cos(8.08 t), where the coefficient inside the trigonometric function has units of rad/s.
LaTeX: x(t)=5\cos{(8.08\textrm{ }t)}x(t)=5cos(8.08 t), where the coefficient inside the trigonometric function has units of rad/s.
LaTeX: x(t)=5\cos{(1.29\textrm{ }t)}x(t)=5cos(1.29 t)x(t)=5cos(1.29 t), where the coefficient inside the trigonometric function has units of rad/s.
LaTeX: x(t)=15\cos{(1.29\textrm{ }t)}x(t)=15cos(1.29 t)x(t)=15cos(1.29 t), where the coefficient inside the trigonometric function has units of rad/s.
Answer:
The equation of the time-dependent function of the position is [tex]x(t)=5\cos(8.08t)[/tex]
(b) is correct option.
Explanation:
Given that,
Length = 12 cm
Mass = 200 g
Extend distance = 27 cm
Distance = 5 cm
Phase angle =0°
We need to calculate the spring constant
Using formula of restoring force
[tex]F=kx[/tex]
[tex]mg=kx[/tex]
[tex]k=\dfrac{mg}{x}[/tex]
[tex]k=\dfrac{200\times10^{-3}\times9.8}{(27-12)\times10^{2}}[/tex]
[tex]k=13.06\ N/m[/tex]
We need to calculate the time period
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Put the value into the formula
[tex]T=2\pi\sqrt{\dfrac{0.2}{13.6}}[/tex]
[tex]T=0.777\ sec[/tex]
At t = 0, the maximum displacement was 5 cm
So, The equation of the time-dependent function of the position
[tex]x(t)=A\cos(\omega t)[/tex]
Put the value into the formula
[tex]x(t)=5\cos(2\pi\times f\times t)[/tex]
[tex]x(t)=5\cos(2\pi\times\dfrac{1}{T}\times t)[/tex]
[tex]x(t)=5\cos(2\pi\times\dfrac{1}{0.777}\times t)[/tex]
[tex]x(t)=5\cos(8.08t)[/tex]
Hence, The equation of the time-dependent function of the position is [tex]x(t)=5\cos(8.08t)[/tex]
A bungee jumper of mass m = 65 kg jumps from a bridge.
A bungee cord with unstretched length L is attached to her leg.
After falling a distance L, the jumper is moving at speed v = 26 m/s and the bungee cord begins stretching and acts like an ideal spring with spring constant k.
Neglect air resistance and the mass of the bungee cord.
Which choice best represents the unstretched length of the bungee cord, L?A. 5.9 m
B. 26 m
C. 34 m
D. 65 m
E. None of these
To solve the problem it is necessary to apply energy conservation.
By definition we know that kinetic energy is equal to potential energy, therefore
PE = KE
[tex]mgh = \frac{1}{2}mv^2[/tex]
Where,
m = mass
g = gravitaty constat
v = velocity
h = height
Re-arrange to find h,
[tex]h=\frac{v^2}{2g}[/tex]
Replacing with our values
[tex]h=\frac{26^2}{2*9.8}[/tex]
[tex]h = 34.498\approx 34.5m[/tex]
Therefore the correct answer is C.