An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99543c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 43.0 km .
(a) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the surface of the earth?
(b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle’s frame.
(c) In the particle’s frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?
Answers
Answer:
Check attachment
The question have two distance
I decided to use the one in the question "a" in attachment and I will use the other one here
Explanation:
Given that,
Speed of particle relative to the earth is
V = 0.99543c
Where c is speed of light
c = 3 × 10^8 m/s
Particle height as detected by scientist is 43km
The initial length is 43km
Lo = 43km
Lo = 43,000m
A. Time taken for the particle to reach the earth surface?
Speed = distance / Time
Time = distance / speed
t = L / V
t = 43,000 / 0.99543c
t = 43,000 / (0.99543 × 3 × 10^8)
t = 1.4399 × 10^-4 seconds
b. Initial Lenght is given as Lo = 43km
Using length contraction formula
L = Lo√(1 — u² / c²)
L = 43√[1 — (0.99543c)² / c²]
L = 43√[1 — 0.990881c² / c²]
L = 43√[1 — 0.990881]
L = 43 × √(9.1191 × 10^-3)
L = 43 × 0.095494
L = 4.1062 km
c. Using time dilation formula
∆to = ∆t√(1 — u² / c²)
∆t is gotten from question a
∆t = 1.4399 × 10^-4 seconds
∆to = ∆t√[1 — (0.99543c)² / c²]
∆to = ∆t√[1 — 0.990881c² / c²]
∆to = ∆t√[1 — 0.990881]
∆to = ∆t × √(9.1191 × 10^-3)
∆to = 1.4399 × 10^-4 × 0.095494
∆to = 1.375 × 10^-5 seconds
To check if the time dilation agree
t = L / V
t = 4.1062 × 1000 / 0.99543c
t = 4.1062 × 1000 / 0.99543 × 3 × 10^8
t = 1.375 × 10^-5 seconds
The time dilation agreed
Related Questions
A 1.0-kg standard cart A collides with a 0.10-kg cart B. The x component of the velocity of cart A is +0.70 m/s before the collision and +0.50 m/s after the collision. Cart B is initially traveling toward cart A at 0.40 m/s , and after the collision the x component of its velocity is +1.6 m/s .
Part A: What is the x component of the change in the momentum of A?
Part B: What is the x component of the change in the momentum of B?
Part C: What is the sum of these two x components of the changes in momentum?
Answers
The x component of the change in momentum of cart A is -0.20 kg·m/s, while for cart B, it is +0.20 kg·m/s. The sum of these changes in momentum is 0 kg·m/s, showing conservation of momentum in the system.
Explanation:The student has provided information about a collision between two carts, which involves concepts of momentum and conservation of momentum from mechanics in physics.
Part A: Change in Momentum of Cart A
The x component of the change in momentum of cart A can be calculated using the formula Δp = m(v_final - v_initial), where m is mass and v is velocity. Here, Δp = 1.0 kg (0.50 m/s - 0.70 m/s) = -0.20 kg·m/s. The negative sign indicates that the momentum decreased.
Part B: Change in Momentum of Cart B
Similarly, for cart B, Δp = 0.10 kg (1.6 m/s - (-0.40 m/s)) = 0.20 kg·m/s. The positive sign signifies an increase in momentum.
Part C: Sum of Changes in Momentum
The total change in momentum for both carts is the sum of their individual momentum changes, which is -0.20 kg·m/s + 0.20 kg·m/s = 0 kg·m/s. This means that the total momentum of the system is conserved.
You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it. Imagine that you slowly pull down with a spring scale so the block is now below the equilibrium position where it was hanging at rest. The scale reading when you let go of the block is 3.0 N.
Part A
Where was the block when you let go? Assume y0 is the equilibrium position of the block and that "down" is the positive direction.
Part B
Determine the work you did stretching the spring.
Express your answer to two significant figures and include the appropriate units.
Part C
What was the energy of the spring-Earth system when you let go (assume that zero potential energy corresponds to the equilibrium position of the block)?
Express your answer to two significant figures and include the appropriate units.
Part D
How far will the block rise after you release it?
Express your answer to two significant figures and include the appropriate units.
Answers
Answer:
a) x = 0.144 m
b) W = 0.15 J
c) E = 0.14 J
d) The block will rise 0.07m after it is released
Explanation:
a) The elastic force equals the gravitational force
F = kx = mg
x = 0.07, m = 0.1 kg, g = 9.81 m/s²
0.07k = 0.1 * 9.8
k = (0.1*9.8)/0.07
k = 14 N/m
When the force, F = 3N
F = kx
3 = 14x
x = 3/14
x = 0.214 m
The position of the block = 0.214 - 0.07 = 0.144m
B) Determine the work you did stretching the spring.
Energy stored in the spring when x = 0.07
E = 0.5 kx²
E = 0.5 * 14 * 0.07²
E = 0.0343 J
Energy stored in the spring when x = 0.214
E = 0.5 kx²
E = 0.5 * 14 * 0.214²
E = 0.32 J
Potential energy lost due to gravity = mgh
PE = 0.1 * 9.81 * 0.144
PE = 0.141 J
So to calculate the work done:
0.0343 + W = 0.32 - 0.141
W = 0.15 J
c) Energy in the spring
E = 0.32 - 0.0343 - 0.15
E = 0.1357 = 0.14 J
d)
[tex]1/2 *k *0.214^{2} = 1/2 kx^{2} + mg(0.214+x)\\0.32 = 7x^{2} + 0.1*9.8(0.214+x)\\[/tex]
Solving for x, x = 0.07 m
The block will rise 0.07m after it is released
A student builds and calibrates an accelerometer and uses it to determine the speed of her car around a certain unbanked highway curve. The accelerometer is a plumb bob with a protractor that she attaches to the roof of her car. A friend riding in the car with the student observes that the plumb bob hangs at an angle of 15.0° from the vertical when the car has a speed of 21.5 m/s. (a) What is the centripetal acceleration of the car rounding the curve? m/s2 (b) What is the radius of the curve? m (c) What is the speed of the car if the plumb bob deflection is 7.00° while rounding the same curve? m/s
Answers
(a) [tex]\(a_c \approx 39.8 \, \text{m/s}^2\)[/tex]
(b) [tex]\(r \approx 11.6 \, \text{m}\)[/tex]
(c) [tex]\(v_2 \approx 28.4 \, \text{m/s}\)[/tex]
(a) calculation of the centripetal acceleration [tex](\(a_c\))[/tex],
[tex]\[ a_c = \frac{g}{\sin(\theta_1)} \][/tex]
Given [tex]\(g \approx 9.8 \, \text{m/s}^2\) and \(\theta_1 = 15.0°\)[/tex], we find:
[tex]\[ a_c = \frac{9.8 \, \text{m/s}^2}{\sin(15.0°)} \][/tex]
Calculating this gives [tex]\(a_c \approx 39.8 \, \text{m/s}^2\).[/tex]
(b) The radius ((r)) of the curve can be found using the formula:
[tex]\[ r = \frac{v^2}{a_c} \][/tex]
Substituting [tex]\(v = 21.5 \, \text{m/s}\) and \(a_c \approx 39.8 \, \text{m/s}^2\),[/tex]we get:
[tex]\[ r = \frac{(21.5 \, \text{m/s})^2}{39.8 \, \text{m/s}^2} \][/tex]
This calculation yields [tex]\(r \approx 11.6 \, \text{m}\).[/tex]
(c) For a new deflection angle [tex]\(\theta_2 = 7.00°\)[/tex], we find the new centripetal acceleration [tex](\(a_{c2}\))[/tex] using:
[tex]\[ a_{c2} = \frac{g}{\sin(\theta_2)} \][/tex]
Substituting [tex]\(g \approx 9.8 \, \text{m/s}^2\) and \(\theta_2 = 7.00°\)[/tex], we get:
[tex]\[ a_{c2} = \frac{9.8 \, \text{m/s}^2}{\sin(7.00°)} \][/tex]
Calculating this gives [tex]\(a_{c2} \approx 84.7 \, \text{m/s}^2\).[/tex]
Then, using [tex]\(a_{c2} = \frac{v_2^2}{r}\)[/tex], we find the new speed [tex](\(v_2\)):[/tex]
[tex]\[ v_2 = \sqrt{a_{c2} \cdot r} \][/tex]
Substituting [tex]\(a_{c2} \approx 84.7 \, \text{m/s}^2\) and \(r \approx 11.6 \, \text{m}\), we get \(v_2 \approx 28.4 \, \text{m/s}\).[/tex]
A projectile with a mass of 0.100 kg is fired at and embeds itself in a stationary target with a mass of 2.57 kg. The target and embedded projectile fly off after being struck. Determine the percent of the projectile's incident kinetic energy carried off by the target-projectile combination.
Answers
Answer:
3.74 %
Explanation:
Given,
Mass of the fired projectile, m = 0.1 Kg
Mass of the stationary target, M = 2.57 Kg
Speed of the particle after collision
Using conservation of momentum
m v = (m + M) V
0.1 x v = (0.1 + 2.57) V
V = 0.0374 v
Initial KE = [tex]\dfrac{1}{2}mv^2 = \dfrac{1}{2}\times 0.1\times v^2[/tex]
Final KE = [tex]\dfrac{1}{2} (M + m) V^2 = \dfrac{1}{2}\times 2.67\times (0.0374 v)^2[/tex]
Now,
the percent of the projectile's KE carried out by target
[tex]=\dfrac{\dfrac{1}{2}\times 2.67\times (0.0374 v)^2}{ \dfrac{1}{2}\times 0.1\times v^2}\times 100[/tex]
[tex]= 3.74\ \%[/tex]
A pilot flies in a straight path for 1 hour 30 minutes. She then makes a course correction, heading 45 degrees to the right of her original course, and flies 2 hours 15 minutes in the new direction. If she maintains a constant speed of 345 mi/h, how far is she from her starting position? Give your answer to the nearest mile.
Answers
Answer:
1199 miles
Explanation:
1 hour 30 minutes = 1 + 30/60 = 1.5 hours
2 hours 15 minutes = 2 + 15/60 = 2.25 hours
The distance she flew in the 1st segment is:
1.5*345 = 517.5 miles
The distance she flew in the 2nd segment is:
2.25 * 345 = 776.25 miles
Since the 2nd segment is 45 degree with respect to the 1st segment, this means that she has flown
776.25 * cos(45) = 549 miles in-line with the 1st segment and
776.25* sin(45) = 549 miles perpendicular to the 1st segment:
So the distance from the end to her starting position is
[tex]\sqrt{(517.5 + 549)^2 + 549^2} = 1199 miles[/tex]
Rotational dynamics about a fixed axis: A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center. Starting with an angular velocity of 2.40 rev/s, it stops after turning through 18.2 rev with uniform acceleration. The net torque acting on this sphere as it is slowing down is closest to:A) 0.149 N m. B) 0.0620N m. C) 0.00593 N m. D) 0.0372 N m. E) 0.0466 N·m
Answers
Answer:
D) 0.0372 N m
Explanation:
r = 45/2 cm = 22.5 cm = 0.225 m
As 1 revolution = 2π rad we can convert to radian unit
2.4 rev/s = 2.4 * 2π = 15.1 rad/s
18.2 rev = 18.2 * 2π = 114.35 rad
We can calculate the angular (de)acceleration using the following equation of motion
[tex]-\omega^2 = 2\alpha \theta [/tex]
[tex]- 15.1^2 = 2*\alpha * 114.35[/tex]
[tex]\alpha = \frac{-15.1^2}{2*114.35} = -0.994 rad/s^2[/tex]
The moment of inertia of the solid uniform sphere is
[tex]2mr^2/5 = 2*1.85*0.225^2/5 = 0.0375 kgm^2[/tex]
The net torque acting on this according to Newton's 2nd law is
[tex]T = I\alpha = 0.0375 * 0.994 = 0.0372 Nm[/tex]
Answer:
(D) The net torque acting on this sphere as it is slowing down is closest to 0.0372 N.m
Explanation:
Given;
mass of the solid sphere, m = 1.85 kg
radius of the sphere, r = ¹/₂ of diameter = 22.5 cm
initial angular velocity, ω = 2.40 rev/s = 15.08 rad/s
angular revolution, θ = 18.2 rev = 114.37 rad
Torque on the sphere, τ = Iα
Where;
I is moment of inertia
α is angular acceleration
Angular acceleration is calculated as;
[tex]\omega_f^2 = \omega_i^2 +2 \alpha \theta\\\\0 = 15.08^2 + (2*114.37)\alpha\\\\\alpha = \frac{-15.08^2}{(2*114.37)} = -0.994 \ rad/s^2\\\\\alpha = 0.994 \ rad/s^2 \ (in \ opposite \ direction)[/tex]
moment of inertia of solid sphere, I = ²/₅mr²
= ²/₅(1.85)(0.225)²
= 0.03746 kg.m²
Finally, the net torque on the sphere is calculated as;
τ = Iα
τ = 0.03746 x 0.994
τ = 0.0372 N.m
Therefore, the net torque acting on this sphere as it is slowing down is closest to 0.0372 N.m
Find the corresponding initial value problem of the following spring-mass systems, that if solved, would give the position u of the mass at any time t. Use 9.8 m/s 2 for the acceleration due to gravity. DO NOT SOLVE THE IVP. A mass of 60 kg stretches a spring 19.6 m. The mass is acted on by an external force of 6 sin(4t) N and moves in a medium that imparts a viscous force of 8 N when the speed of the mass is 16 cm/s. The mass is pushed up 5 cm and set in motion an initial downward velocity of 2 cm/s. Write the IVP so that u would be in meters if solved.
Answers
Answer:
Explanation:
The picture attached shows the full explanation
A string is wrapped tightly around a fixed pulley that has a moment of inertia of 0.0352 kgm2 and a radius of 12.5 cm. A mass of 423 grams is attached to the free end of the string. With the string vertical and taut, the mass is gently released so it can descend under the influence of gravity. As the mass descends, the string unwinds and causes the pulley to rotate, but does not slip on the pulley. What is the speed (in m/s) of the mass after it has fallen through 1.25 m
Answers
Answer:
the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s
Explanation:
Given that :
Mass attached to the free end of the string, m = 423 g = 0.423 kg
Moment of inertia of pulley, I = 0.0352 kg m²
Radius of the pulley, r = 12.5 cm = 0.125 meters
Depth of fallen mass, h = 1.25 m
Acceleration due to gravity, g = 9.8 m/s²
Change in potential energy = mgh
= 0.423 × 9.8 × 1.25
=5.18175 J
From the question, we understand that the change in potential energy is used to raise and increase the kinetic energy of hanging mass and the rotational kinetic energy of pulley.
As such;
[tex]5.18175 \ J= \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2[/tex]
where;
[tex]\omega[/tex] is the angular velocity of the pulley
v is the velocity of the mass after falling 1.25 m
where:
[tex]v = r \omega[/tex]
[tex]\omega = \frac{v}{r}[/tex]
replacing [tex]\omega = \frac{v}{r}[/tex] into above equation; we have:
[tex]5.18175 \ J= \frac{1}{2}mv^2 + \frac{1}{2} I( \frac{v}{r})^2[/tex]
[tex]5.18175= (\frac{1}{2} m + \frac{1}{2}* (\frac{I}{r^2})v^2 \\ \\ v^2 = \frac{5.18175}{0.5*0.423+0.5*\frac{0.0352}{0.1252}} \\ \\ v^2 = 3.873047 \\ \\ v = 1.968 \ m/s[/tex]
Thus, the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s
the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s
SpeedWhat all information we have?
Mass attached to the free end of the string, m = 423 g = 0.423 kg
Moment of inertia of pulley, I = 0.0352 kg m²
Radius of the pulley, r = 12.5 cm = 0.125 meters
Depth of fallen mass, h = 1.25 m
Acceleration due to gravity, g = 9.8 m/s²
Change in potential energyChange in potential energy = mgh
Change in potential energy = 0.423 × 9.8 × 1.25
Change in potential energy =5.18175 J
As such:
5.18175 J= mv²+1w²
where;
w is the angular velocity of the pulley
v is the velocity of the mass after falling 1.25 m
v=rw
w=v/r
Replacing into above equation;
5.18175 J=mv² + 1/2 (w/r²) ²
5.18175 = (m + /* (4) v²
v² = 3.873047
v² =1.968 m/s
Thus, the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s.
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German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity ( Δ v ) Δ x ≥ h 4 π m Δ v where h is Planck's constant and m is the mass of the object. The mass of an electron is 9.11 × 10 − 31 kg. What is the uncertainty in the position of an electron moving at 2.00 × 10 6 m/s with an uncertainty of Δ v = 0.01 × 10 6 m/s ?
Answers
According to the information given, the Heisenberg uncertainty principle would be given by the relationship
[tex]\Delta x \Delta v \geq \frac{h}{4\pi m}[/tex]
Here,
h = Planck's constant
[tex]\Delta v[/tex] = Uncertainty in velocity of object
[tex]\Delta x[/tex] = Uncertainty in position of object
m = Mass of object
Rearranging to find the position
[tex]\Delta x \geq \frac{h}{4\pi m\Delta v}[/tex]
Replacing with our values we have,
[tex]\Delta x \geq \frac{6.625*10^{-34}m^2\cdot kg/s}{4\pi (9.1*10^{-31}kg)(0.01*10^6m/s)}[/tex]
[tex]\Delta x \geq 5.79*10^{-9}m[/tex]
Therefore the uncertainty in position of electron is [tex]5.79*10^{-9}m[/tex]
In 1958, Meselson and Stahl conducted an experiment to determine which of the three proposed methods of DNA replication was correct. Identify the three proposed models for DNA replication. Conservative Semiconservative Dispersive Answer Bank The Meselson and Stahl experiment starts with E. coli containing 15 N / 15 N labeled DNA grown in 14 N media. Which result did Meselson and Stahl observe by sedimentation equilibrium centrifugation to provide strong evidence for the semiconservative model of DNA replication? Both the first and second generation have both 15 N / 15 N DNA and 14 N / 14 N DNA. No hybrid 15 N / 14 N DNA was observed. The first generation has hybrid 15 N / 14 N DNA and the second generation has both hybrid 15 N / 14 N DNA and 14 N / 14 N DNA. No 15 N / 15 N DNA was observed. The first generation has hybrid 15 N / 14 N DNA and the second generation has hybrid 15 N / 14 N DNA. No 15 N / 15 N DNA nor 14 N / 14 N DNA was observed.
Answers
Answer:
Explanation:
The original has hybrid 15N/14N DNA, and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed. In this experiment:
Nitrogen is a significant component of DNA. 14N is the most bounteous isotope of nitrogen, however, DNA with the heavier yet non-radioactive and 15N isotope is likewise practical.
E. coli was developed for several generations in a medium containing NH4Cl with 15N. When DNA is extracted from these cells and centrifuged on a salt density gradient, the DNA separates at which its density equals to the salt arrangement. The DNA of the cells developed in 15N medium had a higher density than cells developed in typical 14N medium. After that, E. coli cells with just 15N in their DNA were transferred to a 14N medium.
DNA was removed and compared to pure 14N DNA and 15N DNA. Immediately after only one replication, the DNA was found to have an intermediate density. Since conservative replication would result in equal measures of DNA of the higher and lower densities yet no DNA of an intermediate density, conservative replication was eliminated. Moreso, this result was consistent with both semi-conservative and dispersive replication. Semi conservative replication would result in double-stranded DNA with one strand of 15N DNA, and one of 14N DNA, while dispersive replication would result in double-stranded DNA with the two strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density.
The DNA from cells after two replications had been completed and found to comprise of equal measures of DNA with two different densities, one corresponding to the intermediate density of DNA of cells developed for just a single division in 14N medium, the other corresponding to DNA from cells developed completely in 14N medium. This was inconsistent with dispersive replication, which would have resulted in a single density, lower than the intermediate density of the one-generation cells, yet at the same time higher than cells become distinctly in 14N DNA medium, as the first 15N DNA would have been part evenly among all DNA strands. The result was steady with the semi-conservative replication hypothesis. The semi conservative hypothesis calculates that each molecule after replication will contain one old and one new strand. The dispersive model suggests that each strand of each new molecule will possess a mixture of old and new DNA.
One consequence of turbulence is mixing. Different layers of fluid flow cross over one another very easily, and get blended together. This is another kind of "transport", which lets atoms which might have started out all in one place get uniformly mixed around. Would you expect turbulent mixing to happen most easily in:
A. waterB. motor oilC. airD. honey
Answers
Answer:
Turbulence mixing will happen mostly in air
Explanation:
When the temperature is at 30∘C, the A-36 steel pipe fits snugly between the two fuel tanks. When fuel flows through the pipe, the temperature at ends A and B rise to 130∘C and 80∘C, respectively. If the temperarture drop along the pipe is linear, determine the average normal stress developed in the pipe. Assume the walls of each tank acts as a spring, each having a stiffness of k=900 MN/m.
Answers
In a system where an A-36 steel pipe is snugly fit between two fuel tanks, the rise in fuel temperature will cause thermal stresses in the pipe due to the restraint from the fuel tanks. This is related to the thermal properties of the material and the rate of change due to temperature rising, from which the average normal stress can be calculated.
Explanation:Determining the average normal stress developed in an A-36 steel pipe when fuel flows through it at varying temperatures requires knowledge of thermal expansion and associated stress in materials. This is related to thermal properties and the rate of change due to temperature rise.
When temperature rises along the pipe such as mentioned, from room temperature to 130∘C and 80∘C, the steel pipe expands. However, the pipe is restrained by the fuel tanks acting as springs, leading to development of stress within the pipe. The average normal stress can be calculated by dividing the force exerted by the expansion (or contraction) by the cross-sectional area of the pipe:
F/A = σ
Where, F is the force and A is the cross-sectional area of the pipe. The force can be obtained from Hooke's law for springs (F = k Δx), where k is the stiffness of the tank walls acting as springs, and Δx is the change in length due to thermal expansion.
The average normal stress is a measure of the extent to which the pipe is going through physical changes due to thermal variations.
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An electron follows a helical path in a uniform magnetic field given by:B =(20i^−50j^−30k^)mTAt time t = 0, the electron's velocity is given by:⃗v=(40i^−30j^+50k^)m/sa. What is the angle ϕ between v and B. The electron's velocity changes with time. Do b. its speed c. the angled. What is the radius of the helical path?
Answers
Answer:
a) 1.38°
b) 7.53*10^11 m/s/s
c) 6.52*10^-9m
Explanation:
a) to find the angle you can use the dot product between two vectors:
[tex]\vec{v}\cdot\vec{B}=vBcos\theta\\\\\theta=cos^{1}(\frac{\vec{v}\cdot\vec{B}}{vB})[/tex]
v: velocity of the electron
B: magnetic field
By calculating the norm of the vectors and the dot product and by replacing you obtain:
[tex]B=\sqrt{(20)^2+(50)^2+(30)^2}=61.64mT\\\\v=\sqrt{(40)^2+(30)^2+(50)^2}=70.71m/s\\\\\vec{v}\cdot\vec{B}=[(20)(40)+(50)(30)-(30)(50)]mTm/s=800mTm/s\\\\\theta=cos^{-1}(\frac{800*10^{-3}Tm/s}{(70.71m/s)(61.64*10^{-3}T)})=cos^{-1}(0.183)=1.38\°[/tex]
the angle between v and B vectors is 1.38°
b) the change in the speed of the electron can be calculated by the change in the momentum in the following way:
[tex]\frac{dp}{dt}=F_e=qvBsin\theta\\\\\frac{dp}{dt}=(1.6*10^{-19}C)(70.71m/s)(61.64*10^{-3}T)(sin(1.38\°))=6.85*10^{-19}N[/tex]
due to the mass of the electron is a constant you have:
[tex]\frac{dp}{dt}=\frac{mdv}{dt}=6.85*10^{-19}N\\\\\frac{dv}{dt}=\frac{6.85*10^{-19}N}{9.1*10^{-31}kg}=7.53*10^{11}(m/s)/s[/tex]
the change in the speed is 7.53*10^{11}m/s/s
c) the radius of the helical path is given by:
[tex]r=\frac{m_ev}{qB}=\frac{(9.1*10^{-31}kg)(70.71m/s)}{(1.6*10^{-19}C)(61.64*10^{-3}T)}=6.52*10^{-9}m[/tex]
the radius is 6.52*10^{-9}m
Magnetic fields and electric fields are identical in that they both-
Answers
Answer:
Similarities between magnetic fields and electric fields: Electric fields are produced by two kinds of charges, positive and negative. Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges). Like poles repel; unlike poles attract.
they attract each other
Explanation:
The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horizontally from north to south. At the electron's latitude the vertical component of the Earth's magnetic field points down with a magnitude of 42.3 μT. What is the direction of the force on the electrons due to this component of the magnetic field?
Answers
Answer:
The direction is due south
Explanation:
From the question we are told that
The energy of the electron is [tex]E = 19.0keV = 19.0 *10^3 eV[/tex]
The earths magnetic field is [tex]B = 42.3 \muT = 42.3 *10^{-6} T[/tex]
Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as
[tex]\= F = q (\= v * \=B)[/tex]
On the first uploaded image is an illustration of the movement of the electron
Looking at the diagram we can see that in terms of direction the magnetic force is
[tex]\= F =q(\=v * \= B)= -( -\r i * - \r k)[/tex]
[tex]= -(- (\r i * \r k))[/tex]
generally i cross k = -j
so the equation above becomes
[tex]\= F = -(-(- \r j))[/tex]
[tex]= - \r j[/tex]
This show that the direction is towards the south
A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\text{m}})x - (4.64 ~\frac{\text{rad}}{\text{sec}})t +(1.33 ~\text{rad}) \big)y=(5.26 m)⋅sin((1.65 m rad )x−(4.64 sec rad )t+(1.33 rad)) How much time will it take for a peak on this traveling wave to propagate a distance of 5.00 meters along the length of the string?
Answers
Answer:
t = 1.77 s
Explanation:
The equation of a traveling wave is
y = A sin [2π (x /λ -t /T)]
where A is the oscillation amplitude, λ the wavelength and T the period
the speed of the wave is constant and is given by
v = λ f
Where the frequency and period are related
f = 1 / T
we substitute
v = λ / T
let's develop the initial equation
y = A sin [(2π / λ) x - (2π / T) t +Ф]
where Ф is a phase constant given by the initial conditions
the equation given in the problem is
y = 5.26 sin (1.65 x - 4.64 t + 1.33)
if we compare the terms of the two equations
2π /λ = 1.65
λ = 2π / 1.65
λ = 3.81 m
2π / T = 4.64
T = 2π / 4.64
T = 1.35 s
we seek the speed of the wave
v = 3.81 / 1.35
v = 2.82 m / s
Since this speed is constant, we use the uniformly moving ratios
v = d / t
t = d / v
t = 5 / 2.82
t = 1.77 s
Ideal incompressible water is flowing in a drainage channel of rectangular cross-section. At one point, the width of the channel is 12 m, the depth of water is 6.0 m, and the speed of the flow is 2.5 m/s. At a point downstream, the width has narrowed to 9.0 m, and the depth of water is 8.0 m. What is the speed of the flow at the second point?a. 3.0 m/s b. 2.0 m/s c. 4.0 m/s d. 2.5 m/s e. 3.5 m/s
Answers
Answer:
(D) The speed of the flow at the second point is 2.5 m/s
Explanation:
Given;
at upstream;
the width of the channel, w = 12 m
the depth of water, d = 6.0 m
the speed of the flow, v = 2.5 m/s
at downstream;
the width of the channel, w = 9 m
the depth of water, d = 8.0 m
the speed of the flow, v = ?
Volumetric flow rate for an Ideal incompressible water is given as;
Q = Av
where;
A is the area of the channel of flow
v is velocity of flow
For a constant volumetric flow rate;
A₁v₁ = A₂v₂
A₁ = area of rectangle = L x d = 12 x 6 = 72 m²
A₂ = area of rectangle = L x d = 9 x 8 = 72 m²
(72 x 2.5) = 72v₂
v₂ = 2.5 m/s
Therefore, the speed of the flow at the second point is 2.5 m/s
Consider the steel spring in the illustration.
(a) Find the pitch, solid height, and number of active turns.
(b) Find the spring rate. Assume the material is A227 HD steel.
(c) Find the force Fs required to close the spring solid.
(d) Find the shear stress in the spring due to the force Fs.
Answers
Answer:
Explanation:
find the answer below
8. In the procedure for measuring the frequency of oscillation, we are instructed to pull downward on the hanging mass about 10 cm. If we performed the experiment a second time and deliberately pulled down on the hanging mass by 15 cm, would our period change? Justify your answer. (5 points)
Answers
Answer:
No
Explanation:
The frequency of oscillation of spring mass system is independent of its amplitude.
The frequency of oscillation will not change because it is not a function of amplitude.
What is the spring-mass system?In a real spring-mass system, the spring has a non-negligible mass m. Since not all of the spring's length moves at the same velocity v as the suspended mass M,
To determine the frequency of oscillation, the effective mass of the spring is defined as the mass that needs to be added to M to correctly predict the behaviour of the system.
For vertical springs, however, we need to remember that gravity stretches or compresses the spring beyond its natural length to the equilibrium position. After we find the displaced position
Thus the frequency of oscillation will not change because it is not a function of amplitude.
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In an alcohol-in-glass thermometer, the alcohol column has length 12.66 cm at 0.0 ∘C and length 22.49 cm at 100.0 ∘C. Part A What is the temperature if the column has length 18.77 cm ? Express your answer using four significant figures. T = nothing ∘C Request Answer Part B What is the temperature if the column has length 14.23 cm ? Express your answer using four significant figures.
Answers
Answer:
[tex]62.1566632757\ ^{\circ}C[/tex]
[tex]15.9715157681\ ^{\circ}C[/tex]
Explanation:
[tex]\Delta T[/tex] = Change in termperature
[tex]\Delta L[/tex] = Change in length
We have the relation
[tex]\dfrac{\Delta L}{\Delta T}=\dfrac{22.49-12.66}{100-0}=\dfrac{18.77-12.66}{t-0}\\\Rightarrow t=\dfrac{18.77-12.66}{0.0983}\\\Rightarrow t=62.1566632757\ ^{\circ}C[/tex]
The temperature is [tex]62.1566632757\ ^{\circ}C[/tex]
[tex]\dfrac{\Delta L}{\Delta T}=\dfrac{22.49-12.66}{100-0}=\dfrac{14.23-12.66}{t-0}\\\Rightarrow t=\dfrac{14.23-12.66}{0.0983}\\\Rightarrow t=15.9715157681\ ^{\circ}C[/tex]
The temperature is [tex]15.9715157681\ ^{\circ}C[/tex]
(A)
According to the question,
The change in length will be:
= [tex]l_2-l_1[/tex]
= [tex]22.49-12.66[/tex]
= [tex]9.83 \ cm[/tex]
The change per degree will be:
= [tex]\frac{Change \ in \ length}{Temperature}[/tex]
= [tex]\frac{9.83}{100}[/tex]
= [tex]0.0983 \ cm/deg[/tex]
Now,
The change in length,
= [tex]18.77-12.66[/tex]
= [tex]6.11 \ cm[/tex]
hence,
The temperature,
= [tex]\frac{6.11}{0.0983}[/tex]
= [tex]62.2^{\circ} C[/tex]
(B)
The change in length,
= [tex]14.23-12.66[/tex]
= [tex]1.57 \ cm[/tex]
hence,
The temperature will be:
= [tex]\frac{1.57}{0.0983}[/tex]
= [tex]15.98^{\circ} C[/tex]
Thus the above answers are correct.
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A standard door into a house rotates about a vertical axis through one side, as defined by the door's hinges. A uniform magnetic field is parallel to the ground and perpendicular to the axis. Through what angle must the door rotate so that the magnetic flux that passes through it decreases from its maximum value to 1/4 of its maximum value?
Answers
Answer:
75.5degrees
Explanation:
Magnitude of magnetic flux= BA
If rotated through an angle= BAcos theta
= (0.25)BA=BAcos theta
= costheta= 0.25
= theta= cos^-1 0.25
=75.5degrees
Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post. Throcky holds the other end loosely in his hand, so that the speed of waves on the clothesline is a relatively slow 0.740m/s . He finds several frequencies at which he can oscillate his end of the clothesline so that a light clothespin 43.0cm from the post doesn't move. What are these frequencies?
Answers
Answer:
Frequencies are 0.86Hz
Explanation:
Given v=0.74m/s
d= 0.43m
Lambda/2=0.43m
Thus lambda= 0.86m
And F= v/lambda
0.74/0.86= 0.86Hz
Answer:
The obsrved frquencies are [tex]f= n (0.86) Hz[/tex] n = 1,2,3,...,n.
Explanation:
From the question we are told that
The speed of the wave is v = 0.740 m/s
The distance from the post is [tex]d = 43.0cm = \frac{43}{10} = 0.43m[/tex]
Generally frequency is mathematically represented as
[tex]f = \frac{v}{\lambda }[/tex]
so for the first frequency which the the fundamental frequency(first harmonic frequency) the clothesline(string) would form only one loop and hence the length between the vertical post and throcky hands is mathematically represented as
[tex]L = \frac{\lambda}{2}[/tex]
=> [tex]\lambda = 2L[/tex]
So
[tex]f_f =f_1= \frac{v}{2L}[/tex]
Substituting values
[tex]f_f = f_1 = \frac{0.740}{2* 0.43}[/tex]
[tex]= 0.860 Hz[/tex]
For the second harmonic frequency i.e is when the clothesline(string) forms two loops the length is mathematically represented as
[tex]L = \lambda[/tex]
So the second harmonic frequency is
[tex]f_2 = \frac{v}{L}[/tex]
[tex]f_2 = 2 * \frac{v}{2L}[/tex]
[tex]f_2 =\frac{0.740}{0.430}[/tex]
[tex]= 1.72Hz[/tex]
For the third harmonic frequency i.e when the clothesline(string) forms three loops the length is mathematically represented as
[tex]L = 3 * \frac{\lambda }{2}[/tex]
So the wavelength is
[tex]\lambda = \frac{2L}{3}[/tex]
And the third harmonic frequency is mathematically evaluated as
[tex]f_3 = \frac{v}{\frac{2L}{3} }[/tex]
[tex]= 3 * \frac{v}{2L}[/tex]
[tex]f_3=2.58Hz[/tex]
Looking at the above calculation we can conclude that the harmonic frequency of a vibrating string(clothesline) can be mathematically represented as an integer multiple of the fundamental frequency
i.e
[tex]f = n f_f[/tex]
[tex]f = n \frac{v}{2L}[/tex]
[tex]f= n (0.86) Hz[/tex]
Where n denotes integer values i.e n = 1,2,3,....,n.
Note : The length that exist between two nodes which are successive is equivalent to half of the wavelength so when one loop is form the number of nodes would be 2 and on anti-node
A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the ends. (a) What is the current in amperes in the wire? (b) What is the magnitude of the current density? (c) Calculate the resistivity of the material of which the wire is made.
Answers
Answer:
(a) Current is 2831.93 A
(b) [tex]8.40A/m^2[/tex]
(c) [tex]\rho =15.52\times 10^{-9}ohm-m[/tex]
Explanation:
Length of wire l = 3.22 m
Diameter of wire d = 7.32 mm = 0.00732 m
Cross sectional area of wire
[tex]A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2[/tex]
Resistance [tex]R=11.9mohm=11.9\times 10^{-3}ohm[/tex]
Potential difference V = 33.7 volt
(A) current is equal to
[tex]i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A[/tex]
(B) Current density is equal to
[tex]J=\frac{i}{A}[/tex]
[tex]J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2[/tex]
(c) Resistance is equal to
[tex]R=\frac{\rho l}{A}[/tex]
[tex]11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}[/tex]
[tex]\rho =15.52\times 10^{-9}ohm-m[/tex]
A simple pendulum is made my attaching a rod of negligible mass to a 2.0 kg pendulum bob at the end. It is observed that on Earth, the period of small-angle oscillations is 1.0 second. It is also observed that on Planet X this same pendulum has a period of 1.8 seconds. How much does the pendulum bob weigh on Planet X
Answers
Final answer:
The weight of the pendulum bob on Planet X remains 2.0 kg, the same as on Earth, because the mass of the pendulum bob does not affect the period of a simple pendulum. The difference in periods indicates a change in gravitational strength, not mass.
Explanation:
The question revolves around understanding how the period of oscillation of a simple pendulum changes with gravity on different planets. Since the mass of the pendulum bob does not affect the period of a simple pendulum, which depends only on the length of the pendulum and the gravitational acceleration (g), the weight of the pendulum bob on Planet X would still be 2.0 kg, the same as on Earth. However, the difference in periods between Earth and Planet X indicates a difference in gravitational acceleration, implying that g on Planet X is weaker than on Earth. The formula for the period (T) of a simple pendulum is T = 2π√(l/g), where l is the length of the pendulum and g is the gravitational acceleration. Since the mass of the pendulum bob does not factor into this equation, the weight of the pendulum bob on Planet X remains the same, but its apparent weight will change according to the planet's gravitational pull.
Point charges 1 mC and −2 mC are located at (3, 2, −1) and (−1, −1, 4), respectively. Calculate the electric force on a 10 nC charge located at (0, 3, 1) and the electric field intensity at that point.
Answers
Answer:
See attached handwritten document for answer
Explanation:
Answer:
Explanation:
a) You can compute the force by using the expression:
[tex]F=k\frac{q_1q_2}{[(x-x_o)^2+(y-y_o)^2+(z-z_o)^2]^{\frac{1}{2}}}[/tex]
where k=8.98*10^9Nm^2/C^2 and q1, q2 are the charges. By replacing for the forces you obtain:
[tex]F_T=k[\frac{(1*10^{-3}C)(10*10^{-9}C)}{[(3-0)^2+(2-3)^2+(-1-1)^2]}}][(3-0)\hat{i}+(2-3)\hat{j}+(-1-1)\hat{k}]\\\\ \ \ \ \ +k[\frac{(-2*10^{-3}C)(10*10^{-9}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}}][(-1-0)\hat{i}+(-1-3)\hat{j}+(4-1)\hat{k}]\\\\F_T=6.41*10^{-3}N[3i-j-2k]-6.9*10^{-3}N[-1i-4j+3k]\\\\=0.026N\hat{i}-0.034N\hat{j}-0.033\hat{k}[/tex]
b)
[tex]E=k[\frac{1*10^{-3}C}{[(3-0)^2+(2-3)^2+(-1-1)^2]}]+k[\frac{(-2*10^{-3}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}]\\\\E=641428.5N/C+690769.23N/C=1332197.73N/C[/tex]
[tex]E=k[\frac{1*10^{-3}C}{[(3-0)^2+(2-3)^2+(-1-1)^2]}][3i-2j-2k]+\\\\k[\frac{(-2*10^{-3}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}][-1i-4j+3k]\\\\E=641428.5N/C[3i-2j-2k]-690769.23N/C[-1i-4j+3k]=2615054.7i+1480219.92j-4973626.23k\\\\|E|=\sqrt{(E_x)^2+(E_y)^2+(E_z)^2}=5810896.56N/C[/tex]
where we you have used that E=kq/r^2
A soft-drink manufacturer purchases aluminum cans from an outside vendor. A random sample of 70 cans is selected from a large shipment, and each is tested for strength by applying an increasing load to the side of the can until it punctures. Of the 70 cans, 53 meet the specification for puncture resistance.
Answers
Answer:
lol..WHAT IS THE QUESTION?!
Explanation:
How does the body maintain homeostasis when its inner temperature becomes higher than the normal body
temperature?
by shivering
by sweating
by producing goose bumps
Answers
Answer:
by sweating
Explanation:
During a track and field event, a metal javelin (not infinitesimally thin) is thrown due east and parallel to the ground. At the field where the event took place, the magnetic field is running straight up and perpendicular to the ground. In which direction will the current flow in the metal object?
Answers
Answer:
There is no induced current
Explanation:
[Find the attachment]
The resistance of physiological tissues is quite variable. The resistance of the internal tissues of humans, primarily composed of salty solutions, is quite low. Here the resistance between two internal points in the body is on the order of 100 ohms. Dry skin, however, can have a very high resistance, with values ranging from thousands to hundreds of thousands of ohms. However, if skin is wet, it is far more conductive, and so even contact with small voltages can create large, dangerous currents though a human body. (For example, although there is no specific minimum current that is lethal, currents generally exceeding a couple tenths of Amps may be deadly.)
Assuming that electrocution can be prevented if currents are kept below 0.1 A, and assuming the resistance of dry skin is 100,000 ohms, what is the maximum voltage with which a person could come into contact while avoiding electrocution? (Of course, all bets are off and things become far more dangerous if this person's skin is wet, which can reduce the resistance by more than a factor of 100.)
Answers
Given Information:
Current = I = 0.1 A
Resistance = R = 100 kΩ
Required Information:
Voltage = V = ?
Answer:
Voltage = V = 1000 V
Step-by-step explanation:
We know that electrocution depends upon the amount of current flowing through the body and the voltage across the body.
V = IR
Where I is the current flowing through the body and R is the resistance of body.
If electrocution can be avoided when the current is below 0.1 A then
V = 0.1*10×10³
V = 1000 Volts
Therefore, 1000 V is the maximum voltage with which a person could come into contact while avoiding electrocution, any voltage more than 1000 V may result in fatal electrocution.
Also note that human body has very low resistance when the body is wet therefore, above calculated value would not be applicable in such case.
A solid cylinder with radius 0.160 m is mounted on a frictionless, stationary axle that lies along the cylinder axis. The cylinder is initially at rest. Then starting at t = 0 a constant horizontal force of 5.00 N is applied tangentially to the surface of the cylinder. You measure the angular displacement θ−θ0 of the cylinder as a function of the time t since the force was first applied. When you plot θ−θ0 (in radians) as a function of t2 (in s2), your data lie close to a straight line.
What is the moment of inertia of the wheel?
Answers
The moment of inertia of the solid cylinder can be found by calculating the torque and the angular acceleration of the cylinder, and then relating these quantities using the equation torque equals moment of inertia times angular acceleration.
Explanation:The moment of inertia of the solid cylinder can be found using the relationship between torque acting on the cylinder and its angular acceleration, both of which can be inferred from the given information. The torque τ on the cylinder is given by the product of the force applied and the radius of the cylinder, τ = Fr. Since this force is causing the cylinder to undergo rotational motion, it is also causing an angular acceleration α, which can be calculated from the slope of plot relating angular displacement θ and time t squared, as α is directly proportional to slope of θ vs t² plot.
By relating these variables using the equation τ = Iα, where I is the moment of inertia, we can solve for I as: I = τ/α = Fr/α. By substituting given values, we can compute specific value for moment of inertia.
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