Answer:
[tex]0.2\; {\rm T}[/tex], assuming that the speed of the electron stays the same.
Explanation:
Let [tex]v[/tex] denote the speed of this electron. Let [tex]q[/tex] denote the electric charge on this electron. Let [tex]m[/tex] denote the mass of this electron.
Since the path of this electron is a circle (not a helix,) this path would be in a plane normal to the magnetic field.
Let [tex]B[/tex] denote the strength of this magnetic field. The size of the magnetic force on this electron would be:
[tex]F = q\, v\, B[/tex].
Assuming that there is no other force on this electron. The net force on this electron would be [tex]F = q\, v\, B[/tex]. By Newton's Second Law of motion, the acceleration of this electron would be:
[tex]\begin{aligned}a &= \frac{F}{m} \\ &= \frac{q\, v\, B}{m}\end{aligned}[/tex].
On the other hand, since this electron is in a circular motion with a constant speed:
[tex]\begin{aligned} a = \frac{v^{2}}{r} \end{aligned}[/tex].
Combine the two equations to obtain a relationship between [tex]r[/tex] (radius of the path of the electron) and [tex]B[/tex] (strength of the magnetic field:)
[tex]\begin{aligned}\frac{q\, v\, B}{m} = \frac{v^{2}}{r}\end{aligned}[/tex].
Simplify to obtain:
[tex]\begin{aligned}r &= \frac{m\, v^{2}}{q\, v\, B} \\ &= \frac{m\, v}{q\, B} \\ &= \left(\frac{m\, v}{q}\right)\, \frac{1}{B}\end{aligned}[/tex].
In other words, if the speed [tex]v[/tex] of this electron stays the same, the radius [tex]r[/tex] of the path of this electron would be inversely proportional to the strength [tex]B[/tex] of the magnetic field. Doubling the radius of this path would require halving the strength of the magnetic field (to [tex]0.2\; {\rm T}[/tex].)
A book is at rest on a flat table. The book is pushed across the table, causing it to move
and eventually come to rest again. How does the potential and kinetic energy change
during this process? (1 point)
The potential and kinetic energy will stay the same unless the book speeds
up or slows down
The potential energy increases as the book is pushed across the table.
The kinetic energy stays the same.
The potential energy will stay the same as long as the book stays on the
table. The kinetic energy increases as the book is pushed across the table.
The potential and kinetic energy decreases as the book comes to rest
again
Answer:
Explanation:
As no altitude or mass changed, the potential energy remained constant.
The potential energy will stay the same as long as the book stays on the
table.
Kinetic energy went from zero at rest to a positive value while moving, and then returned to zero when the book was at rest again.
The kinetic energy increases as the book is pushed across the table.
The potential difference between the (high-voltage) inner shell of a Van de Graaff accelerator and a point where the charge is injected into the moving belt is 3.4 MV. What is the minimum power required to pull the belt when the belt transfers the charge to the case at 2.8 mC/s?
Explanation:
this is the answer to the questions
True or False. Isaac at first resisted his father's command to be the sacrifice.
Would you die if you take the car keys out of the ignition and throw them in the back seat while you coast down a mountain-side in an attempt to save fuel and look cool for the girl in said back seat.
Answer:
bro what.
Explanation:
... is this an actual physics question?
What are the two types of force
Answer:
Forces can be divided into primarily into two types of forces:
Contact Forces. Non-contact Forces.Explanation:
identify the origins of breakdown when using a spectrum analyzer
Four regions of the electromagnetic spectrum that astronomers use when observing objects in the space are the following enumerated answers.
1. First is Ultraviolet
2. Next is Infrared
3. Then the radio
4. Lastly the Visible lights.
These are the answers to the problem.
Transverse waves move particles back and forth along the same direction in which the waves travel.
True false question.
True
False
Answer:
False, I am pretty sure.
Explanation:
Transverse waves are always characterized by particle motion being perpendicular to wave motion. ... As a sound wave moves from the lips of a speaker to the ear of a listener, particles of air vibrate back and forth in the same direction and the opposite direction of energy transport.
Blaine and her sister are identical twins riding roller coasters at Kinetic Kars. They each ride the
roller coaster on their own once. Next time, they ride the roller coaster together. On which ride
do you think they have the most kinetic energy? Explain your answer using information from class
activities.
A rollercoaster accelerates from 10 m/s to 100 m/s2 for 25 seconds. What is the acceleration?
Answer:
A roller coasters accelerates from an initial velocity of of 6.0 m/s to a final velocity of 70 m/s over 4 seconds. What's the acceleration? Q. Acceleration only takes place when things speed up. Q. A drag racer accelerated from 0 m/s to 200 m/s in 5 s.
Explanation:
when do we say a curvilinear motion is accelerated?
Answer:
There always is an acceleration in a curvilinear motion, as the velocity vector changes, so always.
How is probability used in blood typing
Blood Type O AB
Probability 0.45 0.04
As a truck rounds a curve, a box in the bed of the truck slides to the side farthest from the center of the curve. This movement of the box is a result of
Answer:
inertia .
because yes
compare and contrast a transverse wave and a compressional wave Give an example for each type
Answer:
Transverse waves oscillate perpendicular to the direction of the wave (e.g. any electromagnetic wave like radiowaves, x-rays...) whilst compressional waves oscillate in the same direction of the wave (e.g. sound waves)
Explanation:
1. The acceleration equation of a certain particle is a=2t, after 4s, its velocity reaches 20m/s, then the initial velocity of the particle movement is ) ms B Sms 2.
Answer:
Change in velocity is the integral of all the differential accelerations acting over the period of acceleration.
[tex]20 = u + \int\limits^4_0 {2t} \, dt[/tex]
20 = u + t²[tex]\left \{ {{4} \atop {0}} \right.[/tex]
20 = u + 4² - 0²
20 = u + 16
u = 4 m/s
A box of mass 7.7 - kg is accelerated from rest across a floor at a rate of 2.6 m/s2 for 18.5 s. Find the net work done on the box
Answer:
Explanation:
The net work will change the kinetic energy
W = ½mv² = ½m(at)² = ½ma²t²
W = ½(7.7)2.6²(18.5²) = 8907.3985 = 89 kJ
If a force of 20kn acts on a circular rod of a diameter 10mm, calculate the stress of the rod
σ = F/A = F/ (πD²/4) = 20000 / (π0.10²/4) = 2.55 MN/m²
Q1) Assertion : a vector can be resolved into maximum two components
Reason : in projectile motion the velocity is resolved into vertical and horizontal components
a) Both A and R are correct; R is the correct explanation of A
b) Both A and R are correct; R is not correct explanation of A
c) A is correct; R is not correct
d) A is not correct; R is correct
Q2) Assertion : speed is constant in uniform circular motion
Reason : acceleration is constant in uniform circular motion
a) Both A and R are correct; R is the correct explanation of A
b) Both A and R are correct; R is not correct explanation of A
c) A is correct; R is not correct
d) A is not correct; R is correct
Answer:
Explanation:
Q1) d) A is not correct; R is correct (sort of) but is not the reason for A.
A vector may be resolved into an infinite number of components.
Breaking a projectile motion into the vertical and horizontal components is the EASIEST way to solve it because acceleration can be assumed to be constant in both directions. Zero for horizontal and g(ravity) in the vertical. However it is not the ONLY way to do it. Other ways are much more complex.
Q2) a) Both A and R are correct; R is the correct explanation of A (sort of)
A is correct. Speed is constant in uniform circular motion.
R is correct if you call CENTRIPETAL acceleration constant. It is only constant in magnitude. The direction is always changing as the vector always points toward the center of rotation.
3. How much power does an electric light installation draw at 120 V if it creates a current of 0.8 amps?
O A. 32 watts
O B. 120 watts
Q C. 48 watts
D. 96 watts
Answer:D
Explanation:
R =
V
I
=
120 volt
0.8 ampere
= 150 ohm (Ω)
P = V × I
= 120 volt × 0.8 ampere
= 96 watt (W)
An electric light installation can draw Power of 96 Watt at 120 V if it creates a current of 0.8 amps
What is electric power ?Electric power is the rate of electric energy transfer by an electric circuit per unit time . It is denoted by P and measured using the SI unit of power that is Watt (W)
since, Electric Power = voltage * current
Power = 120 V * 0.8 A
Power = 96 Watts
An electric light installation can draw Power of 96 Watt at 120 V if it creates a current of 0.8 amps
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explain how gravitational erosion changes landforms
Answer:
Gravity can pull soil, mud, and rocks down cliffs and hillsides.
Explanation:
Gravity is responsible for erosion by flowing water and glaciers. But gravity also causes erosion directly. This type of erosion and deposition is called mass wasting.
Answer:
Erosion is another geological process that creates landforms. When mechanical and chemical weathering breaks up materials on the Earth's surface, erosion can move them to new locations. For example, wind, water or ice can create a valley by removing material. This can create new landforms.Explanation:
Hope this helps !!Use the equation of motion to answer the question. Use the equation of motion to answer the question.
x=x0+v0t+12at2
An object has a starting position of x = 2 m, a starting velocity of 4.5 m/s, and no acceleration. Which option shows the final position of the object after 2 s?
The final position of the object after 2 s is 11 m.
Motion: This can be defined as the change in position of a body.
⇒ Formula:
x = x₀+v₀t+1/2(at²)........................ Equation 1⇒ Where:
x = Final position of the objectx₀ = Starting positionv₀ = Starting velocityt = timea = accelerationFrom the question,
⇒ Given:
x₀ = 4.5 m/st = 2 sx₀ = 2ma = 0 m/s²⇒ Substitute these values into equation 1
x = 2+(4.5×2)+1/2(0²×2)x = 2+9+0x = 11 mHence, The final position of the object after 2 s is 11 m
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An LED is useful because when a current passes through it, it gives out... what?
Enter your answer
An LED is useful because when a current passes through it, it gives out light.
An LED is useful because when a current passes through it, it gives out Light.
What is an LED?
LED, in full light-emitting diode, in electronics, a semiconductor device that emits infrared or visible light when charged with an electric current.
A light-emitting diode (LED) emits light by applying a forward current to the pn junction of a compound semiconductor.
When forward current is passed through the light-emitting diode, carriers (electrons and holes) move. The holes in the p-type region move to the n-type region and the electrons in the n-type region move to the p-type region. The injected carriers recombine, and the energy difference before and after recombination is released as light. The emitted light depends on the energy band gap (Eg) of the compound semiconductor.
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Which type of muscle cell can have multiple nuclei
Answer:
Skeletal Muscle cells
Skeletal muscle cells are long, cylindrical, and striated. They are multi-nucleated meaning that they have more than one nucleus. This is because they are formed from the fusion of embryonic myoblasts.
Answer:
skeletal muscle cells can have multiple nuclei
Explanation:
This is because they are formed from the fusion of embryonic myoblasts.
14. Earthworms are crucial for forming soil. As they search for food by digging tunnels,
they expose rocks and minerals to the effects of weathering. Over time, this process
creates new soil. Worms are not the only living things that help to create soil. Plants
also play a part in the weathering process. As the roots of plants grow and seek out
water and nutrients, they help to break large rock fragments into smaller ones. Have
you ever seen a plant growing in a sidewalk? As the plants grows, its roots spread into
tiny cracks in the sidewalk. These roots apply pressure to the cracks, and, over time,
the cracks become larger, ice wedging can occur more readily. As the cracks expand,
more water can flow into them. When the water freezes, it expands and presses against
the walls of the cracks, which makes the cracks larger. Over time, the weathering
caused by water, plants, and worms help to form soil. QUESTION: Ice wedging, as
described in the passage, is an example of which of the following?
A mechanical weathering
B. Oxidation
C. chemical weathering
D. Hydrolysis
Ice wedging, as described in the passage, is an example of mechanical
weathering.
Mechanical weathering is also known as physical weathering and it
involves the breaking of rock into smaller particles without causing changes
in the chemical properties.Mechanical weathering is usually carried through
physical processes such as freezing and thawing etc.
In this scenario, we were told that water freezes, expands and presses
against the walls of the crack thereby breaking into smaller parts which is a
physical process hence mechanical weathering being present.
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Consider the following force:
A horse is pulling forward on a wagon with a force of 200 N.
According to Newton's third law, what other force must be happening?
The horse is pushing backward on the wagon with a force of 200 N.
The ground is pushing backward on the horse with a force of 200 N.
The wagon is pulling backward on the horse with a force of 200 N.
Suppose a bulldozer is being hauled at 50 km/h to a construction site on the back of a flatbed truck. From which reference point would the bulldozer not appear to be moving?
Answer:
from someone watching the bulldozer from the cab of the truck or from elsewhere on the flatbed.
Explanation:
A 3.1 kg ball is dropped from the top of a 38 m tall building. What is the speed of the ball when it is halfway from the building to the ground? Round your answer to 2 decimal places.
Answer:
19.3m/s
Explanation:
Use third equation of motion
[tex]v^2-u^2=2gh[/tex]
where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
[tex]v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s[/tex]
the book has a mass of 2.5 kg. What net force must act on the book to mak it accelerate to the left at a rate of 7.0m/s2?
Answer:
17.5 N
Explanation:
Given :Mass (m) = 2.5 kgAcceleration (a) = 7.0 m/s²To calculate :Force exerted (F)Calculation :• F = ma→ F = (2.5 × 7.0) N
→ F = 25/10 × 7 N
→ F = 5/2 × 7 N
→ F = (5 × 7)/2 N
→ F = 35/2 N
→ F = 17.5 N towards left
Hence, 17.5 N of net force must act on the book to make it accelerate to the left.
Which combination of three concurrent forces acting on a body could not produce equilibrium?
1
1 N, 3N, EN
2
2 N, 2N, 2N
.
3.
3 N, 4N, EN
4.
4N, 4N, 5N
All the three concurrent forces acting on a body will not produce equilibrium.
The given parameters:
1. 1 N, 3 N and 5 N
2. 2N, 2N and 2 N
3. 3N, 4N and 5 N
4. 4N, 4N and 5 N
Concurrent forces lie on the same plane and their line of action pass through a common point.
A body under concurrent forces is in equilibrium if the resultant of the forces on the body is zero.
[tex]\Sigma F = 0\\\\F_1 + F_2 + F_3 = 0\\\\F_1 + F_ 2 = - F_3[/tex]
where;
[tex]F_3[/tex] is the equilibrant force
First set of concurrent forces;
[tex]1 \ N \ + \ 3\ N = 4 \ N\\\\F_ 3 = 5 \ N\\\\5 \ N > 4 \ N[/tex]
Second set of concurrent forces;
[tex]2 \ N \ + \ 2 \ N = 4 \ N\\\\F_ 3 = 2 \ N\\\\4 \ N > 2 \ N[/tex]
Third set of concurrent forces;
[tex]3 \ N \ + \ 4 \ N = 7 \ N\\\\F_ 3 = 5 \ N\\\\7 \ N > 5 \ N[/tex]
Fourth set of concurrent forces;
[tex]4 \ N \ + \ 4 \ N = 8 \ N\\\\F_ 3 = 5 \ N\\\\8 \ N > 5 \ N[/tex]
Thus, we can conclude that all the three concurrent forces acting on a body will not produce equilibrium.
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Cats are groomed often by their owners to remove hair. What logical inference can be made based on this statement?
Answer:
its keeps the fur straight.....im kinda confused with the question
Explanation:
As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constants. The angular speed changes from 3.70 rad/s at t = 0 to 2.00 rad/s at t = 8.60 s.
a. Use this information to determine σ and ω0.
σ = _______s−1
ωo = ______rad/s
b. Determine the magnitude of the angular acceleration at t = 3.00 s.
______rad/s2
c. Determine the number of revolutions the wheel makes in the first 2.50 s
_______rev
d. Determine the number of revolutions it makes before coming to rest.
_______rev
Hi there!
a.
We can use the initial conditions to solve for w₀.
It is given that:
[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]
We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:
[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]
Now, we can solve for sigma using the other given condition:
[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]
b.
The angular acceleration is the DERIVATIVE of the angular velocity function, so:
[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]
c.
The angular displacement is the INTEGRAL of the angular velocity function.
[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]
[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]
[tex]\theta = 8.471 rad[/tex]
Convert this to rev:
[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]
d.
We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.
[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]
Evaluate the improper integral:
[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]
Convert to rev:
[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]