A crate with a mass of M = 62.5 kg is suspended by a rope from the endpoint of a uniform boom. The boom has a mass of m = 116 kg and a length of l = 7.65 m. The midpoint of the boom is supported by another rope which is horizontal and is attached to the wall as shown in the figure.
1. The boom makes an angle of θ = 57.7° with the vertical wall. Calculate the tension in the vertical rope.
2. What is the tension in the horizontal rope?
Answers
The boom makes an angle of θ = 60.2° with the vertical wall and the tension in the horizontal rope is mathematically given as
[tex]T_1=730.85 \mathrm{~N}[/tex]
[tex]T_1'=\frac{1980.51 \mathrm{~N}}{}[/tex]
What is the tension in the vertical rope.?Generally, the equation for is mathematically given as
1) Tension in vertical rope,
[tex]T_{1}=74.5 \times 9.81\\T_1=730.85 \mathrm{~N}[/tex]
2) Tension in horizontal rope,
[tex]\sum{Mg} =0\\Mg\frac{l}{2} \sin \theta+T_{1} \frac{l}{2} \sin \theta &=T_{2} l \cos \theta \\[/tex]
[tex]T_{2} &=\frac{M g(l / 2) \sin \theta+T_{1}(l / 2) \sin \theta}{l \cos \theta} \\\\&=\frac{M g \tan \theta}{2}+\frac{T_{1} \tan \theta}{2} \\\\T_{2} &=\frac{142 \times 9.81 \tan 61.8}{2}+\frac{730.85 \times \tan 61.8}{2} \\\\=& 1299+681.51[/tex]
[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]
In conclusion, the tension in the horizontal rope is
[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]
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Related Questions
A rock of mass 200 g is attached to a 0.75 m long string and swung in a vertical plane.
a) What is the slowest speed that the rock can travel and still maintain a circular path?
b) What is the tension in the string at the bottom of the swing?
Answers
Hello!
a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.
At the top of the circle, we have:
- Force of gravity (downward)
*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.
Therefore, only the force of gravity produces the net centripetal force:
[tex]\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg[/tex]
We can simplify and rearrange the equation to solve for 'v'.
[tex]\frac{v^2}{r} = g\\\\v^2 = gr\\\\v = \sqrt{gr}[/tex]
Plugging in values:
[tex]v = \sqrt{9.8 * 0.75} = \boxed{2.711 m/s^}[/tex]
b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)
- Tension force (upward, +)
The sum of these forces produces a centripetal force, upward (+).
[tex]\Sigma F = T - F_g\\\\F_c = T - F_g\\\\\frac{mv^2}{r} = T - mg[/tex]
Rearranging for 'T":
[tex]T = \frac{mv^2}{r} + mg\\\\[/tex]
Plugging in the appropriate values:
[tex]T = \frac{(0.2)(2.711^2)}{(0.75)} + 0.2(9.8) = \boxed{3.92 N}[/tex]
The correct answer would be number B)!! I took the test hope it helps
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A. (You'll need to get the various positions from the graph. The ends of the strings are exactly on one of the tic marks.)
331.35 N is incorrect.
Answers
The magnitude of the tension in the string marked A and B is mathematically given as
A = 52.5 N
What is the magnitude of the tension in the string marked A?Generally, the equation for is mathematically given as
The angle at A is...
[tex]tan\theta = 3/8[/tex]
When below the negative x
B= tan\phi
[tex]tan\phi = 5/4[/tex]
When below the negative x
C=
[tex]tan \rho = 1/6[/tex]
Hence, considering the Horizontal components
74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)
A = 78.9 - 0.668B
Hence, considering the Vertical components
74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)
40.07 = 1.015B
B = 39.5 N
In conclusion, Sub the value of B is the equation of A
A = 78.9 - 0.668B
Sub
A = 78.9 - 0.668( 39.5 N)
A = 52.5 N
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What is the size of the image on the retina of a size object 1.5 cm, placed at a distance (120 cm) away? Take the .lens-to-retina distance to be 2 cm
Answers
Answer: 0.025 cm
Explanation:
The image distance must match the distance between the lens and the retina for clear vision. Consequently, the image distance is v = 2 cm
The magnification of the lens is given by
m = v/u = h'/h
Where h' is the height of the imagev/u = h'/ h
h' = ( v/u ) × h
h' = ( 2cm /-120cm ) × 1.5cm
h' = - 0.025cm
Therefore, the height of the image is 0.025 cm
Three ropes A, B and C are tied together in one single knot K. (See figure.)
If the tension in rope A is 65.3 N, then what is the tension in rope B?
46.17 N is incorrect.
Answers
The tension in the rope B is determined as 10.9 N.
Vertical angle of cable B
tanθ = (6 - 4)/(5 - 0)
tan θ = (2)/(5)
tan θ = 0.4
θ = arc tan(0.4) = 21.8 ⁰
Angle between B and Cθ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰
Apply cosine rule to determine the tension in rope B;
A² = B² + C² - 2BC(cos A)
B = C
A² = B² + B² - (2B²)(cos A)
A² = 2B² - 2B²(cos 43.6)
A² = 0.55B²
B² = A²/0.55
B² = 65.3/0.55
B² = 118.73
B = √(118.73)
B = 10.9 N
Thus, the tension in the rope B is determined as 10.9 N.
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What states the purpose of an experiment
Answers
Answer:
The purpose of an experiment is to test out your hypothesis. If your hypothesis is correct, then it is a theory that could work every single time the experiment has been performed by scientists.
Melanie gets out of her car at the park. She walks 25 m to the trail entrance.
She jogs around the trail until she reaches a pond, where she stops briefly.
She then continues to follow the trail around the pond. Which reference point
should be used to describe her motion?
A. Her car
B. The trail entrance
C. The pond
D. The trail
Answers
The reference point that should be used to describe Melanie motion is the pond (option C).
What is reference point?Reference point is a particular point in space which is used as an endpoint to measure a distance from or chart a map from.
According to this question, Melanie gets out of her car at the park and walks 25 m to the trail entrance. She jogs around the trail until she reaches a pond, where she stops briefly.
However, she then continues to follow the trail around the pond. This suggests that the pond should serve as the reference point for Melanie's motion.
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Question Completion Status: 1 points QUESTION 1 The following excerpt is taken from the Governors Highway Safety Association (GHSA) website (https://www.ghsa.org): Setting speed limits has traditionally been the responsibility of states, except for the period of 1973-1994. During that time, the federal government enacted mandatory speed limit ceilings on interstate highways and similar limited access roads through a National Maximum Speed Limit. ✓ Sav Congress repealed the National Maximum Speed Limit in 1995. Since then, 41 states have raised speed limits to 70 mph or higher on some portion of their roadway systems. In many states, maximum speeds vary depending on vehicle type (car or truck), roadway location (urban or rural), or time of day. GHSA tracks state maximum speed limits for both urban and rural interstates, as well as other limited access roads. In a few states, speed limits are not set by law. Your textbook includes the following physics problem: If the speed of a car is increased by 50%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver's reaction time. To complete the Civic Engagement component of this course, write an essay about the National Speed Limit. In this essay, do the following: 1. Solve the above physics problem. 2. Use your solution to the above physics problem and the information contained in the above excerpt to argue for the reinstatement of a National Maximum Speed Limit. Your orrny should be 3000 to 5000 characters long (not including spaces) You are expected to format all variables/equations using the math editor, which you access
Answers
There should be a reinstatement of the National Maximum Speed Limit for the safety of road users because the minimum braking distance is 2,25m.
Calculations and ParametersHence, we can see that the minimum braking distance for the car would be:
d'/d= (v'1/v1)^2 = (1.5/1)^2
= 2.25m
The speed limit ensures that drivers do not exceed a set speed limit in order to reduce road accidents and keep road users safe and also the pedestrians would be able to navigate easier.
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A 520 Hz tone is sounded at the same time as a 516 Hz tone. What is the beat frequency?
Answers
A 520 Hz tone is sounded at the same time as a 516 Hz tone now the beat frequency is 4 Hz.
What is a frequency?The pace of direction changes in current per second is known as frequency. It is expressed in hertz (Hz), a unit of measurement that is used internationally. One hertz is equal to one cycle per second. One hertz (Hz) is equivalent to one cycle per second. A complete alternating current or voltage wave is referred to as a cycle.
f0=f2=f1 i the formula for frequency .then 520 Hz- 516 Hz= 4 HHz.Read more about the Frequency :
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physical properties of matter help describe a substance which of the following is not a physical property of a soil sample?
A. the container in which the soul is held
B. the color of the soil
C. the volume of the soil
D. the texture of the soil
Answer asap this is an exam il give brainlist
Answers
WHATS THE CORRECT ANSWER ANSWER ASAP
Answers
Answer:
speed and velocity
Explanation:
Hope this helped :)
An object moves in uniform circular motion at 50 m/s and takes 1.0 second to go a quarter circle. Calculate the
centripetal acceleration.
Answers
Answer:
The centripetal acceleration (ac)=314m/s²
Explanation:
look at the attachment ☝️
the sound of a lamb is ....... due to its pitch depending upon its ....... frequency?
Answers
The sound of a lamb is grave due to its high pitch depending on its low frequency. Option B is correct.
What is the frequency of the sound?A sound pressure wave's frequency is the number of times it repeats itself every second.
The frequency of the sound is the inverse of the period. If the wavelength of a wave is short. The wave will indeed have a lower frequency. A longer wavelength denotes a lower frequency.
Pitch and the frequency of sound are inverse to each other. A lamb's sound is grave because of its high pitch and low frequency.
The sound of a lamb is grave due to its high pitch depending on its low frequency.
Hence, option B is correct.
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In the first seconds of flight, the Saturn V rocket achieved an altitude of m, and a velocity of m/s. The rocket weighed approximately kg. What was the average power produced by the rocket?
Answers
The average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.
Average power produced by the rocketThe average power produced by the rocket is calculated as follows;
P = FV
where;
P is the average powerF is the force exerted by the rocketV is the velocity of the rocketThus, the average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.
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Q1) Write True or False for each statement:
Equal volumes of the two different substances have equal masses.
Answers
Answer:
True
Explanation:
it is true because blabla bla
A rocket takes off from Earth's surface, accelerating straight up at 47.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 80.9 kg, including her space suit. (Assume the rocket's initial motion parallel to the +y-direction. Indicate the direction with the sign of your answer.)
HINT
Answers
Answer:
Approximately [tex]4.61\times 10^{3}\; {\rm N}[/tex] upwards (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
External forces on this astronaut:
Weight (gravitational attraction) from the earth (downwards,) andNormal force from the floor (upwards.)Let [tex](\text{normal force})[/tex] denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:
[tex]\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}[/tex].
Let [tex]m[/tex] denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be [tex](\text{weight}) = m\, g[/tex].
Let [tex]a[/tex] denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be [tex](\text{net force}) = m\, a[/tex].
Rearrange [tex]\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}[/tex] to obtain an expression for the magnitude of the normal force on this astronaut:
[tex]\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}[/tex].
You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The bucket is attached to a heavy duty steel chain. The mass of the chain is 19.3 kg.
How much work do you perform during the lifting process?
If it takes 1.75 minutes for you to raise the bucket of water out of the well, then what was your average power?
Answers
The work done is 8427 J while the energy expended is 80.25 W.
What is work done?Work done is defined as the product of the force and the distance. We know that the work done in a gravitational field is given as;
W = mgh
Total mass of the water bucket and chain = 13.9 kg + 19.3 kg = 33.2Kg
Distance covered = 25.9 m
W = 33.2Kg * 9.8 m/s^2 * 25.9 m
W = 8427 J
Recall that the work done = Energy expended
Power = Energy expended/ Time
Power = 8427 J/ 1.75 * 60 seconds
Power = 80.25 W
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importance of states of matter
Answers
Answer:
STATES OF MATTER The three important states of the matter are (i) Solid state (ii) Liquid state (iii) Gaseous state, which can exist together at a particular temperature and pressure e.g. water has three states in equilibrium at 4.58 mm and 0.0098ºC.
Explanation:
You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The bucket is attached to a heavy duty steel chain. The mass of the chain is 19.3 kg.
How much work do you perform during the lifting process?
8427 J is Incorrect.
If it takes 1.75 minutes for you to raise the bucket of water out of the well, then what was your average power?
80.25 W is Incorrect.
Answers
The total work done is 5980 Joules and the power expended is 57 Watts.
What is work done?
The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;
Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules
Height of the center of mass of chain = 25.9 / 2 = 12.95 m
Work done by the chain Wc;
Wc = 12.95 * 19.3 * 9.8 = 2450 Joules
Total work = 3530 + 2450 = 5980 Joules
Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts
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(15 points) A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
1. What is the horizontal component of the force exerted by the hi.nge on the beam? (Use the `to the right' as + for the horizontal direction.)
2. What is the magnitude of the force that the beam exerts on the hi.nge?
Answers
The horizontal component of the force is 336 N
The magnitude of the force that the beam exerts on the hi_nge is 537.9 Newtons
What is the horizontal component of the force exerted by the hi_nge on the beam?The system of the beam, hi_nge and cable are in equilibrium.
The horizontal component of the force exerted by the beam is given below as:
Horizontal component of the force = mg × cos 31°
Horizontal component of the force = 40 × 9.8 × cos 31°
Horizontal component of the force = 336 Newtons
The magnitude of the force that the beam exerts on the hi_nge is given as:
F = mg × cos 31° + mg × sin 31°
F = 40 × 9.8 × cos 31° + 40 × 9.8 × sin 31°
F = 537.9 Newtons
Hence, the magnitude of the force that the beam exerts on the hi_nge is 537.9 Newtons.
In conclusion, the system of forces acting on the hi_nge and the beam are in equilibrium.
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A hammer of mass m = 0.46 kg is moving horizontally at a velocity of v = 6.5 m/s when it strikes a nail and comes to rest after driving the nail a distance Δx = 1.1 cm into a board.
What is the duration of the impact, assuming the acceleration is constant during this time period, in terms of the given variables? It is not accepting numerical values
What was the average force exerted on the nail, in terms of the mass, initial velocity, and distance traveled?
What was the average force, in newtons, exerted on the nail?
Answers
The impact will last for 3.3 milliseconds, the average force applied to the nail will be [m (v² - u²)]/2s, and the average force applied to the nail will be 883.2 N, respectively.
What is force?Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
Force is defined as the product of mass and acceleration. Its unit is Newton.
Given data;
Average force,F = ?
Mass,m = 0.46 kg
Acceleration,a
Final velocity,v = 0 m/sec
Initial velocity,u = 6.5 m/sec
The duration of the impact, assuming the acceleration is constant is found with the help of Newton's third equation of motion as;
v²=u²+2as
0 = (6.5)² + 2 ×a × 1.1 × 10⁻²
- 42.25 = 2 × a × 1.1 × 10⁻²
a = - 1920 m/s²
a = (v-u)/t
-1920 = (0-6.5)/t
t = 3 × 10⁻³ sec
t = 3.3 milli second
From Newton's third equation of motion;
v² = u² +2as
a = (v² - u² ) /2s
The average force exerted on the nail, in terms of the mass, initial velocity, and distance traveled is;
F = ma
F = [m (v² - u² )] /2s
The average force, in newtons, exerted on the nail is;
F = [m (v² - u² )] /2s
F= ma
Substitute the given value;
F = 0.469 × 1920
F = 883.2 N
Hence the duration of the impact, the average force exerted on the nail and the average force, in newtons, exerted on the nail will be 3.3 milliseconds, F = [m (v² - u² )] /2s and 883.2 N respectively.
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A person of mass 100kg runs up a staircase with a vertical height of 10m. If the trips takes 5 seconds to complete, calculate the person's power ( acceleration due to gravity =10m/s²)
Answers
Answer:
Power = 2000J/s
Explanation:
Power= Total Work done/Time
Given
mass = 100kg
distance(height) = 10m
time = 5s
acceleration due to gravity = 10m/s²
Work done = force × distance
Finding Force
Force = mass × acceleration due to gravity
Force = 100kg × 10m/s²
Force = 1,000N
Finding work done
Work done = Force × distance
Work done = 1,000N × 10m
Work done = 10,000J
Finding Power
Power = work done/time
Power = 10,000J/5s
Power = 2,000J/s
Therefore Power is 2,000J/s
Theory of uniform accelerated motion lab reports
Answers
The theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.
What is the theory of uniform accelerated motion?The theory of uniform accelerated motion is a scientific statement regarding the acceleration of an object in steady conditions.
This theory (uniform accelerated motion) is fundamentally applied in physic and astrophysics.
In conclusion, the theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.
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What is the magnitude of the displacement?
Answers
A) 20 m
What is called magnitude of displacement?
The magnitude of displacement is defined as the shortest distance between the initial and final position of the object.
For a particle in motion, the magnitude is either less than or equal to the distance travelled.
Also,
Velocity is simply defined as the rate of change of displacement with time.
Mathematically, it can be expressed as:
Velocity = displacement / time
According to the question,
The following data were obtained:
Time = 5 s
Velocity = 4 m/s
Displacement =?
Velocity = displacement / time
4 = d / 5
D = 4 × 5
D= 20 m
Hence,
20 m is the magnitude of the displacement of the object after it travelled for five seconds.
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The minimum takeoff speed for a certain airplane is 50 m/s. What minimum acceleration is required if the plane must leave a runway of length 2000 m? assume the plane starts from rest at one end of the runway.
Answers
Assuming constant acceleration, the requisite magnitude is [tex]a[/tex] such that
[tex]\left(50\dfrac{\rm m}{\rm s}\right)^2 - 0^2 = 2a (2000\,\mathrm m)[/tex]
Solve for [tex]a[/tex] :
[tex]a = \dfrac{\left(50\frac{\rm m}{\rm s}\right)^2}{4000\,\mathrm m} = \boxed{0.625 \dfrac{\rm m}{\mathrm s^2}}[/tex]
A figure skater glides along a circular path of radius 6.70 m. If she coasts around one half of the circle, find the following.
(a)her distance (in m) from the starting location
(b)the length (in m) of the path she skated
Answers
(a) Her distance from the starting location is 21.05 m.
(b) The length of the path she skated is 21.05 m.
Distance of the skater from the starting positionThe distance around a complete circular path is calculated as 2πr.
The distance for a half circle is calculated as ¹/₂ x 2πr = πr
Distance from the starting location = π x 6.7 m = 21.05 m
The length of the path she skated is the same as her distance from the starting location = 21.05 m.
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please help me there
Answers
Answer:
when switch is off no electricity will flow and then the circuit is called an open circuit
Explanation:
Electricity will not flow in open circuit
A container of PS5s with a mass of 451kg is loaded onto a Walmart truck using a ramp. The ramp is 6.12m long and
the bed of the truck is 1.53m above the ground. A force of 2025N applied parallel to the ramp moves the precious
consoles at a constant speed up the ramp. Find the efficiency of the ramp..
a) 54.6%
b) 32%
c) 46.4%
d) 54.65%
Answers
A. The efficiency of the ramp is 54.6 %.
Velocity ratio of the rampV.R = distance moved by effort/distance moved by load = L/h
V.R = (6.12)/(1.53) = 4
Mechanical advantage of the rampM.A = Load/Effort
M.A = (451 x 9.8)/(2025)
M.A = 2.183
Efficiency of the rampE = (M.A / V.R) x 100%
E = (2.183 / 4) x 100%
E = 54.6 %
Thus, the efficiency of the ramp is 54.6 %.
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Which expression is equivalent to 6(a - 3)?
Answers
Answer:
it's 6a - 18
Explanation:
because 6 is multiplied witk a and -3
Kim has her coffee cup on her car's dash when she takes a corner with radius 4 m and 20 km/h.
What is the minimum coefficient of static friction which would allow the coffee cup to stay there
without slipping?
Answers
The minimum coefficient of static friction is 0.807
What is static friction?Static friction is friction between two or more solid objects that are not moving relative to each other.
According to the question,
When an object turns on a curve path at a specific speed, two forces will be acting on the object which balance each other:
Centripetal force and friction force.
The friction coefficient between the object and the path is independent of the object's mass.
Given,
The radius of a corner is, r=4.16m
The turning speed of the car is, v
[tex]20km/h * (\frac{\frac{5}{18}m/s }{1km/h} )[/tex]
= 5.72 m/s
To avoid slipping a coffee cup on the car's dashboard, the centripetal force would balance the friction force.
So, Fc=Ff
[tex]\frac{mv^2}{r}[/tex]=μmg
μ=[tex]\frac{v^2}{rg}[/tex]
Here,
Fc represents the centripetal force.
Ff represents the friction force.
μ represents the coefficient of static friction.
g represents the gravitational acceleration whose value is 9.8m/s²
By Substituting the values in the above formula, we get:
μ =[tex]\frac{(5.72m/s)^2}{(4m)(9.8m/s^2)} \\[/tex] ≈ 0.807
Therefore,
The minimum coefficient of static friction is 0.807
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28. An electron with a speed of 4.0 x 10° m/s enters a uniform magnetic field of magnitude 0.040 T at an angle of 35 degrees to the magnetic field lines. The electron will follow a helical path. a) Determine the radius of the helical path. b) How far forward will the electron have moved after completing one circle?
Answers
The radius of the helical path and the period of the electron are 3. 26* 10^-10m and 8.94*10^-10m respectively.
How to determine the radiusUsing the formula:
r = [tex]\frac{mvsin \alpha }{Bq}[/tex]
m = 9.109 × 10-31 kg
α = 35°
B = 0. 040 T
q = 1.6 × 10-19 C
Substitute values into the formula
[tex]r = \frac{9. 109 * 10^-31 * 4.0 * 10^0 * sin 35}{0. 040 * 1.6 * 10^-19 }[/tex]
[tex]r = \frac{2. 089 * 10^-30}{6. 4* 10^-21}[/tex]
[tex]r = 3. 26 * 10^-10[/tex] m
b. [tex]T = \frac{2\pi m}{qB}[/tex]
[tex]T = \frac{2 * 3.142 * 9.109 *10^-31}{1.6* 10^-19* 0.040}[/tex]
[tex]T = \frac{5. 72* 10^-30}{6. 4* 10^-21}[/tex]
[tex]T = 8. 94* 10^-10[/tex]
Therefore, the radius of the helical path and the period of the electron are 3. 26* 10^-10m and 8.94*10^-10m respectively.
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