Answer:
c) Elastic Modulus
Explanation
As we know that when deformation is under elastic limit then stress applied to the given material is proportional to the strain developed in it
So here we can say that since they both are directly proportional to each other so the proportionality constant here is known as Modulus of elasticity.
So we can say it is given as
[tex]stress = E (strain)[/tex]
so now if we draw a graph between between stress and strain then it must be a straight line and the the slope of this straight line is given as
[tex]Slope = \frac{Stress}{Strain}[/tex]
So here correct answer will be
c) Elastic Modulus
Final answer:
The slope of the stress-strain curve in the elastic deformation region is known as the Elastic Modulus or Young's Modulus, which measures the stiffness of the material. The correct answer to the given question is 'c. Elastic Modulus'.
Explanation:
The slope of the stress-strain curve in the elastic deformation region represents the relationship between stress and strain under elastic conditions. This slope is in fact the Elastic Modulus, which is also known as Young's Modulus when it's in tension or compression. It serves as a measure of the stiffness or rigidity of the material, indicating how much stress is required to achieve a certain amount of strain.
In the context of a stress-strain curve, the plastic modulus is associated with plastic deformation, not the elastic region. Poisson's ratio is another material property that describes the ratio of transverse strain to axial strain, and is not the slope of the curve. Hence, the correct answer to the question is 'c. Elastic Modulus'.
What is the frequency of oscillation for a mass on the end of spring with a period of motion of 2.6 seconds? Calculate answer to one decimal place.
Answer:
Frequency, f = 0.38 Hz
Explanation:
Time period of the spring, T = 2.6 seconds
We need to find the frequency of oscillation for a mass on the end of spring. The relation between the time period and the frequency is given by :
Let f is the frequency of oscillation. So,
[tex]f=\dfrac{1}{T}[/tex]
[tex]f=\dfrac{1}{2.6\ s}[/tex]
f = 0.38 Hz
or
f = 0.4 Hz
So, the frequency of oscillation for a mass on the end of a spring is 0.38 hertz. Hence, this is the required solution.
An electron is confined in a harmonic oscillator potential well. What is the longest wavelength of light that the electron can absorb if the net force on the electron behaves as though it has a spring constant of 74 N/m? (m el = 9.11 × 10-31 kg, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J, ℏ = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s)
Answer:
The longest wavelength of light is 209 nm.
Explanation:
Given that,
Spring constant = 74 N/m
Mass of electron [tex]m= 9.11\times10^{-31}\ kg[/tex]
Speed of light [tex]c= 3\times10^{8}\ m/s[/tex]
We need to calculate the frequency
Using formula of frequency
[tex]f =\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
Where, k= spring constant
m = mass of the particle
Put the value into the formula
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{74}{9.11\times10^{-31}}}[/tex]
[tex]f=1.434\times10^{15}\ Hz[/tex]
We need to calculate the longest wavelength that the electron can absorb
[tex]\lambda=\dfrac{c}{f}[/tex]
Where, c = speed of light
f = frequency
Put the value into the formula
[tex]\lambda =\dfrac{3\times10^{8}}{1.434\times10^{15}}[/tex]
[tex]\lambda=2.092\times10^{-7}\ m[/tex]
[tex]\lambda=209\ nm[/tex]
Hence, The longest wavelength of light is 209 nm.
Two metal spheres of identical mass m = 4.20 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a charge of 0.785 µC, and the right-hand sphere carries a charge of 1.47 µC. What is the equilibrium separation between the centers of the two spheres?
Answer:
Explanation:
Answer:
0.632 m
Explanation:
let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.
According to the diagram,
d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)
Let T be the tension in the string.
Resolve the components of T.
T Sinθ = k q1 q2 / d^2
T Sinθ = k q1 q2 / (2LSinθ)² .....(2)
T Cosθ = mg .....(3)
Dividing equation (2) by equation (3), we get
tanθ = k q1 q2 / (4 L² Sin²θ x mg)
tan θ Sin²θ = k q1 q2 / (4 L² m g)
For small value of θ, tan θ = Sin θ
So,
Sin³θ = k q1 q2 / (4 L² m g)
Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)
Sin³θ = 0.2523
Sinθ = 0.632
θ = 39.2 degree
So, the separation between the two charges, d = 2 x L x Sin θ
d = 2 x 0.5 x 0.632 = 0.632 m
A proton is moving at 105 m/s at a point where the potential is 10 V. Later, it is at a place where the potential is 5 V. What is its speed there, assuming energy is conserved?
Answer:
The speed is [tex]7.07\times10^{4}\ m/s[/tex]
Explanation:
Given that,
Speed of proton [tex]v= 10^{5}\ m/s[/tex]
Final potential = 10 v
Initial potential = 5 V
We need to calculate the speed
Using formula of energy
[tex]\dfrac{1}{2}mv^2=eV[/tex]
[tex]v^2=\dfrac{2eV}{m}[/tex]
The speed of the particle is directly proportional to the potential.
[tex]v^2\propto V[/tex]
Put the value into the formula
[tex](10^{5})\propto 10[/tex]....(I)
For 5 V,
[tex]v^2\propto 5[/tex].....(II)
From equation (I) and (II)
[tex]\dfrac{(10^{5})^2}{v^2}=\dfrac{10}{5}[/tex]
[tex]v=70710.67\ m/s[/tex]
[tex]v=7.07\times10^{4}\ m/s[/tex]
Hence, The speed is [tex]7.07\times10^{4}\ m/s[/tex]
The speed of the proton in the second place is 74.3 m/s.
To calculate the speed of the proton in the second place, first, we need to find the mass of the proton.
Using,
P.E = mv²/2............ Equation 1Where:
P.E = potential energy of the protonm = mass of the protonv = speed of the proton.Make m the subject of the equation
m = 2P.E/v²............. Equation 2Given:
P.E = 10 Vv = 105 m/sSubstitute these values into equation 2
m = 2×10/(105²)m = 1.81×10⁻³ kg.Finally, to calculate the speed in the second place, we make v the subject of equation 1
v = √(2P.E/m)................. Equation 3Given:
P.E = 5 Vm = 1.81×10⁻³ kgSubstitute these values into equation 3
v = √[(2×5)/(1.81×10⁻³)]v = 74.3 m/sHence, The speed of the proton in second place is 74.3 m/s.
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A star of uniform mass with a radius R rotates about its diameter with angular momentum L. Under the action of internal forces the star collapses until its radius is one fourth of its initial size. The magnitude of its new angular momentum is:
L
16L
L/16
4L
L/4
Answer:
L / 16
Explanation:
Mass = m, Radius = R, angular momentum = L
Now, new radius, R' = R/4, mass = m, angular momentum, L' = ?
By the law of conservation of angular momentum
If there is no external torque is applied, the angular momentum of the system remains conserved.
L = I x w
Moment of inertia I depends on the mass and the square of radius of the star.
If the radius is one fourth, the angular momentum becomes one sixteenth.
So, L' = L / 16
An infinite plane of charge has surface charge density 7.2 μC/m^2. How far apart are the equipotential surfaces whose potentials differ by 100 V?
Answer:
so the distance between two points are
[tex]d = 0.246 \times 10^{-3} m[/tex]
Explanation:
Surface charge density of the charged plane is given as
[tex]\sigma = 7.2 \mu C/m^2[/tex]
now we have electric field due to charged planed is given as
[tex]E = \frac{\sigma}{2\epsilon_0}[/tex]
now we have
[tex]E = \frac{7.2 \times 10^{-6}}{2(8.85 \times 10^{-12})}[/tex]
[tex]E = 4.07 \times 10^5 N/C[/tex]
now for the potential difference of 100 Volts we can have the relation as
[tex]E.d = \Delta V[/tex]
[tex]4.07 \times 10^5 (d) = 100[/tex]
[tex]d = \frac{100}{4.07 \times 10^5}[/tex]
[tex]d = 0.246 \times 10^{-3} m[/tex]
Two small plastic spheres are given positive electrical charges. When they are 30.0 cm apart, the repulsive force between them has magnitude 0.130 N. If one sphere has four times the charge of the other, what is the charge of the least charged sphere? Give the answer in nanocoulomb (nC).
Answer:
Charge on least sphere, q = 570 nC
Explanation:
It is given that,
Two small plastic spheres are given positive electrical charges. The distance between the spheres, r = 30 cm = 0.3 m
The repulsive force acting on the spheres, F = 0.13 N
If one sphere has four times the charge of the other.
Let charge on other sphere is, q₁ = q. So, the charge on first sphere is, q₂ = 4 q. The electrostatic force is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]0.13=9\times 10^9\times \dfrac{q\times 4q}{(0.3\ m)^2}[/tex]
[tex]q^2=\dfrac{0.13\times (0.3)^2}{9\times 10^9\times 4}[/tex]
[tex]q=5.7\times 10^{-7}\ C[/tex]
q = 570 nC
So, the charge on the least sphere is 570 nC. Hence, this is the required solution.
A cylindrical specimen of some metal alloy having an elastic modulus of 102 GPa and an original cross-sectional diameter of 3.8 mm will experience only elastic deformation when a tensile load of 2440 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.47 mm.
Answer:
[tex]l=222.803mm[/tex]
Explanation:
Given:
Elastic modulus, E = 102 GPa
Diameter, d = 3.8mm = 0.0038 m
Applied tensile load = 2440N
Maximum allowable elongation, = 0.47mm = 0.00047
Now,
The cross-sectional area of the specimen,[tex]A_o=\frac{\pi d^2}{4}[/tex]
substituting the values in the above equation we get
[tex]A_o=\frac{\pi 0.0038^2}{4}[/tex]
or
[tex]A_o=1.134\times 10^{-5}[/tex]
now
the stress (σ) is given as:
[tex]\sigma=\frac{Force}{Area}[/tex]
and[tex]E=\frac{\sigma}{\epsilon}[/tex]
where,
[tex]\epsilon =\ Strain[/tex]
also,
[tex]\epsilon=\frac{\Delta l}{l}[/tex]
where,
[tex]l=initial \ length[/tex]
thus,
[tex]E=\frac{\frac{F}{A_o}}{\frac{\Delta l}{l}}[/tex]
or on rearranging we get,
[tex]l=\frac{E\times \Delta l\times A}{F}[/tex]
substituting the values in the above equation we get
[tex]l=\frac{102\times 10^9\times 0.00047\times 1.134\times 10^{-5}}{2440}[/tex]
or
[tex]l=0.222803m[/tex]
or
[tex]l=222.803mm[/tex]
A young man walks daily through a gridded city section to visit his girlfriend, who lives m blocks East and nblocks North of where the young man resides. Because the young man is anxious to see his girlfriend, his route to her never doubles back—he always approaches her location. In terms of m and n, how many different routes are there for the young man to take?
Answer:
The man ate eggs.
Explanation:
He should brush his teeth before seeing his girlfriend.
Which of the following characterizes the earth’s revolution? a) it takes approximately 24 hours b) it is responsible for creating the day/night relationship c) it determines the timing of seasons and the length of the year d) it is clockwise when viewed from above the North Pole.
Explanation:
The revolutions of the Earth (also called translation movement), consist of the elliptical orbit that describes the Earth around the Sun.
In this sense, a complete revolution around the Sun occurs every 365 days, 5 hours, 48 minutes and about 46 seconds. It is thanks to this movement and that the Earth's axis is tilted with respect to the plane of its orbit about [tex]23\º[/tex], that the four seasons of the year exist.
For this reason, some regions receive different amounts of sunlight according to the seasons of the year. These variations are more evident near the poles and softer or imperceptible in the tropics (near the equator). Because near the equator the temperature tends to be more stable, with only two seasons: rain and drought.
The Earth's revolution around the Sun, taking approximately 365.24 days, determines the timing of seasons and the length of the year, and when viewed from the North Pole, the revolution is counterclockwise.
The revolution of the Earth around the Sun characterizes the Earth's journey through space as it orbits the Sun. This movement takes approximately 365.24 days, which equals one year.
The revolution is responsible for the timing of seasons and the length of the year. When observed from above the North Pole, the Earth's revolution around the Sun occurs counterclockwise, which is different from the rotation of the Earth on its own axis, the latter causing day and night cycles. Therefore, the correct answer to the student's question about what characterizes the Earth's revolution is (c) it determines the timing of seasons and the length of the year.
A block of ice with a mass of 2.50 kg is moving on a frictionless, horizontal surface. At time t = 0, the block is moving to the right with a velocity of magnitude 8.00 m/s. Calculate the velocity of the block after a force of 7.00 N directed to the left has been applied for 5.00 s.
Answer:
The velocity of the block is 22 m/s.
Explanation:
Given that,
Mass = 2.50 kg
Velocity = 8 .00 m/s
Force = 7.00 N
Time t = 5.00
We need to calculate the change in velocity it means acceleration
Using newton's law
[tex]F = ma[/tex]
Where,
m = mass
a = acceleration
Put the value into the formula
[tex]a=\dfrac{F}{m}[/tex]
[tex]a = \dfrac{7.00}{2.50}[/tex]
[tex]a= 2.8m/s^2[/tex]
We need to calculate the velocity of the block
Using equation of motion
[tex]v = u+at[/tex]
Where,
v = final velocity
u = initial velocity
a = acceleration
t =time
Put the value in the equation
[tex]v= 8.00+2.8\times5.00[/tex]
[tex]v=22\ m/s[/tex]
Hence, The velocity of the block is 22 m/s.
The final velocity of the block after applying a force of 7.00 N for 5.00 s is approximately -6.00 m/s.
Explanation:To calculate the velocity of the block after a force of 7.00 N directed to the left has been applied for 5.00 s, we can use Newton's second law of motion.
Newton's second law states that the force applied to an object is equal to the mass of the object multiplied by its acceleration.
In this case, the mass of the block is given as 2.50 kg and the force applied is 7.00 N. We can calculate the acceleration using the formula:
acceleration = force/mass
Substituting the given values, we get:
acceleration = 7.00 N / 2.50 kg
Calculating, the acceleration is approximately 2.80 m/s² to the left. Since the block initially had a velocity of 8.00 m/s to the right, we subtract the acceleration from the initial velocity to get the final velocity:
final velocity = initial velocity - acceleration * time
Substituting the given values:
final velocity = 8.00 m/s - 2.80 m/s² * 5.00 s
Calculating, the final velocity is approximately 8.00 m/s - 14.00 m/s = -6.00 m/s.
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Consider a Cassegrain-focus, reflecting telescope. Images recorded at Cassegrain-focus will be:
A. Oriented the same as in the sky
B. Flipped compared to what is in the sky
C. Rotated clockwise compared to what is in the sky
D. Rotated counter-clockwise compared to what is in the sky
High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 42 m/s. Find the speed of the golf ball just after impact. m/s
Explanation:
It is given that,
Mass of golf club, m₁ = 210 g = 0.21 kg
Initial velocity of golf club, u₁ = 56 m/s
Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg
After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.
Initial momentum of golf ball, [tex]p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s[/tex]
After the collision, final momentum [tex]p_f=0.21\ kg\times 42\ m/s+0.046v[/tex]
Using the conservation of momentum as :
[tex]p_i=p_f[/tex]
[tex]11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v[/tex]
v = 63.91 m/s
So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.
A new particle, the joelon, has just been discovered! Careful measurements show that the joelon has an average lifetime (at rest) of 37 ns. How fast must an average joelon be moving to travel 24 m (as viewed from the lab frame) before it decays?
Well [tex]s=\dfrac{d}{t}[/tex] where s is speed, d is distance and t is time.
We have distance and time so we can calculate speed.
[tex]s=\dfrac{24}{37\cdot10^{-9}}\approx6.5\cdot10^8\frac{\mathbf{m}}{\mathbf{s}}\approx\boxed{6.5\cdot10^2\frac{\mathbf{Mm}}{\mathbf{s}}}[/tex]
Hope this helps.
r3t40
Charge 1 of +5 micro-coulombs is placed at the origin, charge 2 of +24 micro-coulombs is placed at x = +0.23 m, y = -0.69 m, charge 3 of -5 micro-coulombs is placed at x = -0.27 m, y = 0 m. What is the magnitude of the total electric force on charge 1 in Newtons?
Answer:
[tex]F_{net} = 4.22 N[/tex]
Explanation:
Since charge 1 and charge 2 are positive in nature so here we will have repulsion type of force between them
It is given as
[tex]F_{12} = \frac{kq_1q_2}{r^2}[/tex]
[tex]F_{12} = \frac{(9\times 10^9)(5 \mu C)(24 \mu C)}{0.23^2 + 0.69^2}\frac{-0.23\hat i + 0.69 \hat j}{\sqrt{0.23^2 + 0.69^2}}[/tex]
[tex]F_{12} = 2.81(-0.23\hat i + 0.69\hat j)[/tex]
Since charge three is a negative charge so the force between charge 1 and charge 3 is attraction type of force
[tex]F_{13} = \frac{(9\times 10^9)(5 \mu C)(5 \mu C)}{0.27^2 + 0^2} (-\hat i)[/tex]
[tex]F_{13} = 3.1(- \hat i)[/tex]
Now we will have net force on charge 1 as
[tex]F_{net} = F_{12} + F_{13}[/tex]
[tex]F_{net} = (-0.65 \hat i + 1.94 \hat j) + (-3.1 \hat i)[/tex]
[tex]F_{net} = (-3.75 \hat i + 1.94 \hat j)[/tex]
now magnitude of total force on the charge is given as
[tex]F_{net} = 4.22 N[/tex]
A jet turbine rotates at a velocity of 7,500 rpm. Calculate the stress acting on the turbine blades if the turbine disc radius is 70 cm and the cross-sectional area is 15 cm2. Take the length to be 10 cm and the alloy density to be 8.5 g/cm3.
Answer:
stress = 366515913.6 Pa
Explanation:
given data:
density of alloy = 8.5 g/cm^3 = 8500 kg/m^3
length turbine blade = 10 cm = 0.1 m
cross sectional area = 15 cm^2 = 15*10^-4 m^2
disc radius = 70 cm = 0.7 m
angular velocity = 7500 rpm = 7500/60 rotation per sec
we know that
stress = force/ area
force = m*a
where a_{c} is centripetal acceleration =
[tex]a_{c} =r*\omega ^{2}= r*(2*\pi*\omega)^{2}[/tex]
=[tex]0.70*(2*\pi*\frac{7500}{60})^{2}[/tex]
= 431795.19 m/s^2
mass = [tex]\rho* V[/tex]
Volume = area* length = 15*10^{-5} m^3
[tex]mass = m = \rho*V = 8500*15*10^{-5} kg[/tex]
force = m*a_{c}
[tex]=8500*15*10^{-5}*0.70*(2*\pi*\frac{7500}{60})^{2}[/tex]
force = 549773.87 N
stress = force/ area
= [tex]\frac{549773.87}{15*10^{-5}}[/tex]
stress = 366515913.6 Pa
Final answer:
The question requires the calculation of stress on jet turbine blades using physics principles involving centrifugal force and material stress.
Explanation:
The question involves calculating the stress on the turbine blades of a jet engine, which requires a knowledge of physics concepts, particularly mechanics and dynamics. The given data include the turbine's rotational velocity (7,500 rpm), radius of the turbine disc (70 cm), the cross-sectional area of the blades (15 cm2), the length of the blades (10 cm), and the alloy density (8.5 g/cm3). To solve this, one would need to calculate the centrifugal force acting on the blades due to rotation and then divide that force by the cross-sectional area to find the stress. However, the calculation involves steps and concepts not provided in the information above, so the direct calculation cannot be completed without additional physics formulae and explanation.
An object is oscillating on a spring with a period of 4.60 s. At time t = 0.00 s the object has zero speed and is at x = 8.30 cm. What is the acceleration of the object at t = 2.50 s?
The acceleration of the object at t = 2.50 s in simple harmonic motion can be found using the equation a = -ω²x, where ω is the angular frequency and x is the displacement from the equilibrium position.
Explanation:The acceleration of the object at t = 2.50 s can be found using the equation for simple harmonic motion:
a = -ω²x
where ω is the angular frequency and x is the displacement from the equilibrium position.
The period of the oscillation is related to the angular frequency by the equation:
T = 2π/ω
Substituting the given period (T = 4.60 s) into the equation and solving for ω, we get:
ω = 2π/T = 2π/4.60 s
Now, substituting the values we have, ω = 2π/4.60 s and x = 8.30 cm, into the acceleration equation:
a = -ω²x = -(2π/4.60 s)² * 8.30 cm
Calculate the value of a to find the acceleration of the object at t = 2.50 s using the given equation for acceleration.
The brakes of a car moving at 14m/s are applied, and the car comes to a stop in 4s. (a) What was the cars acceleration? (b) How long would the car take to come to a stop starting from 20m/s with the same acceleration? (c) How long would the car take to slow down from 20m/s to 10m/s with the same acceleration?
(a) [tex]-3.5 m/s^2[/tex]
The car's acceleration is given by
[tex]a=\frac{v-u}{t}[/tex]
where
v = 0 is the final velocity
u = 14 m/s is the initial velocity
t = 4 s is the time elapsed
Substituting,
[tex]a=\frac{0-14}{4}=-3.5 m/s^2[/tex]
where the negative sign means the car is slowing down.
(b) 5.7 s
We can use again the same equation
[tex]a=\frac{v-u}{t}[/tex]
where in this case we have
[tex]a=-3.5 m/s^2[/tex] is again the acceleration of the car
v = 0 is the final velocity
u = 20 m/s is the initial velocity
Re-arranging the equation and solving for t, we find the time the car takes to come to a stop:
[tex]t=\frac{v-u}{a}=\frac{0-20}{-3.5}=5.7 s[/tex]
(c) [tex]2.9 s[/tex]
As before, we can use the equation
[tex]a=\frac{v-u}{t}[/tex]
Here we have
[tex]a=-3.5 m/s^2[/tex] is again the acceleration of the car
v = 10 is the final velocity
u = 20 m/s is the initial velocity
Re-arranging the equation and solving for t, we find
[tex]t=\frac{v-u}{a}=\frac{10-20}{-3.5}=2.9 s[/tex]
(1) The acceleration of the car will be [tex]a=-3.5\frac{m}{s^2}[/tex]
(2) The time taken [tex]t=5.7s[/tex]
(3) The time is taken by the car to slow down from 20m/s to 10m/s [tex]t=2.9s[/tex]
What will be the acceleration and time of the car?(1) The acceleration of the car will be calculated as
[tex]a=\dfrac{v-u}{t}[/tex]
Here
u= 14 [tex]\frac{m}{s}[/tex]
[tex]a=\dfrac{0-14}{4} =-3.5\dfrac{m}{s^2}[/tex]
(2) The time is taken for the same acceleration to 20[tex]\frac{m}{s}[/tex]
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]t=\dfrac{v-u}{a}[/tex]
u=20[tex]\frac{m}{s}[/tex]
[tex]t=\dfrac{0-20}{-3.5} =5.7s[/tex]
(3) The time is taken to slow down from 20m/s to 10m/s with the same acceleration
From same formula
[tex]t=\dfrac{v-u}{a}[/tex]
v=10[tex]\frac{m}{s}[/tex]
u=20[tex]\frac{m}{s}[/tex]
[tex]t=\dfrac{10-20}{-3.5} =2.9s[/tex]
Thus
(1) The acceleration of the car will be [tex]a=-3.5\frac{m}{s^2}[/tex]
(2) The time taken [tex]t=5.7s[/tex]
(3) The time is taken by the car to slow down from 20m/s to 10m/s [tex]t=2.9s[/tex]
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The magnetic field due to a 2-A current flowing in a long, straight wire is 8 μT at a point P, a certain distance away from the wire. How far is this point from the center of the wire?
Answer:
Distance from the center of wire is 0.05 meters.
Explanation:
It is given that,
Current flowing in the wire, I = 2 A
Magnetic field, [tex]B=8\ \mu T=8\times 10^{-6}\ T[/tex]
Let d is the distance from the center of the wire. The magnetic field at a distance d from the wire is given by :
[tex]B=\dfrac{\mu_oI}{2\pi d}[/tex]
[tex]d=\dfrac{\mu_oI}{2\pi B}[/tex]
[tex]d=\dfrac{4\pi \times 10^{-7}\times 2\ A}{2\pi \times 8\times 10^{-6}\ T}[/tex]
d = 0.05 meters
So, the distance from the wire is 0.05 meters. Hence, this is the required solution.
To find the distance from the wire to point P where the magnetic field is 8 μT due to a 2-A current, the formula B = μ₀I/(2πR) is used, and the calculation reveals that R = 0.1 mm.
To, calculating the distance from the center of the wire to a point P where the magnetic field due to a 2-A current flowing in a long, straight wire is 8 μT.
To find this distance, we use the formula for the magnetic field around a long straight wire, given by B = μ₀I/(2πR), where B is the magnetic field strength, μ₀ is the permeability of free space (4π x 10⁻⁷ Tm/A), I is the current, and R is the distance from the wire.
Plugging the given values into this equation, we get:
8 x 10⁻⁶ T = (4π x 10⁻⁷ Tm/A)(2A) / (2πR)
Simplifying, we find that R = (4π x 10⁻⁷ Tm/A)(2A) / (2π x 8 x 10⁻⁶ T) = 0.0001 m or 0.1 mm.
A particular wire has a resistivity of 6.47×10-8 Ωm and a cross-sectional area of 2.32 mm2. A length of this wire is to be used as a resistor that will develop 130 W of power when connected to a 9.00 V battery. What length of wire is required?
Using the formula for power:
P = V^2 / R
130 W = (9.00 V)^2 / R
Solve for r:
R = 81/130
R = 0.623 ohms
Now solve for the length of wire:
R = rho L / A
A must be in m^2 - 2.32 mm^2 * 1 m^2/10^6 mm^2 = 2.132x10^-6 m^2
Now you have:
0.623 = (6.47x10^-8) L / (2.32x10^-6)
L = 22.34 m (Round answer as needed.)
Answer:
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During a tennis serve, a racket is given an angular acceleration of magnitude 155 rad/s2. At the top of the serve, the racket has an angular speed of 20.0 rad/s. If the distance between the top of the racket and the shoulder is 1.40 m, find the magnitude of the total acceleration of the top of the racket.
Answer:
600.6 m/s^2
Explanation:
α = 155 rad/s^2
ω = 20 rad/s
r = 1.4 m
Tangential acceleration, aT = r x α = 1.4 x 155 = 217 m/s^2
Centripetal acceleration, ac = rω^2 = 1.4 x 20 x 20 = 560 m/s^2
The tangential acceleration and the centripetal acceleration both are perpendicular to each other. Let a be the resultant acceleration.
a^2 = aT^2 + ac^2
a^2 = 217^2 + 560^2
a = 600.6 m/s^2
The total acceleration of the top of the racket during the tennis serve is approximately 580 m/s². This is determined by considering both the centripetal and tangential accelerations as perpendicular components and using the Pythagorean theorem for calculations.
Explanation:In this physics problem, we're given the angular acceleration, angular speed, and the distance between the top of the racket and shoulder (radius) to determine the total acceleration of the racket top during a tennis serve. To find the total acceleration, we must take into account both the centripetal (or radial) acceleration and the tangential acceleration (due to the change in speed).
First, let's calculate the centripetal acceleration, given by the formula ac=ω²r, where ω is the angular speed and r is the radius of the motion (in this case, the length of the arm). So, ac = (20.0 rad/s)² x 1.4m = 560 rad/s².
The tangential acceleration (at) is simply equal to the angular acceleration, which is 155 rad/s² (as provided in the question).
To find the total acceleration, we consider these two accelerations as perpendicular components and use the Pythagorean theorem: a = sqrt(ac² + at²). Substituting the values, we get a = sqrt((560 m/s²)² + (155 m/s²)²) ≈ 580 m/s².
Therefore, the total acceleration of the top of the racket is approximately 580 m/s².
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What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?
The weight of an 8 kg substance can be calculated in various units using the weight equation w = mg and the appropriate conversion factors. The weight is 78.4 N, 0.0784 kN, 78.4 kg·m/s², 8 kgf, 17.64 lbf, and 10.83 lbm·ft/s².
Explanation:To calculate the weight of an object in different units, we need to use the equation for weight: w = mg, where m is the mass of the object and g is the acceleration due to gravity. In this case, the mass (m) of the substance is given as 8 kg, and the value of g on Earth is approximately 9.80 m/s².
Therefore, the weight of the substance in various units is:
Newtons (N): w = mg = (8 kg)(9.80 m/s²) = 78.4 N.KiloNewtons (kN): 78.4 N = 0.0784 kN (as 1kN = 1000 N).kg·m/s²: This is just another term for Newton, so the weight is 78.4 kg·m/s².Kilogram-force (kgf): Here, 1 kgf equals the gravitational force exerted on a 1 kg mass, so 8 kgf.Pound-force (lbf): Since 1 N = 0.225 lbf, the weight in lbf is 78.4 N * 0.225 lb/N = 17.64 lbf.Pound-mass feet per second squared (lbm·ft/s²): We can use the conversion factor 1 lbm·ft/s² = 0.13825 N, so the weight is 78.4 N * 0.13825 lbm·ft/s²/N = 10.83 lbm·ft/s².Learn more about Weight Conversion here:https://brainly.com/question/11429990
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Benny wants to estimate the mean lifetime of Energizer batteries, with a confidence level of 97%, and with a margin of error not exceeding ±10 hours. If the standard deviation of the lifetime is known to be 55 hours, how many batteries does Benny need to sample?
Answer:
143 batteries does Benny need to sample
Explanation:
Given data
confidence level = 97%
error = ±10 hours
standard deviation SD = 55 hours
to find out
how many batteries does Benny need to sample
solution
confidence level is 97%
so a will be 1 - 0.97 = 0.03
the value of Z will be for a 0.03 is 2.17 from standard table
so now we calculate no of sample i.e
no of sample = (Z× SD/ error)²
no of sample = (2.16 × 55 / 10)²
no of sample = 142.44
so 143 batteries does Benny need to sample
An AC generator consists of eight turns of wire, each of area 0.0775 m2 , and total resistance of 8.53 Ω. The loop rotates in the external magnetic field of strength 0.222 T at a constant frequency of 51 Hz. What is the maximum induced emf? Answer in units of V.
Answer:
44.08 Volt
Explanation:
N = 8, A = 0.0775 m^2, R = 8.53 ohm, B = 0.222 T, f = 51 Hz
e0 = N B A w
e0 = 8 x 0.222 x 0.0775 x 2 x 3.14 x 51
e0 = 44.08 Volt
The question pertains to the calculation of the maximum induced emf (voltage) in an AC generator. This can be calculated using the generator's specifications and the formula ε_max = NBAω.
Explanation:The subject of your question pertains to electromagnetic induction in physics. The induction of emf in an AC generator is described by the equation ε = NBAω sin ωt, where ε is the induced emf, N is the number of turns of wire, B is the magnetic field strength, A is the cross-sectional area of the coil, and ω is the angular velocity of the generator. However, considering you are asking specifically for the maximum induced emf, we calculate this using the equation ε_max = NBAω, as sin ωt=1 at its peak. In your case, the generator consists of 8 turns of wire (N=8), an area of 0.0775 m^2 (A=0.0775), the strength of the magnetic field is 0.222 T (B=0.222), and a frequency of 51 Hz (f=51) which converts to angular velocity (ω) using the formula ω = 2πf. Substituting these values into the equation will give you the maximum induced emf.
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A critical part has a manufacturing specification (in cm) of 0.325 ± 0.010. Based on this information, if this measurement is larger than 0.335 or smaller than 0.315, the product fails at a cost of $120. Determine the Taguchi loss function in the given scenario.
Answer:
[tex]L(y)=12\times 10^{5}(y-0.325)^2[/tex]
Explanation:
We know that Taguchi loss function given as
[tex]L(y)=k(y-m)^2[/tex]
Where L is the loss when quality will deviate from target(m) ,y is the performance characteristics and k is the quality loss coefficient.
Given that 0.325±0.010 ,Here over target is m=0.325 .
When y=0.335 then L=$120,or when y=0.315 then L=$120.
Now to find value of k we will use above condition
[tex]L(y)=k(y-m)^2[/tex]
[tex]120=k(0.335-0.325)^2[/tex]
[tex]k=12\times 10^{5}[/tex]
So Taguchi loss function given as
[tex]L(y)=12\times 10^{5}(y-0.325)^2[/tex]
Answer:
Explanation:
Manufacturing specification
0.325 ± 0.010 I'm
The quality characteristic is 0.325
Functional tolerance is $120
The lost function is given
λ = C (x—t)²
Where, C is a constant
t is quality characteristic
And x is target value
Constant’ is the coefficient of the Taguchi Loss, or the ratio of functional tolerance and customer loss.
Then, C= tolerance / loss²
Measurement loss is
Loss = 0.335-0.315
Loss =0.01cm
Therefore,
C = 120/0.01²
C = 1,200,000
λ = C (x —t)²
λ = 1,200,00 (x—0.325)²
A proton moves perpendicular to a uniform magnetic field B S at a speed of 1.00 3 107 m/s and experiences an acceleration of 2.00 3 1013 m/s2 in the positive x direction when its velocity is in the positive z direction. Determine the magnitude and direction of the field.
Explanation:
It is given that,
Speed of proton, [tex]v=10^7\ m/s[/tex]
Acceleration of the proton, [tex]a=2\times 10^{13}\ \ m/s^2[/tex]
The force acting on the proton is balanced by the magnetic force. So,
[tex]ma=qvB\ sin(90)[/tex]
[tex]B=\dfrac{ma}{qv}[/tex]
m is the mass of proton
[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 2\times 10^{13}\ \ m/s^2}{1.6\times 10^{-19}\times 10^7\ m/s}[/tex]
B = 0.020875
or
B = 0.021 T
So, the magnitude of magnetic field is 0.021 T. As the acceleration in +x direction, velocity in +z direction. So, using right hand rule, the magnitude of B must be in -y direction.
An irrigation channel has a rectangular cross section of 1.5 ft deep x 11 ft wide on the input side. On the far end of the channel, the channel expands to 6 ft wide while maintaining the same depth. If the water flowing into the channel has a speed of 30 ft/sec, calculate the velocity of the water flow on the far end of the channel.
Answer:
55 ft/s
Explanation:
A₁ = Area of rectangular cross-section at input side = 1.5 x 11 = 16.5 ft²
A₂ = Area of rectangular cross-section at far end = 1.5 x 6 = 9 ft²
v₁ = speed of water at the input side of channel = 30 ft/s
v₂ = speed of water at the input side of channel = ?
Using equation of continuity
A₁ v₁ = A₂ v₂
(16.5) (30) = (9) v₂
v₂ = 55 ft/s
A runner of mass 56.1 kg starts from rest and accelerates with a constant acceleration of 1.2 m/s^2 until she reaches a velocity of 5.3 m She then continues running at this constant velocity. How long in seconds does the runner take to travel 118 m? A) 19.08 sec B) 24.47 sec C) 53.9 sec D) 15.733333 sec E) 31.152 sec
Answer:
Option B is the correct answer.
Explanation:
Final velocity = 5.3 m/s
Acceleration till 5.3 m/s = 1.2 m/s²
Time taken for this
[tex]t_1=\frac{5.3}{1.2}=4.42s[/tex]
Distance traveled in 4.42 s can be calculated
s = ut + 0.5 at²
s = 0 x 4.42 + 0.5 x 1.2 x 4.42² = 11.72 m
Remaining distance = 118 - 11.72 = 106.28 m
Uniform velocity = 5.3 m/s
Time taken
[tex]t_2=\frac{106.28}{5.3}=20.05s[/tex]
Total time, t = t₁ + t₂ = 4.42 + 20.05 = 24.47 s
Option B is the correct answer.
The runner takes approximately 24.47 seconds to travel 118 meters, considering the time spent accelerating and then running at constant velocity. Therefore, the correct answer is option B.
To determine the time it takes for the runner to travel 118 meters, we need to consider two phases of her motion: acceleration and constant velocity.
Phase 1: Acceleration
Initially, the runner starts from rest (initial velocity, u = 0) and accelerates at a constant rate of 1.2 m/s² until she reaches a velocity of 5.3 m/s.
Step 1: Calculate the time (t1) taken to reach the velocity of 5.3 m/s using the formula v = u + at.
v = 5.3 m/s, u = 0, a = 1.2 m/s²
t1 = (v - u) / a = (5.3 - 0) / 1.2 ≈ 4.417 s
Step 2: Calculate the distance (s1) covered during this acceleration phase using the formula s = ut + 0.5at².
s1 = 0 + 0.5 * 1.2 * (4.417)² ≈ 11.7 m
Phase 2: Constant Velocity
After reaching 5.3 m/s, the runner continues at this constant velocity. We need to find the distance she covers in this phase and the total time taken.
Step 3: Calculate the remaining distance (s2) that needs to be covered at constant velocity.
s1 = 11.7 m, Total distance = 118 m
s2 = 118 - 11.7 = 106.3 m
Step 4: Calculate the time (t2) taken to cover the distance s2 at the constant velocity using the formula t = s / v.
t2 = 106.3 m / 5.3 m/s ≈ 20.075 s
Total Time
Step 5: Add the time taken in both phases to find the total time.
Total time = t1 + t2 ≈ 4.417 s + 20.075 s ≈ 24.492 s
Therefore, the runner takes approximately 24.47 seconds to travel 118 meters. The correct answer is option B.
In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 27000 V. The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.
Answer:
9.74 x 10^7 m/s
Explanation:
V = 27000 V
energy of electrons = e x V
K = 1.6 x 10^-19 x 27000 = 43200 x 10^-19 J
Energy = 1/2 m v^2
43200 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2
v^2 = 9.495 x 10^15
v = 9.74 x 10^7 m/s
The half-life of1 is 8.04 days. (a) Calculate the decay constant for this isotope. (b) Find the number of 1311 nuclei necessary to of 0.5 uCi produce a sample with an activity
Explanation:
Given that,
[tex]T_{\frac{1}{2}}=8.04\ days[/tex]
We need to calculate the decay constant
Using formula of decay constant
[tex]\lambda=\dfrac{0.693}{t_{\frac{1}{2}}}[/tex]
[tex]\lambda=\dfrac{0.693}{8.04\times24\times3600}[/tex]
[tex]\lambda=9.97\times10^{-7}\ sec^{-1}[/tex]
We need to calculate the number of [tex]^{131}I[/tex] nuclei
[tex]N=\dfrac{A\ ci}{\lambda}[/tex]
Where,
A= activity
ci = disintegration
[tex]N=\dfrac{0.5\times10^{-6}\times3.7\times10^{10}}{9.97\times10^{-7}}[/tex]
[tex]N=1.855\times10^{10}[/tex]
Hence, This is the required solution.