Answer:
Explanation:
Let E be the strength of electric field and m be the mass of plastic sphere.
charge on plastic sphere, q = - 3 n C = - 3 x 10^-9 C
The force due to electric field on the charged plastic sphere = q E
The weight of the plastic sphere = m g
where, g is the acceleration due to gravity.
Now the force due to electrostatic field is balanced by teh weight of the plastic sphere.
q E = m g
E = m g / q
E = m x 9.8 / (3 x 10^-9) = 3.26 x 10^9 x m
Substitute the value of m and get the value E.
The magnitude of the vertical electric field required to balance the weight of the charged plastic sphere with a mass of 2.1 g and a charge of -3.0 nC is [tex]6.86 * 10^6 N/C[/tex].
To find the magnitude of the vertical electric field that will balance the weight of a charged plastic sphere, we need to equate the electrostatic force on the sphere due to the electric field with the weight of the sphere.
Given information:
- Mass of the plastic sphere, m = 2.1 g = [tex]2.1 * 10^-3 kg[/tex]
- Charge on the plastic sphere, q = -3.0 nC = [tex]-3.0 * 10^-9 C[/tex]
- Acceleration due to gravity, g = [tex]9.8 m/s^2[/tex]
- Coulomb's constant, k = 1/(4πε₀) =[tex]9.0 * 10^9 N.m^2/C^2[/tex]
Step 1: Calculate the weight of the plastic sphere.
Weight of the sphere = mg
Weight = [tex](2.1 * 10^-3 kg)[/tex] × [tex](9.8 m/s^2)[/tex]
Weight = 2.058 × [tex]10^{-2}[/tex] N
Step 2: Equate the electrostatic force on the sphere to its weight.
Electrostatic force = qE
Weight = qE
[tex]2.058 * 10^{-2} N = (-3.0 * 10^{-9} C) * E[/tex]
Step 3: Rearrange the equation to find the electric field magnitude E.
Equation to find the electric field magnitude E.
E = (2.058 × [tex]10^{-2}[/tex] N) / (-3.0 × [tex]10^{-9}[/tex]C)
[tex]E = 6.86 * 10^6 N/C[/tex]
Complete question:
what's the magnitude of a vertical electric field that will balance the weight of a plastic sphere of mass 2.1g that has been charged to -3.0nC? (k=[tex]1/4\pi[/tex], [tex]E_0[/tex]=[tex]9.0*10^9 N*m^2/C^2[/tex])
During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket has an angular speed of 12.0 rad/s. If the distance between the top of the racket and the shoulder is 1.30 m, find the magnitude of the total acceleration of the top of the racket.
Final answer:
The magnitude of the total acceleration of the top of the racket during the serve, considering both tangential and radial components, is approximately 270.5 m/s^2.
Explanation:
To find the magnitude of the total acceleration of the top of the racket, we need to consider both tangential and radial (centripetal) acceleration components. The tangential acceleration (at) is directly provided by the angular acceleration (α), and is calculated by multiplying the angular acceleration by the radius (r), so at = α × r. Here, α = 150 rad/s2 and r = 1.30 m, giving at = 195 m/s2. The radial acceleration (ar), also known as centripetal acceleration, depends on the angular speed (ω) and the radius (r), and is calculated with the formula ar = ω2 × r. Given ω = 12.0 rad/s and r = 1.30 m, we find ar = 187.2 m/s2. Finally, the total acceleration (a) is the square root of the sum of the squares of at and ar, resulting in a = √(at2 + ar2). This gives us a = √(1952 + 187.22) m/s2, and by calculating, we find the magnitude of the total acceleration to be approximately 270.5 m/s2.
Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open pneumothorax. b. empyema. c. pleural effusion. d. tension pneumothorax.
Answer:
The correct answer is d. tension pneumothorax.
Explanation:
The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.
Tension pneumothorax is the condition where air that enters the pleural space during inspiration but is unable to exit during expiration, causing a collapse of the lung and compression of surrounding structures.
Explanation:Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called tension pneumothorax. In this condition, there is a buildup of air in the pleural space, leading to a collapse of the lung and compression of surrounding structures. It is a medical emergency that requires immediate treatment.
Learn more about Conditions affecting the pleural space here:https://brainly.com/question/10615855
#SPJ12
A copper wire has a circular cross section with a radius of 1.25 mm. If the wire carries a current of 3.70 A, find the drift speed in mm/s of the electrons in this wire. (Note: the density of charge carriers (electrons) in a copper wire is n = 8.46 × 1028 electrons/m3)
Answer:
0.0557 mm / s
Explanation:
r = 1.25 mm = 1.25 x 10^-3 m, i = 3.7 A, n = 8.46 x 10^28 per cubic metre,
e = 1.6 x 10^-19 C
Let vd be the drift velocity
use the formula for the current in terms of drift velocity
i = n e A vd
3.7 = 8.46 x 10^28 x 1.6 x 10^-19 x 3.14 x 1.25 x 1.25 x 10^-6 x vd
vd = 5.57 x 10^-5 m/s
vd = 0.0557 mm / s
An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of 3.5× 104 kg, and the force causes its speed to change from 6600 to 5700m/s. (a) What is the work done by the force? (b) If the asteroid slows down over a distance of 1.5× 106 m determine the magnitude of the force.
Try this option; answers are marked with red colour: a) 19.215 GJ; b) 12.81 kN.
All the details are in the attached picture.
A soccer player swings his leg forward to kick the ball. If his leg has a moment of inertia of 0.9 kg m2, and is initially at rest, how much torque must he generate with his hip flexor muscle in order to reach an angular velocity of -8 rad/s in 0.5 seconds?
Answer:
14.4 Nm
Explanation:
Moment of Inertia, I = 0.9 kg m^2, w0 = 0, w = 8 rad/s, t = 0.5 second
Use first equation of motion for rotational motion
w = w0 + α t
where, α be the angular acceleration
8 = 0 + α x 0.5
α = 16 rad/s^2
Now Torque = Moment of inertia x angular acceleration
τ = I x α
τ = 0.9 x 16 = 14.4 Nm
An electron and a proton are each placed at rest in an electric field of 500 N/C. Calculate the speed (and indicate the direction) of each particle 54.0 ns after being released. electron m/s O in the same direction as the field O in a direction opposite to the field proton m/s O in a direction opposite to the field O in the same direction as the field
Answer:
For proton: 2592 m/s In the same direction of electric field.
For electron: 4752000 m/s In the opposite direction of electric field.
Explanation:
E = 500 N/C, t = 54 ns = 54 x 10^-9 s,
Acceleration = Force /mass
Acceleration of proton, ap = q E / mp
ap = (1.6 x 10^-19 x 500) / (1.67 x 10^-27) = 4.8 x 10^10 m/s^2
Acceleration of electron, ae = q E / me
ae = (1.6 x 10^-19 x 500) / (9.1 x 10^-31) = 8.8 x 10^13 m/s^2
For proton:
u = 0, ap = 4.8 x 10^10 m/s^2, t = 54 x 10^-9 s
use first equation of motion
v = u + at
vp = 0 + 4.8 x 10^10 x 54 x 10^-9 = 2592 m/s In the same direction of electric field.
For electron:
u = 0, ae = 8.8 x 10^13 m/s^2, t = 54 x 10^-9 s
use first equation of motion
v = u + at
vp = 0 + 8.8 x 10^13 x 54 x 10^-9 = 4752000 m/s In the opposite direction of electric field.
The cheetah is the fastest land animal, reaching speeds as high as 33 m/s, or roughly 75 mph. What is the kinetic energy of a 65 kg cheetah running at top speed? Express your answer in SI units.
Answer:
Ek= 35.392 kJ or 35.392 10³ J
Explanation:
Formula for Kinetic energy:
Ek= [tex]\frac{1}{2}[/tex] × m × [tex]v^{2}[/tex]
since m= 65kg and v= 33m/s
Ek= [tex]\frac{1}{2}[/tex]× 65×[tex]33^{2}[/tex]
Ek= 35.392 kJ or 35.392 10³ J
Final answer:
The kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is calculated using the formula KE = 1/2 mv², resulting in a kinetic energy of 35442.5 Joules.
Explanation:
The question asks for the kinetic energy of a 65 kg cheetah running at its top speed, which is given to be 33 m/s. To calculate kinetic energy, we use the formula KE = 1/2 mv², where m is the mass of the object (in this case, the cheetah) and v is the velocity of the object. Substituting the given values, we get:
KE = 1/2 × 65 kg × (33 m/s)²
KE = 1/2 × 65 × 1089
KE = 1/2 × 70885
KE = 35442.5 Joules
Therefore, the kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is 35442.5 Joules.
A proton, traveling with a velocity of 4.5 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 8.0 × 10−14 N and a direction of due south. What are the magnitude and direction of the magnetic field causing the force?
Answer:
Magnetic field, B = 0.11 i (in k direction)
Explanation:
It is given that,
Velocity of the proton, [tex]v=4.5\times 10^6\ m/s[/tex]
Magnetic force, [tex]F=8\times 10^{-14}\ N[/tex] (due south)
The magnetic force acting on the electron is given by :
[tex]F=qvB[/tex]
(-j) = (i) × (B)
[tex]B=\dfrac{F}{qv}[/tex]
q is the charge on proton
[tex]B=\dfrac{8\times 10^{-14}\ N}{1.6\times 10^{-19}\ C\times 4.5\times 10^6\ m/s}[/tex]
B = 0.11
So, the magnitude of 0.11 T is acting on the proton in and is acting directed upward above the plane. Hence, this is the required solution.
Final answer:
To calculate the magnitude of the magnetic field that causes a force on a moving proton, we use the rearranged Lorentz force law, with inputs of proton charge, velocity, and the force experienced. The direction of the magnetic field is determined by the right-hand rule, given the directions of the force and proton's velocity.
Explanation:
The student's question involves the effect of a magnetic field on a moving charged particle, specifically a proton, which falls under the subject of Physics. The Lorentz force law states that the force (F) on a charged particle is directly proportional to the charge (q), the velocity (v), and the magnetic field (B), and is given by the equation F = q(v x B), where v x B denotes the cross product of the velocity vector and the magnetic field vector. The force is perpendicular to both the velocity and the magnetic field.
To find the magnitude of the magnetic field, we can rearrange this equation to B = F / (q * v * sin(θ)), where θ is the angle between the velocity of the proton and the direction of the magnetic field. In the case where force is maximum, the angle is 90 degrees, and the sine of 90 degrees is 1. Given the maximum magnetic force (8.0 × 10⁻¹⁴ N), the charge of a proton (approximately 1.6 × 10⁻¹⁹ C), and the velocity of the proton (4.5 × 10⁶ m/s), we can calculate the magnitude of the magnetic field. As the force experienced by the proton is directed due south and the velocity is due east, the magnetic field must be pointing downwards to produce a force in that direction according to the right-hand rule.
An angle has a value of 2.2 radians. Find its value in degrees.
Answer:
The value of 2.2 radians in degree is 126.11°.
Explanation:
Given that,
Angle = 2.2 radians
We need to calculate the value in degrees
We know that,
[tex]1\ radians = \dfrac{180}{\pi}[/tex]
[tex]2.2\ radians = \dfrac{180}{3.14}\times2.2[/tex]
[tex]2.2\ radians = 126.11^{\circ}[/tex]
Hence, The value of 2.2 radians in degree is 126.11°.
A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subjected to an axial load of 200 N. Determine the deflection of the spring. C = 80 kN/mm2.
Answer:
The deflection of the spring is 34.56 mm.
Explanation:
Given that,
Diameter = 10 mm
Number of turns = 10
[tex]Radius_{mean} = 60\ mm[/tex]
[tex]Diameter_{mean} = 120\ mm[/tex]
Load = 200 N
We need to calculate the deflection
Using formula of deflection
[tex]\delta=\dfrac{8pD^3n}{Cd^4}[/tex]
Put the value into the formula
[tex]\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}[/tex]
[tex]\delta =34.56\ mm[/tex]
Hence, The deflection of the spring is 34.56 mm.
Calculate the force (in pounds) ona 4.0 ft by 7.0 ft door due to a 0.4 psi pressure difference.
Answer:
7174.09 N
Explanation:
Area = 7 ft x 4 x ft = 28 ft^2 = 2.60129 m^2
Pressure = 0.4 psi = 2757.9 Pa
Force = Pressure x Area
Force = 2757.9 x 2.60129 = 7174.09 N
Consider two objects on the Moon's surface that can just be resolved by one of the 10.0 m telescopes at the Keck Observatory. What is the separation of the two objects? Assume the resolving power is only limited by diffraction effects. The distance from the Earth to the Moon is 3.8 x 105km, assume the wavelength of the light is 550 nm.
Answer: 25 m
Explanation:
The resolving power of a telescope is given by:
[tex] \theta = 1.22\frac{\lambda}{diameter}[/tex]
Diameter of the telescope = 10.0 m
Wavelength of the telescope = 550 nm = 550 ×10⁻⁹m
Thus, the resolving power of one of telescopes at the Keck Observatory is:
[tex] \theta = 1.22 \frac{550\times 10^{-9}}{10.0} =671 \times 10^{-10} rad = 3.85 \times 10^{-6} deg[/tex]
Also, [tex] tan\theta = \frac {\text {separation between two objects}}{\text{distance to the objects}} [/tex]
separation between objects = tan (3.85×10⁻⁶)° × 3.8×10⁵km = 2.5 × 10⁻²km =25 m
While doing an experiment you measure the length of an object to be 32.5cm. The measuring device that you are using measures to a millimeter. How can you decrease the error associated with your measurements
Answer and Explanation:
By measuring in millimeter we can decrease the associated error with the measurement because when we measure in smaller unit the measurement is more precise and accurate rather than when we measure in larger unit so when we measure 325 millimeter instead of 32.5 cm then there is chance of less error produced.
A thin conducing plate 2.3 m on a side is given a total charge of −20.0 µC. (Assume the upward direction is positive.) (a) What is the electric field (in N/C) 1.0 cm above the plate? (Indicate the direction with the sign of your answer.)
Answer:
The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Distance = 2.3 m
Charge [tex]q= -20.0\ \muC[/tex]
We need to calculate the electric field
Using formula of electric field
[tex]E=\dfrac{\sigma}{2\epsilon_{0}}[/tex]
Where, [tex]\sigma=\dfrac{Q}{A}[/tex]
Q = charge
A = area
Put the value into the formula
[tex]E=\dfrac{Q}{2A\epsilon_{0}}[/tex]
[tex]E=\dfrac{-20.0\times10^{-6}}{2\times2.3\times2.3\times8.85\times10^{-12}}[/tex]
[tex]E=-213599.91\ N/C[/tex]
[tex]E=-2.14\times10^{5}\ N/C[/tex]
Negative sign shows the direction of the electric field.
The direction of electric field is toward the plates.
Hence, The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]
Final answer:
The electric field 1.0 cm above a thin conducting plate with a side length of 2.3 m and a total charge of -20.0 µC is -213,559.32 N/C, indicating a downward direction.
Explanation:
A thin conducting plate 2.3 m on a side is given a total charge of -20.0 µC. To find the electric field 1.0 cm above the plate, we use the concept of surface charge density (σ) and the formula for the electric field near an infinite plane sheet of charge, E = σ / (2ε0), where ε0 is the vacuum permittivity (ε0 = 8.85 x [tex]10^-^1^0[/tex] C2/N·m²). The surface charge density is σ = Q / A, with Q being the charge and A the area of the plate.
The area of the plate, A, is 2.3 m × 2.3 m = 5.29 m2. Therefore, σ = (-20.0 x 10-6 C) / (5.29 m2) = -3.78 x [tex]10^-^6[/tex] C/m². Substituting σ into the electric field equation yields E = -3.78 x [tex]10^-^6[/tex] C/m² / (2 × 8.85 x [tex]10^-^1^2[/tex] C2/N·m2) = -213,559.32 N/C, indicating the direction is downward due to the negative sign.
A sound wave from a siren has an intensity of 100.0 point, what the sound intensity level of the sound?
Answer:
Sound intensity level of the sound = 140 dB
Explanation:
We have expression for sound intensity level
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )[/tex]
We have sound intensity, I = 100 W/m²
I₀ = 10⁻¹² W/m²
We need to find sound intensity level
Substituting
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )=10log_{10}\left ( \frac{100}{10^{-12}}\right )=10x14=140dB[/tex]
Sound intensity level of the sound = 140 dB
An electric heater is rated at 1430 W, a toaster at 890 W, and an electric grill at 1760 W. The three appliances are connected in parallel to a common 120 V circuit. How much total current does this circuit draw?
Answer:
34.01 A
Explanation:
[tex]P_{heater}[/tex] = Power of the electric heater = 1430 W
[tex]P_{toaster}[/tex] = Power of the toaster = 890 W
[tex]P_{grill}[/tex] = Power of the electric grill = 1760 W
V = Voltage of the battery connected in parallel to appliances = 120 volts
Using ohm's law, current drawn by heater is given as
[tex]i_{heater} = \frac{P_{heater}}{V}[/tex]
[tex]i_{heater} = \frac{1430}{120}[/tex]
[tex]i_{heater} [/tex] = 11.92 A
Using ohm's law, current drawn by toaster is given as
[tex]i_{toaster} = \frac{P_{toaster}}{V}[/tex]
[tex]i_{toaster} = \frac{890}{120}[/tex]
[tex]i_{toaster} [/tex] = 7.42 A
Using ohm's law, current drawn by grill is given as
[tex]i_{grill} = \frac{P_{grill}}{V}[/tex]
[tex]i_{grill} = \frac{1760}{120}[/tex]
[tex]i_{grill} [/tex] = 14.67 A
Total circuit drawn is given as
[tex]i_{total} [/tex] = [tex]i_{heater} [/tex] + [tex]i_{toaster} [/tex] + [tex]i_{grill} [/tex]
[tex]i_{total} [/tex] = 11.92 + 7.42 + 14.67
[tex]i_{total} [/tex] = 34.01 A
Two workers are sliding 390 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 200 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
Answer:
The coefficient of kinetic friction [tex]\mu= 0.16989[/tex]
Explanation:
From Newton's second law
[tex]\sum\overset{\rightarrow}{F}=m\cdot\overset{\rightarrow}{a}[/tex]
If the velocity is constant, that means the summation of all forces must be equal to zero. Draw the free-body diagram to obtain the sums of forces in x and y. It must include the Friction Force, in the opposite direction of the displacement, the weight ([tex]W=mg=390*9.81=3825.9N[/tex]), the Normal Force, which is the is the consequence of Newton's third law and the forces from the two workers.
The sum in y is:
[tex]\sum F_{y}=F_{N}-3825.9=0[/tex]
Solving for the [tex]F_{N}[/tex]:
[tex]F_{N}=$ $3825.\,\allowbreak9N[/tex]
The sum in x is:
[tex]\sum F_{x}=450+200-F_{f}=0[/tex]
Solving for the [tex]F_{f}[/tex]:
[tex]$F_{f}=650.0N[/tex]
The formula of the magnitude of the Friction force is
[tex]F_{f}=\mu F_{N}[/tex]
That means the coefficient of friction is:
[tex]\mu=\frac{F_{f}}{F_{N}}=\frac{650.0}{3825.\,\allowbreak9}=\allowbreak0.16989[/tex]
The amplitude of a wave decreases gradually as the wave travels down a long, stretched string. What happens to the energy of the wave when this happens? (The frequency is the same).
Answer:
energy decreases
Explanation:
The energy of a wave is directly proportional to the square of amplitude of a wave.
Amplitude is defined as the maximum displacement of the wave particle from its mean position.
So, as the amplitude goes on decreasing, the energy of the wave is also decreasing.
If the amplitude of a wave decreases gradually as the wave travels down, the energy of the wave increases.
What is a wave?A wave can be defined as a disturbance in a medium that progressively transport energy from a source to another location, especially without the transportation of matter.
The characteristics of a wave.In Science, some of the characteristics of a wave include the following:
FrequencyWavelengthAmplitudeSpeedGenerally, the energy of a wave is directly proportional to the square of wave amplitude and the square of its frequency. Thus, the energy of the wave increases when the amplitude of a wave decreases gradually as the wave travels down.
Read more on waves here: https://brainly.com/question/19250621
The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the station so she accelerates at 1.50 m/s2. What is the magnitude of the acceleration of the space station as the astronaut is pushing off the wall? Give your answer relative to an observer who is space walking and therefore does not accelerate with the space station due to the push.
Answer:
[tex]a = 5.83 \times 10^{-4} m/s^2[/tex]
Explanation:
Since the system is in international space station
so here we can say that net force on the system is zero here
so Force by the astronaut on the space station = Force due to space station on boy
so here we know that
mass of boy = 70 kg
acceleration of boy = [tex]1.50 m/s^2[/tex]
now we know that
[tex]F = ma[/tex]
[tex]F = 70(1.50) = 105 N[/tex]
now for the space station will be same as above force
[tex]F = ma[/tex]
[tex]105 = 1.8 \times 10^5 (a)[/tex]
[tex]a = \frac{105}{1.8 \times 10^5}[/tex]
[tex]a = 5.83 \times 10^{-4} m/s^2[/tex]
The magnitude of the acceleration of the space station is mathematically given as
[tex]a = 5.83 * 10^{-4} m/s^2[/tex]
What is the magnitude of the acceleration of the space station as the astronaut is pushing off the wall?Question Parameter(s):
The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg.
The station pushes off one wall of the station so she accelerates at 1.50 m/s2.
Generally, the equation for the Force is mathematically given as
F = ma
Therefore
F = 70(1.50)
F= 105 N
In conclusion, the space station force
F = ma
105 = 1.8 * 10^5 a
a = 5.83 * 10^{-4} m/s^2
Read more about Speed
https://brainly.com/question/4931057
a football of mass 0.38 kg is traveling towards a boy at 2.8ms-1. the boy kicks the ball with a force of 21N, reversing its direction. if the time of imapct between the football and boy’s foot is 250ms. what is the new speed of trhe ball?
Answer:
Final speed in reverse direction will be 11 m/s
Explanation:
As per impulse momentum theorem we know that
Change in momentum of the object is product of force and time
so we have
[tex]P_f - P_i = F\Delta t[/tex]
now we have
[tex]P_f = mv_f[/tex]
[tex]P_i = mv_i[/tex]
[tex]\Delta t = 250 ms[/tex]
F = 21 N
now we have
[tex]0.38(v_f) - 0.38(-2.8) = 21(250\times 10^{-3})[/tex]
[tex]v_f = 13.82 - 2.8[/tex]
[tex]v_f = 11 m/s[/tex]
A toroid of mean radius 26 cm and circular cross section of radius 1.9 cm is wound with a superconducting wire of length 1000 m that carries a current of 410 A. (a) What is the number of turns on the coil?
Answer:
8377 turns
Explanation:
Given:
Mean radius of toroid = 26 cm
Circular cross-section radius, r = 1.9 cm = 0.019
Length of the wire, l = 1000 m
The number of turns (n) on the coil is given by
[tex]n = \frac{l}{s}[/tex]
where,
l is the length of the wire
s is the circumference of the circular wire
⇒[tex]n = \frac{1000}{2\pi r}[/tex]
or
[tex]n = \frac{1000}{2\pi \times 0.019}[/tex]
or
n = 8376.57 or 8377 turns
hence, the number of turns on the coils is 8377
The number of turns of the coil is determined as 8377 turns.
Number of turnsThe number of turns of the superconducting wire is calculated as follows;
n = L/2πr
where;
L is the length of superconducting wire = 1000 mr is the radius of the circular cross section = 1.9 cm = 0.019 mSubstitute the given parameters and solve for the number of turns of the coil,
n = 1000/(2π x 0.019)
n = 8377 turns
Thus, the number of turns of the coil is determined as 8377 turns.
Learn more about number of turns here: https://brainly.com/question/25886292
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900c to 0.942c MeV
Answer:
The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.
Explanation:
We know that,
Mass of electron [tex]m_{e}=9.11\times10^{-31}\ kg[/tex]
Rest mass energy for electron = 0.511 Mev
(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev
Using formula of rest,
[tex]E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}[/tex]
[tex]E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}[/tex]
[tex]E=0.582\ Mev[/tex]
(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev
Using formula of rest,
[tex]E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}[/tex]
[tex]E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}[/tex]
[tex]E=0.350\ Mev[/tex]
Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.
A gas is equilibrium at T kelvin. If mass of one molecule is m and its component of velocity in x direction is v. Then mean of its v2 is 3kT 2kT 2 kT 3 (4) zero
Answer:
The value of [tex]v_{x}^2[/tex] is [tex]\dfrac{KT}{M}[/tex].
Explanation:
A gas is equilibrium at T kelvin.
Mass = M
We know that,
The average square of the velocity in the x,y and z direction are equal.
[tex]\bar{v}_{x}^2=\bar{v}_{y}^{2}=\bar{v}_{z}^{2}[/tex]
[tex]v_{rms}^2=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}[/tex]
[tex]v_{rms}^2=3v_{x}^2[/tex]
[tex]v_{x}^{2}=\dfrac{v_{rms}^2}{3}[/tex]
Equation of ideal gas
[tex]PV=RT[/tex]
[tex]P=\dfrac{nKT}{V}[/tex]
Here, R = nK
We know that,
[tex]\dfrac{nKT}{V}=\dfrac{nM}{3V}v_{rms}^2[/tex]
[tex]v_{rms}^2=\dfrac{3KT}{M}[/tex]....(I)
Put the value of [tex]v_{rms}^2[/tex] in the equation (I)
[tex]v_{x}^2=\dfrac{1}{3}\times\dfrac{3KT}{M}[/tex]
[tex]v_{x}^2=\dfrac{KT}{M}[/tex]
Hence, The value of [tex]v_{x}^2[/tex] is [tex]\dfrac{KT}{M}[/tex].
A force of 36.0 N is required to start a 3.0-kg box moving across a horizontal concrete floor. Part A) What is the coefficient of static friction between the box and the floor? Express your answer using two significant figures. Part B) If the 36.0-N force continues, the box accelerates at 0.54 m/s^2. What is the coefficient of kinetic friction? Express your answer using two significant figures.
Answer:
A) 1.2
B) 1.1
Explanation:
A)
F = force required to start the box moving = 36.0 N
m = mass of the box = 3 kg
[tex]F_{g}[/tex] = weight of the box = mg = 3 x 9.8 = 29.4 N
[tex]F_{n}[/tex] = normal force acting on the box by the floor
normal force acting on the box by the floor is given as
[tex]F_{n}[/tex] = [tex]F_{g}[/tex] = 29.4
[tex]F_{s}[/tex] = Static frictional force = F = 36.0 N
[tex]\mu _{s}[/tex] = Coefficient of static friction
Static frictional force is given as
[tex]F_{s}[/tex] = [tex]\mu _{s}[/tex] [tex]F_{n}[/tex]
36.0 = [tex]\mu _{s}[/tex] (29.4)
[tex]\mu _{s}[/tex] = 1.2
B)
a = acceleration of the box = 0.54 m/s²
F = force applied = 36.0 N
[tex]f_{k}[/tex] = kinetic frictional force
[tex]\mu _{k}[/tex] = Coefficient of kinetic friction
force equation for the motion of the box is given as
F - [tex]f_{k}[/tex] = ma
36.0 - [tex]f_{k}[/tex] = (3) (0.54)
[tex]f_{k}[/tex] = 34.38 N
Coefficient of kinetic friction is given as
[tex]\mu _{k}=\frac{f_{k}}{F_{n}}[/tex]
[tex]\mu _{k}=\frac{34.38}{29.4}[/tex]
[tex]\mu _{k}[/tex] = 1.1
When four people with a combined mass of 315 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.90 cm. (a) What is the effective force constant of the springs?
Answer:
3.43 x 10⁵ N/m
Explanation:
M = mass of the car alone = 2000 kg
m = mass of the four people combined = 315 kg
x = compression of the spring when there is car alone
x' = compression of the spring when there is car and four people = x + 0.009
k = spring constant
When there is car, the weight of the car is balanced by the spring force and the force equation is given as
k x = Mg
k x = (2000)(9.8)
k x = 19600 eq-1
When there is car and four people, the weight of the car and four people is balanced by the spring force and the force equation is given as
k x' = (M + m) g
k (x + 0.009) = (2000 + 315) (9.8)
k x + 0.009 k = (2000 + 315) (9.8)
using eq-1
19600 + 0.009 k = 22687
k = 3.43 x 10⁵ N/m
Final answer:
The effective force constant of the springs k = 343000 N/m
Explanation:
To solve for the effective force constant of the springs in a car's suspension system, we can apply Hooke's Law, which is stated as F = -kx. Here, F represents the restoring force supplied by the springs, k is the spring constant or force constant, and x is the displacement of the springs from their equilibrium position.
When the four people with a combined mass of 315 kg sit in the car, their weight causes an additional displacement of 0.90 cm, or 0.009 m.
Since weight acts as the restoring force F, we can calculate it as F = mg, where m is the total mass and g is the acceleration due to gravity (9.80 m/s²). So, F = (315 kg)(9.80 m/s²) = 3087 N.
Using Hooke's Law, we can rearrange the formula to solve for k: k = F/x. Plugging in the values gives us k = 3087 N / 0.009 m, which provides the spring constant for the suspension system.
The effective force constant of the springs k = 343000 N/m
Is the following statement true or false? The time constant of an RC circuit is the time it takes to completely charge or discharge the capacitor. O True False
Answer:
FALSE
Explanation:
As we know that equation for charging of a capacitor is given by
[tex]q = CV(1 - e^{-\frac{t}{\tau}})[/tex]
now at t = 0 we can say there is no charge on the capacitor
now as the time will increase the value of charge will increase on the plates of capacitor
now after t = 1 time constant
we will have
[tex]q = CV(1 - e^{-1})[/tex]
[tex]q = 0.63 CV[/tex]
so after one time constant the capacitor will charged upto 63% of its maximum value.
You heat a circular steel sheet of diameter 50.0 cm in an oven. What will be the radius for 50 K increase in temperature?
Answer:
R = 50.016 cm
Explanation:
Coefficient of thermal expansion for steel is given as
[tex]\alpha = 13 \times 10^{-6} per ^0C[/tex]
now as we know that the change in the length due to thermal expansion depends of change in temperature and initial length of the object
so here we will have
[tex]\Delta R = R_o \alpha \Delta T[/tex]
here we know that
[tex]R_o = \frac{50}{2} cm = 25 cm[/tex]
[tex]\Delta T = 50 K[/tex]
now we will have
[tex]\Delta R = (25)(13 \times 10^{-6})(50)[/tex]
[tex]\Delta R = 0.016 cm[/tex]
so final radius of the disc will be R = 50.016 cm
Is finding the distance between two points in the rectangular coordinate system related to the Pythagorean theorem? Why or why not? Provide an example in your answer.
Answer:
Yes it is related
Explanation:
Consider 2 points in a rectangular co-ordinate system
[tex](x_{1},y_{1}),(x_{2},y_{2})[/tex]
The distance between them can be found by
d = [tex]\sqrt{(x_{1}-x_{2})^{2}-(y_{2}-y_{1})^{^{2}}}[/tex]
This is clear from the figure attached below
Optics (Lens) Problem A lens forms an image of an object. The object is 16.0cm from the lens. The image is 12cm from the lens on the same side as the object. a. What is the focal length of the lens? Is the lens converging or diverging? b. If the object is 8.5mm tall, how tall is the image? Is it erect or inverted? c. Draw a principal ray diagram.
Answer:
Part a)
[tex]f = -48 cm[/tex]
Since focal length is negative so its a diverging lens
Part b)
[tex]h_i = 6.375 mm[/tex]
Since the magnification is position for diverging lens so it is ERECT
Explanation:
Part a)
As we know by lens formula
[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]
[tex]-\frac{1}{12cm} + \frac{1}{16} = \frac{1}{f}[/tex]
[tex]f = -48 cm[/tex]
Part b)
Since focal length is negative so its a diverging lens
Part c)
As we know that
[tex]\frac{h_i}{h_o} =\frac{d_i}{d_o}[/tex]
[tex]\frac{h_i}{8.5 mm} = \frac{12}{16}[/tex]
[tex]h_i = 6.375 mm[/tex]
Part d)
Since the magnification is position for diverging lens so it is ERECT
Part e)
A ray of light traveling in air is incident on the flat surface of a piece of glass at an angle of 65.9° with respect to the normal to the surface of the glass. If the ray refracted into the glass makes an angle of 34.8° with respect to the normal, what is the refractive index of the glass?
Answer:
[tex]n_{glass}[/tex] = 1.6
Explanation:
[tex]\theta _{i}[/tex] = Angle of incidence = 65.9°
[tex]\theta _{r}[/tex] = Angle of refraction = 34.8°
[tex]n_{air}[/tex] = Index of refraction of air = 1
[tex]n_{glass}[/tex] = Index of refraction of glass = ?
Using Snell's law
[tex]n_{air}[/tex] Sin[tex]\theta _{i}[/tex] = [tex]n_{glass}[/tex] [tex]\theta _{r}[/tex]
(1) Sin65.9 = [tex]n_{glass}[/tex] Sin34.8
[tex]n_{glass}[/tex] = 1.6
Since the index of refraction of air is 1, the refractive index of the glass is 1.6 approximately
Refractive IndexRefractive Index is the measure of refraction or bending when light passes from one medium to another.
Given that a ray of light traveling in air is incident on the flat surface of a piece of glass at an angle of 65.9° with respect to the normal to the surface of the glass. If the ray refracted into the glass makes an angle of 34.8° with respect to the normal, the refractive index of the glass can be calculated with the formula below
n = sin i / sin r
Where
n = refractive indexi = angle of incidencer = angle of refractionSubstitute all the parameters
n = sin 65.9 / sin 34.8
n = 0.913 / 0.5707
n = 1.599
n = 1.6 approximately
Therefore, the refractive index of the glass is 1.6 approximately.
Learn more about Refractive Index here: https://brainly.com/question/10729741