Answer:
I = 18 x 10⁻⁹ A = 18 nA
Explanation:
The current is defined as the flow of charge per unit time. Therefore,
I = q/t
where,
I = Average Current passing through nerve cell
q = Total flow of charges through nerve cell
t = time period of flow of charges
Here, in our case:
I = ?
q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C
t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s
Therefore,
I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)
I = 18 x 10⁻⁹ A = 18 nA
How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated volume? Which method might be more accurate? Why?
Answer:
The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate
Explanation:
This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error
A solenoid with 500 turns, 0.10 m long, carrying a current of 4.0 A and with a radius of 10-2 m will have what strength magnetic field at its center
Answer:
B = 0.025T
Explanation:
In order to calculate the strength of the magnetic field at the center of the solenoid, you use the following formula:
[tex]B=\frac{\mu N i}{L}[/tex] (1)
μ: magnetic permeability of vacuum = 4π*10^-7 T/A
N: turns of the solenoid = 500
i: current = 4.0A
L: length of the solenoid = 0.10m
You replace the values of the parameters in the equation (1):
[tex]B=\frac{(4\pi*10^{-7}T/A)(500)(4.0A)}{0.10m}=0.025T[/tex]
The strength of the magnetic field at the center of the solenoid = 0.025T
Answer:
Magnetic field strength at the center is 2.51x10^-2T
Explanation:
Pls see attached file for step by step calculation
Nerve impulses in a human body travel at a speed of about 100 m/s. Suppose a woman accidentally steps barefoot on a thumbtack. About how much time does it take the nerve impulse to travel from the foot to the brain (in s)
The gas law for a fixed mass m of an ideal gas at absolute temperature T, pressure P, and volume V is PV=mRT, where R is the gas constant. Find the partial derivatives
Answer: Question for edmentum : Part D
Next, consider a scientific law known as the ideal gas law. This law deals with the relationship between temperature and pressure of a fixed mass of gas. The law states that the product of the pressure (P) and volume (V) is proportional to the temperature (T) of the mass of gas. The equation can be expressed this way, with k as a constant value:
PV = kT
Based on this equation, what are the two possible outcomes of heating up a certain mass of air?
Explanation: edmentum's sample Answer
The pressure of the gas or the volume of the gas will increase as the temperature increases. That’s because, according to the ideal gas law, pressure and volume are directly proportional to temperature. The value of k does not change.
Ideal gas law states the relation between pressure, volume, temperature and the no.of moles of a gas. The partial derivatives of this equation can be written as [tex](\frac{\partial P}{\partial T})_{n,V} = (\frac{\partial }{\partial T}\frac{nRT}{V})_{n,V}[/tex].
What is ideal gas law?According to ideal gas law the product of pressure and volume of an ideal gas is equal to the product of number of moles, temperature and universal gas constant R.
Thus, PV = nRT.
The partial derivatives are used to find the differentials of some variable with respect to other variable provided that some of the variables are constant.
We can differentiate the ideal gas law where n and R are constants. With respect to constant volume and no.of moles the PV can be differentiated as :[tex](\frac{\partial P}{\partial T})_{n,V}[/tex]. Where the constants are provided below the brackets.
Similarly the right side also can be differentiated with respect to change in temperature at constant volume and pressure as [tex](\frac{\partial }{\partial T}\frac{nRT}{V})_{n,V}[/tex].
Therefore, both these differentials are equal and thus can be equated as [tex](\frac{\partial P}{\partial T})_{n,V} = (\frac{\partial }{\partial T}\frac{nRT}{V})_{n,V}[/tex].
The same can be done with respect to change in pressure also.
To find more about ideal gas law, refer the link below:
https://brainly.com/question/13821925
#SPJ2
Will an object present on the surface of the earth possess any energy?
Answer:
An object lying on the surface of the earth has energy like it maybe kinetic or potential energy
Explanation:
Hope it will help you :)
In the child's game of tetherball, a rope attached to the top of a tall pole is tied to a ball. Players hit the ball in opposite directions in an attempt to wrap the ball and rope around the pole. Assume the rope has negligible mass and that resistive forces, such as air resistance and friction, can be neglected. As the ball wraps around the pole between hits, how does the angular speed of the ball change
Answer:
The angular speed of the ball will increase
Explanation:
the angular speed of the ball will increase because the force of hit by the players will sum up in opposite direction to increase the angular speed
A diver running at 2.5 m/s dives out horizontally from the edge of a vertical cliff and 3.0 seconds later reaches the water below. How far from its base did the diver hit the water
Explanation:
u = 2.5 m/s
v = 0
t = 3sec
s = ?
s = (u+v)/t
s = (0+2.5)/3
s = 2.5/3 = 0.83 m
A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I1 is the moment of inertia of this object with respect to an axis passing through the center of the rod and perpendicular to it and I2 is the moment of inertia with respect to an axis passing through one of the masses, it follows that:
a. I1 > I2
b. I2 > I1.
c. I1 = I2.
Answer:
B: I2>I1
Explanation:
See attached file
A boat that has a speed of 6km / h must cross a 200m wide river perpendicular to the current that carries a speed of 1m / s. Calculate a) the final speed of the boat b) displacement experienced by the boat in the direction of the current when making the journey
Answer:
a) 1.94 m/s
b) 120 m
Explanation:
Convert km/h to m/s:
6 km/h = 1.67 m/s
a) The final speed is found with Pythagorean theorem:
v = √((1.67 m/s)² + (1 m/s)²)
v = 1.94 m/s
b) The time it takes the boat to cross the river is:
t = (200 m) / (1.67 m/s)
t = 120 s
The displacement in the direction of the current is:
x = (1 m/s) (120 s)
x = 120 m
A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 344-g rock that displaces 216 cm3 of water
Answer:
density d = 1.59 g/cm^3
The density of the rock is 1.59 g/cm^3
Explanation:
The density of an object can be derived by measuring its mass and then measuring its volume by submerging it in a graduated cylinder.
Density = mass/volume of water displaced
d = m/v ........1
Given;
mass m = 344 g
Volume of water displaced v = 216 cm^3
from equation 1, we can calculate the value of the density;
Substituting the given values;
d = 344/216 g/cm^3
d = 1.592592592592 g/cm^3
d = 1.59 g/cm^3
The density of the rock is 1.59 g/cm^3
A 30.0-g object moving to the right at 19.5 cm/s overtakes and collides elastically with a 13.0-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision. (Take the positive direction to be to the right. Indicate the direction with the sign of your answer.) 30.0-g object Incorrect: Your answer is incorrect. seenKey 16.8 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. cm/s 13.0-g object
Answer:
[tex]V(30g)=1.4\frac{cm}{s} \\V(13g)=5.9\frac{cm}{s}[/tex]
Explanation:
So my calculations may be off (the final step is plugging in a bunch of things and getting a value, which creates room for error [and they honestly seem too small]) but I'm confident that the process is correct. I'll upload my work shortly but here is the method:
Use the Principle of Conservation of Momentum (PCoM) to set up a statement between the momentum of each object before and after the collision. It should be [before]=[after] because the collision is said to be elastic.Do the same thing using the Principle of Conservation of Energy (PCoE). Note that you can start with either one, the problem can't really be done without both anyways.You can chose to either divide the PCoM expression by the PCoE expression or do a series of substitutions. If you do the sub., you'll find that it's the same thing as just dividing them outright. (This is a little time saving trick I picked up from one of my Physics professors, I definitely didn't come up with it on my own.)After that, you should be able to reduce the expressions into something with only the two initial velocities and the two final velocities. Note that in order to reduce, you'll need to use the "difference of squares" relationship in the PCoE expression. Choose either one of the final velocities to solve for first via substitution. In my example, I chose to solve for the 30g mass's final velocty first by subbing out the 13g mass's final velocity. (If you do it the other way around, your work might look different from mine at the end but it should give the same answer either way.)After substituting, you should have an expression where the final velocity equals some combination of the masses and initial velocities of both objects (it's not a "nice" looking equation).With one final velocity found, you can plug it into the expression you found in step 4. to find the other mass's final velocity.And with that, you've got it! This whole process is kinda long and involved so I would try practicing it a lot before any tests/quizzes so it doesn't eat up your time.
Edit: In my work I made u to be the initial velocities and v to be the final velocities because it was easier to keep track of
Answer:
The final velocity of the 30 g object is 16.8 cm/s
The final velocity of the 13 g object is 21.3 cm/s
Explanation:
Let's study the elastic collision with conservation of linear momentum, assigning object 1 to the 30 g object, and object 2 to the 13 gr object:
[tex]p_{1\,i}+p_{2\,i}=p_{1\.f}+p_{2\,f}\\(30)\,(19.5)+(13)\,(15) = (30)\,v_{1\,f} +(13)\,v_{2\,f}\\780 = (30)\,v_{1\,f} +(13)\,v_{2\,f}[/tex]
so we can write one of the unknowns in terms of the other one:
[tex]v_{1\,f}=(780-13\,v_{2\,f})/30[/tex]
Now we analyze the equation for conservation of kinetic energy that verifies in elastic collisions:
[tex]\frac{30}{2} (19.5)^2+\frac{13}{2} \,(15)^2=\frac{30}{2} (v_{1\,f})^2+\frac{13}{2} \,(v_{2\,f})^2\\7166.25=15\, (v_{1\,f})^2+6.5\,(v_{2\,f})^2[/tex]
now we can write this quadratic equation replacing [tex]v_{1\,f}[/tex] with its expression in terms of [tex]v_{2\,f}[/tex] and solve it (with the help of a graphing calculator is simpler by looking for the roots).
We get two answers for [tex]v_{2\,f}[/tex] : one 15 cm/s, and the other one 21.28 cm/s.
We select the 21.28 cm/s answer since otherwise, the situation is the same as the initial one at which the second object was moving at 15 cm/s.
This velocity can be rounded to one decimal to: 21.3 cm/s
Given the value 21,28 for [tex]v_{2\,f}[/tex] , then:
[tex]v_{1\,f}=(780-13\,v_{2\,f})/30\\v_{1\,f}=(780-13\,(21.28))/30=16.78 \,\,cm/s[/tex]
which can be rounded to 16.8 cm/s
A 1,470-N force pushes a 500-kg piano up along a ramp. What is the work done by the 1,470-N pushing force on the piano as it moves 10 m up the ramp
Answer:
W = 14700 J
Explanation:
This is an exercise on Newton's second law.
To solve it we must fix a coordinate system, the most common is an axis parallel to the ramp and the other perpendicular axis, we write Newton's second law
Y Axis . Perpendicular to the ramp
N - Wy = 0
X axis. Parallel to the ramp, we assume it is positive when the ramp is going up
F - Wx = m a
in this case F = 1470 N and it is parallel to the plane.
Work is defined by
W = F .d
boldface indicates vectors
W = F d cos θ
let's calculate
W = 1470 10 cos 0
W = 14700 J
In a two-slit experiment, monochromatic coherent light of wavelength 500 nm passes through a pair of slits separated by 1.30 x 10-5 m. At what angle away from the centerline does the first bright fringe occur
Answer:
2.20°
Explanation:
For the central bright spot, we will use the constructive pattern for a double slit interference,
[tex]m\times w = d \times Sin\beta[/tex]
where w indicates the wavelength
and [tex]\beta[/tex] indicates the angle between the bright spot and center line.
now we will use the given values,
1 × 500 × 10^-9 = 1.3 × 10^-5 × Sin [tex]\beta[/tex]
Solving for [tex]\beta[/tex],
[tex]\beta[/tex] = 2.204° ~ 2.20°
Therefore the correct answer is 2.20°
A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a=a1+Fm where a1=3.00 meter/second2, F=12.0kilogram⋅meter/second2 and m=7.00kilogram. what is the value of a?
Complete Question
Now if you look at the equation for acceleration given in the question i.e
[tex]a = a_1 + F * m[/tex]
We see that evaluating it in terms of dimension it is incorrect instead the equation should be
[tex]a = a_1 + \frac{F}{m}[/tex]
So in the solution below we will be making use of [tex]a = a_1 + \frac{F}{m}[/tex]
Answer:
The values of a is [tex]a = 4.714 \ m/s^2[/tex]
Explanation:
From the question we are told that
The expression for the acceleration is [tex]a = a_1 + \frac{F}{m}[/tex]
The value of [tex]a_ 1 = 3.0 \ m/s^2[/tex]
The values of [tex]F = 12.0 \ kg \cdot m/s^2[/tex]
The values of m is [tex]m = 7.0 \ kg[/tex]
substituting values
[tex]a = 3 + \frac{12}{7}[/tex]
[tex]a = 4.714 \ m/s^2[/tex]
A particle confined to a motion along the x axis moves with a constant acceleration of 2.5m/s2. Its velocity at t=0s is 6m/s. Find its velocity at t=4s.
Answer:
v = 16 m/s
Explanation:
It is given that,
Acceleration of a particle along x -axis is [tex]2.5\ m/s^2[/tex]
At t = 0s, its velocity is 6 m/s
We need to find the velocity at t = 4 s
It means that the initial velocity of the particle is 6 m/s
Let v is the velocity at t = 4 s
So,
v = u + at
[tex]v=6+2.5\times 4\\\\v=16\ m/s[/tex]
So, the velocity at t = 4 s is 16 m/s.
Answer:
v = 16 m/s
Explanation:
It is given that,
Acceleration of a particle along x -axis is
At t = 0s, its velocity is 6 m/s
We need to find the velocity at t = 4 s
It means that the initial velocity of the particle is 6 m/s
Let v is the velocity at t = 4 s
So,
v = u + at
So, the velocity at t = 4 s is 16 m/s.
What is the length (in m) of a tube that has a fundamental frequency of 108 Hz and a first overtone of 216 Hz if the speed of sound is 340 m/s?
Answer:
Length of a tube = 1.574 m
Explanation:
Given:
Fundamental frequency (f1) = 108 Hz
First overtone (f2) = 216 Hz
Speed of sound (v) = 340 m/s
Find:
Length of a tube
Computation:
We know that,
f = v / λ
f = nv / 2L [n = number 1,2,3]
So,
f1 = 1(340) / 2L
f1 = 170 / L
L = 170 / 108 = 1.574 m
f2 = 2(340) / 2L
L = 340 / 216
L = 1.574 m
A commercial diffraction grating has 500 lines per mm. Part A When a student shines a 480 nm laser through this grating, how many bright spots could be seen on a screen behind the grating
Answer:
The number of bright spot is m =4
Explanation:
From the question we are told that
The number of lines is [tex]s = 500 \ lines / mm = 500 \ lines / 10^{-3} m[/tex]
The wavelength of the laser is [tex]\lambda = 480 nm = 480 *10^{-9} \ m[/tex]
Now the the slit is mathematically evaluated as
[tex]d = \frac{1}{s} = \frac{1}{500} * 10^{-3} \ m[/tex]
Generally the diffraction grating is mathematically represented as
[tex]dsin\theta = m \lambda[/tex]
Here m is the order of fringes (bright fringes) and at maximum m [tex]\theta = 90^o[/tex]
So
[tex]\frac{1}{500} * sin (90) = m * (480 *10^{-3})[/tex]
=> [tex]m = 4[/tex]
This implies that the number of bright spot is m =4
Forces that act in pairs are _____ in size and ________ in direction.
A certain lightning bolt moves 40.0 C of charge. How many units ???? of fundamental charge e is this?
q = 40 C
e = 1.6×10^-19 C
n = ?
n = q/e
n = 40/1.6×10^-19 C
= 2.6×10^20
Which of the following changes will increase the frequency of the lowest frequency standing sound wave on a stretching string?Choose all that apply.A. Replacing the string with a thicker stringB. Plucking the string harderC. Doubling the length of the string
Answer:
A, C
Explanation:
Since the frequency is inversely proportional to the length of a string, then I want to increase the frequency of the lowest
A. Replacing the string with a thicker string.
Thicker strings have more density. The more density the string has, the lower the sound.
Mathematically, we can see the proportionality (direct and inverse) by looking at those formulas for Frequency and Speed, when combined:
For:
[tex]f=\frac{v}{\lambda}[/tex]
[tex]f=\frac{v}{\lambda}*\sqrt{\frac{T}{D} }[/tex]
See above, how density (D) and [tex](\lambda)[/tex] wave length are inversely proportional.
C. Doubling the length of the string.
Because the length of the string is inversely proportional to the frequency.
The longer the string, the lower the frequency.
So, if we double string, we'll hear lower sounds in any string instrument
--
In short, for A, and C We can justify both since length and density are inversely proportional to the Frequency, we need longer or thicker string.
A 750 kg car is moving at 20.0 m/s at a height of 5.0 m above the bottom of a hill when it runs out of gas. From there, the car coasts. a. Ignoring frictional forces and air resistance, what is the car’s kinetic energy and velocity at the bottom of the hill
Answer:
Explanation:
Kinetic energy at the height = 1/2 m v²
= 1/2 x 750 x 20²
= 150000 J
Its potential energy = mgh
= 750 x 9.8 x 5
=36750 J
Total energy = 186750 J
Its total kinetic energy will be equal to 186750 J , according to conservation of mechanical energy
If v be its velocity at the bottom
1/2 m v² = 186750
v = √498
= 22.31 m /s
A projectile is launched with V0 = 7.6 m/s and initial angle = 1.27 radians above the horizontal. What is the initial horizontal component of the projectile velocity in miles per hour?
Answer:
The horizontal component is [tex]v_h = 1.7096 \ m/s[/tex]
Explanation:
A diagram illustrating the projection is shown on the first uploaded image (from IB Maths Resources from British international school Phuket )
From the question we are told that
The initial velocity is [tex]v_o = 7.6 \ m/s[/tex]
The angle of projection is [tex]\theta = 1.27 \ rad = 72.77^o[/tex]
The horizontal component of this projectile velocity is mathematically represented as
[tex]v_h = v_o * cos (\theta )[/tex]
substituting values
[tex]v_h = 7.6 * cos (72.77 )[/tex]
[tex]v_h = 1.7096 \ m/s[/tex]
Approximately how many asteroids that are 0.98 km in radius would it take to make a planet which has a radius of 6420.0 km
Answer:
It'd take approximately 6543 asteroids.
Explanation:
In order to calculate the number of asteroids needed to make that planet we first need to determine the volume of each object. To do this we will consider them as spheres and apply the appropriate formula:
[tex]V_{asteroid} = \frac{4*\pi*0.98}{3} = 4.11 \text{ km}^3[/tex]
[tex]V_{planet} = \frac{4*\pi*6420}{3} = 26892.03 \text{ km}^3[/tex]
The number of asteroids needed are given by the division of planet's volume by the asteroid's volume.
[tex]n = \frac{26892.03}{4.11} = 6543.07[/tex]
It'd take approximately 6543 asteroids.
A very large sheet of a conductor carries a uniform charge density of on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Complete Question
A very large sheet of a conductor carries a uniform charge density of [tex]4.00\ pC/mm^2[/tex] on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Answer:
The electric field is [tex]E = 4.5198 *10^{5} \ N/C[/tex]
Explanation:
From the question we are told that
The charge density is [tex]\sigma = 4.00pC /mm^2 = 4.00 * 10^{-12 } * 10^{6} = 4.00 *10^{-6}C/m[/tex]
The position outside the surface is [tex]a = 3.00 \ mm = 0.003 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon _o }[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with values [tex]\epsilon _o = 8.85 *10^{-12} F/m[/tex]
substituting values
[tex]E = \frac{4.0*10^{-6}}{8.85 *10^{-12} }[/tex]
[tex]E = 4.5198 *10^{5} \ N/C[/tex]
A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region is 5.40 mlong and reduces the toboggan's speed by 1.20 m/s .
a) What average friction force did the rough region exert on the toboggan?
b) By what percent did the rough region reduce the toboggan's kinetic energy?
c) By what percent did the rough region reduce the toboggan's speed?
Answer:
a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.
Explanation:
a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:
[tex]K_{1} = K_{2} + W_{f}[/tex]
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] are the initial and final translational kinetic energies of the tobbogan, measured in joules.
[tex]W_{f}[/tex] - Dissipated work due to friction, measured in joules.
By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:
[tex]f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})[/tex]
Where:
[tex]f[/tex] - Friction force, measured in newtons.
[tex]\Delta s[/tex] - Distance travelled by the toboggan in the rough region, measured in meters.
[tex]m[/tex] - Mass of the toboggan, measured in kilograms.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the toboggan, measured in meters per second.
The friction force is cleared:
[tex]f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}[/tex]
If [tex]m = 375\,kg[/tex], [tex]v_{1} = 4.50\,\frac{m}{s}[/tex], [tex]v_{2} = 1.20\,\frac{m}{s}[/tex] and [tex]\Delta s = 5.40 \,m[/tex], then:
[tex]f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}[/tex]
[tex]f = 653.125\,N[/tex]
The average friction force exerted on the toboggan is 653.125 newtons.
b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:
[tex]\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%[/tex]
[tex]\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%[/tex]
[tex]\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%[/tex]
[tex]\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%[/tex]
[tex]\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%[/tex]
If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:
[tex]\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%[/tex]
[tex]\%K_{loss} = 92.889\,\%[/tex]
The rough region reduced the kinetic energy of the toboggan in 92.889 %.
c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:
[tex]\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%[/tex]
[tex]\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%[/tex]
If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:
[tex]\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%[/tex]
[tex]\%v_{loss} = 73.333\,\%[/tex]
The speed of the toboggan is reduced in 73.333 %.
The average frictional force exerted on the toboggan by the rough surface is 661.5 N.
The percentage of the toboggan kinetic energy reduction is 7.11%.
The percentage of the toboggan speed reduction is 26.67%.
The given parameters;
mass of the toboggan, m = 375 kginitial speed of the toboggan, u = 4.5 m/slength of the rough region, d = 5.4 mfinal speed of the toboggan, v = 1.2 m/sThe normal force on the toboggan is calculated as follows;
Fₙ = mg
Fₙ = 375 x 9.8 = 3675 N
The acceleration of the toboggan is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2 }{2s} \\\\a = \frac{(1.2)^2 - (4.5)^2 }{2(5.4)}\\\\a = -1.74 \ m/s^2[/tex]
The coefficient of friction is calculated as follows;
[tex]\mu_k = \frac{a}{g} \\\\\mu_k = \frac{1.74}{9.8} \\\\\mu_k = 0.18[/tex]
The average frictional force exerted on the toboggan by the rough surface;
[tex]F_k = \mu_k F_n\\\\F_k = 0.18 \times 3675\\\\F_k = 661.5 \ N[/tex]
The initial kinetic energy of the toboggan is calculated as follows;
[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 375\times 4.5^2\\\\K.E_i = 3,796.88 \ J[/tex]
The final kinetic energy of the toboggan is calculated as follows;
[tex]K.E_f = \frac{1}{2} mv^2\\\\K.E_f = \frac{1}{2} \times 375\times 1.2^2\\\\K.E_f = 270 \ J[/tex]
The percentage of the toboggan kinetic energy reduction is calculated as follows;
[tex]\frac{K.E_f}{K.E_i} \times 100\% = \frac{270}{3796.88} \times 100\% = 7.11 \%[/tex]
The percentage of the toboggan speed reduction is calculated as follows;
[tex]\frac{1.2}{4.5} \times 100\% = 26.67 \%[/tex]
Learn more here: https://brainly.com/question/14121363
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about _____ years.
Answer:
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.
Explanation:
Given;
orbital period of 3 years, P = 3 years
To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.
Kepler's third law;
P² = a³
where;
P is the orbital period
a is the orbital semi-major axis
(3)² = a³
9 = a³
a = [tex]a = \sqrt[3]{9} \\\\a = 2.08 \ years[/tex]
Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.
A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 1 kilograms is tied to the middle of the clothesline, it sags a distance of 4 meters. What is the magnitude of the tension on the ends of the clothesline
Answer:
The tension on the clotheslines is [tex]T = 8.83 \ N[/tex]
Explanation:
The diagram illustrating this question is shown on the first uploaded image
From the question we are told that
The distance between the two poles is [tex]d = 12 \ m[/tex]
The mass tie to the middle of the clotheslines [tex]m = 1 \ kg[/tex]
The length at which the clotheslines sags is [tex]l = 4 \ m[/tex]
Generally the weight due to gravity at the middle of the clotheslines is mathematically represented as
[tex]W = mg[/tex]
let the angle which the tension on the clotheslines makes with the horizontal be [tex]\theta[/tex] which mathematically evaluated using the SOHCAHTOA as follows
[tex]Tan \theta = \frac{ 4}{6}[/tex]
=> [tex]\theta = tan^{-1}[\frac{4}{6} ][/tex]
=> [tex]\theta = 33.70^o[/tex]
So the vertical component of this tension is mathematically represented a
[tex]T_y = 2* Tsin \theta[/tex]
Now at equilibrium the net horizontal force is zero which implies that
[tex]T_y - mg = 0[/tex]
=> [tex]T sin \theta - mg = 0[/tex]
substituting values
[tex]T = \frac{m*g}{sin (\theta )}[/tex]
substituting values
[tex]T = \frac{1 *9.8}{2 * sin (33.70 )}[/tex]
[tex]T = 8.83 \ N[/tex]
A proton of mass and a charge of is moving through vacuum at a constant velocity of 10000 directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of E =3.62e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction.
Required:
How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field.
Complete Question
A proton of mass mp= 1.67×10−27 kg and a charge of qp= 1.60×10−19 C is moving through vacuum at a constant velocity of 10,000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of E = 3.62e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction. How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field. You may neglect the effects of friction and gravity, and assume that the electric field is zero outside the specified region. Answer is to be in units of meters
Answer:
[tex]s = 0.039 \ m[/tex]
Explanation:
From the question we are told that
The mass of the proton is [tex]m = 1.67 *10^{-27} \ g[/tex]
The charge of on the proton is [tex]q = 1.60 *10^{-19} \ C[/tex]
The speed of the proton is [tex]v = 10000 \ m/s[/tex]
The magnitude of the electric field is [tex]E = 3.62*10^{3 } \ N/C[/tex]
The width covered by the electric field [tex]d = 5mm = 5 *10^{-3} \ m[/tex]
Generally the acceleration of the proton due to the electric toward the south (at the point where the force on the proton is equal to the electric force due to the electric field) is mathematically represented as
[tex]a = \frac{q* E}{m}[/tex]
Substituting values
[tex]a = \frac{1.60*10^{-19 } * 3.26 *10^{3}}{ 1.67*10^{-27}}[/tex]
[tex]a = 3.12*10^{11} \ m/s^2[/tex]
Generally the time it will take the proton to cross the electric field is mathematically represented as
[tex]t = \frac{d}{v}[/tex]
Substituting values
[tex]t = \frac{5 *10^{-3}}{10000}[/tex]
[tex]t = 5 *10^{-7} \ s[/tex]
Generally the the distance covered by the proton toward the south is
[tex]s = ut + \frac{1}{2} * a*t^2[/tex]
Here u = 0 m/s this because before the proton entered the electric field region the it velocity towards the south is zero
So
[tex]s = \frac{1}{2} * a*t^2[/tex]
Substituting values
[tex]s = \frac{1}{2} * 3.12 *10^{11}*(5 *10^{-7})^2[/tex]
[tex]s = 0.039 \ m[/tex]
Five identical cylinders are each acted on by forces of equal magnitude. Which force exerts the biggest torque about the central axes of the cylinders
Answer:
From the image, the force as shown in option A will exert the biggest torque on the cylinder about its central axes.
Explanation:
The image is shown below.
Torque is the product of a force about the center of rotation of a body, and the radius through which the force acts. For a given case such as this, in which the cylinders are identical, and the forces are of equal magnitude, the torque at the maximum radius away from the center will exert the maximum torque. Also, the direction of the force also matters. To generate the maximum torque, the force must be directed tangentially away from the circle formed by the radius through which the force acts away from the center. Option A satisfies both condition and hence will exert the most torque on the cylinder.
When a nerve cell fires, charge is transferred across the cell membrane to change the cell's potential from negative to positive. For a typical nerve cell, 9.2pC of charge flows in a time of 0.52ms .What is the average current through the cell membrane?
Answer:
The average current will be "17.69 nA".
Explanation:
The given values are:
Charge,
q = 9.2 pC
Time,
t = 0.52ms
The equivalent circuit of the cell surface is provided by:
⇒ [tex]i_{avg}=\frac{charge}{t}[/tex]
Or,
⇒ [tex]i_{avg}=\frac{q}{t}[/tex]
On substituting the given values, we get
⇒ [tex]=\frac{9.2\times 10^{-12}}{0.52\times 10^{-3}}[/tex]
⇒ [tex]=17.69^{-9}[/tex]
⇒ [tex]=17.69 \ nA[/tex]