Answer:
a) -z direction, b) +z direction, c) F=0
Explanation:
The magnetic force is given by the expression
F = q v x B
the bold indicate vectors, this equation can be separated in its module
F = a v B sin θ
and where θ is the angles between the speed and the magnetic field.
The direction of the force can be found with the right-hand rule. For a positive charge, the thumb goes in the direction of speed, the fingers extended in the direction of the magnetic field and the palm points in the direction of the force, if the charge is negative the force is in the opposite direction.
a) Let's apply this to our case
the proton is positively charged
moves in the direction of + x
The magnetized field goes in the direction of y
therefore applying the right hand rule the force must be in the direction of the negative part of the z-axis (-z)
The right-hand rule is used to find this address.
b) in this case it indicates that the proton moves in the recode of -y
again we apply the right hand rule and the force is in the direction of + z
c) The proton moves in the x direction
In this case the force is zero because the angle between the field and the speed is zero and the sine is zero, therefore the force is zero
Two 40 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. Part A How much power is dissipated by each bulb
Answer:
The power dissipated by each bulb is [tex]P = 10.0 \ W[/tex]
Explanation:
From the question we are told that
The power rating of both bulbs is [tex]P = 40 \ W[/tex]
The voltage rating of both bulb is [tex]V = 120 \ V[/tex]
The both bulbs are connected a voltage of [tex]V_C = 120 V[/tex]
The amount of power rating of each bulb is mathematically represented as
[tex]P = \frac{V^2}{R }[/tex]
=> [tex]R = \frac{V^2}{P}[/tex]
substituting values
[tex]R = \frac{ (120)^2}{40}[/tex]
[tex]R = 360 \Omega[/tex]
Now given that the bulbs are connected is series, the equivalent resistance is evaluated as
[tex]R_{eq } = R +R[/tex]
substituting values
[tex]R_{eq } = 360 + 360[/tex]
[tex]R_{eq } =720 \ \Omega[/tex]
The current flowing through the bulbs is mathematically evaluated as
[tex]I =\frac{V_C}{R_{eq}}[/tex]
substituting values
[tex]I =\frac{120}{720}[/tex]
[tex]I = 0.1667 \ A[/tex]
Now the power dissipated by both bulbs is mathematically represented as
[tex]P = I ^2 * R[/tex]
substituting values
[tex]P = 0.1668^2 * 360[/tex]
[tex]P = 10.0 \ W[/tex]
The power that should be dissipated by each bulb is P = 10.0 W.
Calculation of the power:
Since
The power rating of both bulbs is P = 40 W.
The voltage rating of both bulbs is V = 120 V.
And, both bulks that should be connected a voltage of Vc = 120V
Now the amount of power that should be rated of each bulb should be
P = V^2/R
So, R = V^2/P
= 120^2/40
= 360Ω
The equivalent resistance should be
I = Vc/Req
= 120/720
= 0.1667 A
Now the power is = 0.1668^2 * 360
= 10.0 W
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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 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
It is known that the gravitational force of attraction between two alpha particles is much weaker than the electrical repulsion. For two alpha particles at a distance d apart, calculate the ratio of the size of the gravitational attraction to that of the electrical repulsion. Specifically, find the magnitude of Gravitational/Electrical.
Answer:
The ratio of gravitational force to electrical force is 3.19 x 10^-36
Explanation:
mass of an alpha particle = 6.64 x [tex]10^{-27}[/tex] kg
charge on an alpha particle = +2e = +2(1.6 x [tex]10^{-19}[/tex] C) = 3.2 x [tex]10^{-19}[/tex] C
distance between particles = d
For gravitational attraction:
The force of gravitational attraction F = [tex]\frac{Gm^{2} }{r^{2} }[/tex]
where G = gravitational constant = 6.67 x [tex]10^{-11}[/tex] m^3 kg^-1 s^-2
r = the distance between the particles = d
m = the mass of each particle
therefore, gravitational force = [tex]\frac{6.67*10^{-11}*(6.64*10^{-27} )^{2} }{d^{2} }[/tex] = [tex]\frac{2.94*10^{-63} }{d^{2} }[/tex] Newton
For electrical repulsion:
Electrical force between the particles = [tex]\frac{-kQ^{2} }{r^{2} }[/tex]
where k is the Coulomb's constant = 9.0 x [tex]10^{9}[/tex] N•m^2/C^2
r = distance between the particles = d
Q = charge on each particle
therefore, electrical force = [tex]\frac{-9*10^{9}*(3.2*10^{-19} )^{2} }{d^{2} }[/tex] = [tex]\frac{-9.216*10^{-28} }{d^{2} }[/tex] Newton
the negative sign implies that there is a repulsion on the particles due to their like charges.
Ratio of the magnitude of gravitation to electrical force = [tex]\frac{2.94*10^{-63} }{9.216*10^{-28} }[/tex]
==> 3.19 x 10^-36
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)
a block of wood is pulled by a horizontal string across a rough surface at a constant velocity with a force of 20N. the coefficient of kinetic friction between the surfaces is 0.3 the force of the friction is
Answer:
6 N
Explanation:
From the laws of friction
F = ¶R = 0.3 × 20 = 6 N
The force of friction opposing the block's motion is 6 N.
The given parameters;
force applied on the block, F = 20 Ncoefficient of kinetic friction = 0.3The force of friction which opposes the motion of the block is obtained by applying Newton's second law of motion.
F = ma
Fₓ = μF
Substitute the given parameters to calculate the frictional force on the object.
Fₓ = 0.3 x 20
Fₓ = 6 N
Thus, the force of friction opposing the block's motion is 6 N.
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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
An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
The new voltage between the plates of the capacitor is 18 V
Explanation:
The charge on parallel plate capacitor is calculated as;
q = CV
Where;
V is the battery voltage
C is the capacitance of the capacitor, calculated as;
[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]
[tex]q = \frac{\epsilon _0A V}{d}[/tex]
where;
ε₀ is permittivity of free space
A is the area of the capacitor
d is the space between the parallel plate capacitors
If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;
[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]
Therefore, the new voltage between the plates of the capacitor is 18 V
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 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 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
An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.38 m/s, and its maximum acceleration is 6.83 m/s2. How much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum
Answer:
t = 0.31s
Explanation:
In order to calculate the time that the object takes to travel from the point with its maximum speed to the point with the maximum acceleration, you first use the following formulas, for the maximum speed and the maximum acceleration:
[tex]v_{max}=\omega A\\\\a_{max}=\omega^2A[/tex]
A: amplitude
v_max = 1.38m/s
a_max = 6.83m/s^2
w: angular frequency
From the previous equations you can obtain the angular frequency w.
You divide vmax and amax, and solve for w:
[tex]\frac{v_{max}}{a_{max}}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega}\\\\\omega=\frac{a_{max}}{v_{max}}=\frac{6.83m/s^2}{1.38m/s^2}=4.94\frac{rad}{s}[/tex]
Next, you take into account that the maximum speed is obtained when the object passes trough the equilibrium point, and the maximum acceleration for the maximum elongation, that is, the amplitude. In such a trajectory the time is T/4 being T the period.
You calculate the period by using the information about the angular frequency:
[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{4.94rad/s}=1.26s[/tex]
Then the required time is:
[tex]t=\frac{T}{4}=\frac{1.26s}{4}=0.31s[/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.
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Answer: Net electrostatic force on C is 24.2×[tex]10^{-2}[/tex] Newtons.
Explanation: Coulomb's Law is used to determine Electrostatic Force. Its formula is:
F = k.[tex]\frac{q_{0}.q_{1}}{r^{2}}[/tex]
where:
k is electrostatic constant (k = 8.987×[tex]10^{9}[/tex] Nm²/C²);
q is the charge of the object in Coulumb;
r is the distance between charges;
The net force is the sum of all the forces acting on C, so:
Force B on C:
They are both positive, so there is a relpusive force acting between them on the y-axis.
[tex]F_{BC} = 8,987.10^{9}.\frac{4.35.10^{-3}.9.67.10^{-4}}{(6.14.10^{2})^{2}}[/tex]
[tex]F_{BC} = 10.03.10^{-2}[/tex] N
Force D on C:
There is an atractive force between them on the x-axis.
[tex]F_{CD} = 8.987.10^{9}.\frac{9.67.10^{-4}.1.92.10^{-3}}{(1.42.10^{3})^{2}}[/tex]
[tex]F_{CD} = 13.64.10^{-4}[/tex] N
Force A on C:
First, find the distance between objects:
The distance is a diagonal line that divides the rectangle into a right triangle. Distance is square of the hypotenuse .
[tex]r^{2} = (6.14.10^2)^{2} + (1.42.10^{3})^{2}[/tex]
[tex]r^{2} = 37.72.10^{4}[/tex]
and hypotenuse: r = [tex]6.14.10^2[/tex]m
There is an atractive force between charges, but there are components of the force in x- and y-axis. So, because of that, force will be:
[tex]F_{CA} = F_{CA}[/tex].sinα + [tex]F_{CA}.[/tex]cosα
[tex]F_{CA} = 8.987.10^{9}.\frac{3.12.10^{-3}.9.67.10^{-4}}{37.72.10^{4}}[/tex]
[tex]F_{CA} = 7.2.10^{-2}[/tex]
The trigonometric relations is taken from the rectangle:
sinα = [tex]\frac{6.14.10^{2}}{6.14.10^{2}}[/tex]
cosα = [tex]\frac{1.42.10^{3}}{6.14.10^{2}}[/tex]
[tex]F_{CA}.[/tex]cosα = [tex]7.2.10^{-2}(\frac{1.42.10^{3}}{6.14.10^{2}})[/tex] = 0.17
[tex]F_{CA}.[/tex]sinα = [tex]7.2.10^{-2}.(\frac{6.14.10^{2}}{6.14.10^{2}} )[/tex] = 0.072
[tex]F_{CA} =[/tex] 0.17î + 0.072^j
Now, sum up all the terms in its respective axis:
X: [tex]13.64.10^{-4} + 0.17 =[/tex] 0.1714
Y: [tex]10.03.10^{-2} + 7.2.10^{-2}[/tex] = 0.1723
These forms another right triangle, whose hypotenuse is the net electrostatic force:
[tex]F_{net} = \sqrt{(0.1714)^{2} + (0.1723)^2}[/tex]
[tex]F_{net} = 24.3.10^{-2}[/tex] N
The net electrostatic force acting on C has magnitude [tex]F_{net} = 24.3.10^{-2}[/tex] N.
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.
How do your results from ray tracing compare to your results from using the thin-lens equation?
What is the focal length of a convex lens that produces an image 10 cm away with a magnification of -0.5? Show all calculations in your answer.
Answer:
f = 6.66 cm
Explanation:
For this exercise we will use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image
the expression for magnification is
m = h '/ h = - q / p
with this we have a system of two equations with two unknowns, in the problem they give us the distance to the image q = 10 cm and a magnification of m = -0.5
-0.5 = - q / p
p = - q / 0.5
p = - 10 / 0.5
p = 20 cm
now we can with the other equation look for the focal length
1 / f = 1/20 + 1/10
1 / f = 0.15
f = 6.66 cm
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 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
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]
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.
Forces that act in pairs are _____ in size and ________ in direction.
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]
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
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 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
A particle moves along a straight line with the acceleration a = (12t - 3t ^ 1/2) feet / s ^ 2, where t is in seconds. Determine your speed and position as a function of time. When t = 0, v = 0 and s = 15 feet.
Answer:
v = 6t² − 2t^³/₂
s = 2t³ − ⅘t^⁵/₂ + 15
Explanation:
a = 12t − 3t^½
Integrate to find velocity.
v = ∫ a dt
v = ∫ (12t − 3t^½) dt
v = 6t² − 2t^³/₂ + C
Use initial condition to find C.
0 = 6(0)² − 2(0)^³/₂ + C
C = 0
v = 6t² − 2t^³/₂
Integrate to find position.
s = ∫ v dt
s = ∫ (6t² − 2t^³/₂) dt
s = 2t³ − ⅘t^⁵/₂ + C
Use initial condition to find C.
15 = 2(0)³ − ⅘(0)^⁵/₂ + C
15 = C
s = 2t³ − ⅘t^⁵/₂ + 15
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.
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 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]
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