Answer:
3924 Pa
Explanation:
Volume of cylinder = 2 L = 0.002 m^3 (1000 L = 1 m^3)
diameter of the inner cylinder = 8 cm = 0.08 m (100 cm = 1 m)
radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m
area of the inner cylinder = [tex]\pi r^{2}[/tex]
where [tex]\pi[/tex] = 3.142,
and r = radius = 0.04 m
area of inner cylinder = 3.142 x [tex]0.04^{2}[/tex] = 0.005 m^2
height h of the water in this cylinder = volume/area
h = 0.002/0.005 = 0.4 m
pressure at the bottom of the cylinder due to the height of water = pgh
where
p = density of water = 1000 kg/m^3
g = acceleration due to gravity = 9.81 m/s^2
h = height of water within this cylinder = 0.4 m
pressure = 1000 x 9.81 x 0.4 = 3924 Pa
Underline your answer for each situation: If you advance the movie one frame, the knot at point A would be a) in the same place b) higher c) lower d) to the right e) to the left If the person generates a new pulse like the first but more quickly, the pulse would be a) same size b) wider c) narrower If the person generates another pulse like the first but he moves his hand further, the pulse would be a) same size b) taller c) shorter If the person generates another pulse like the first but the rope is tightened, the pulse will move a) at the same rate b) faster c) slower Now the person moves his hand back and forth several times to produce several waves. You freeze the movie and get this snapshot. Underline your answer for each situation: If you advance the movie one frame, the knot at point A would be a) in the same place b) higher c) lower d) to the right e) to the left If you advance the movie one frame, the pattern of the waves will be _________relative to the hand. a) in the same place b) shifted right c) shifted left d) shifted up e) shifted down If the person starts over and moves his hand more quickly, the peaks of the waves will be a) the same distance apart b) further apart c) closer together If you lower the frequency of a wave on a string you will lower its speed. b) increase its wavelength. c). lower its amplitude. d) shorten its period.
Answer:
a) correct answer is b higher , b) correct answer is b higher , c) correct answer is b faster , d) traveling wave , e)
Explanation:
A traveling wave is described by the expression
y = A sin (kx - wt)
where k is the wave vector and w is the angular velocity
let's examine every situation presented
a) a new faster pulse is generated
A faster pulse should have a higher angular velocity
equal speed is related to the period and frequency
w = 2π f = 2π / T
therefore in this case the period must decrease so that the angular velocity increases
the correct answer is c narrower
b) Generate a pulse, but move your hand more.
Moving the hand increases the amplitude (A) of the pulse
the correct answer is b higher
c) generates a pulse but the force is tightened
Set means that more tension force is applied to the string, so the velicate changes
v = √ (T /μ)
the correct answer is b faster
d) move your hand back and forth
in this case you would see a pulse series whose sum corresponds to a traveling wave
e) Advance a frame the movie
in this case the wave will be displaced a whole period to the right
the correct answer is b
f) move your hand faster
the waves will have a maximum fast, so they are closer
answer C
g) decrease wave frequency
Since the speed of the wave is a constant m ak, decreasing the frequency must increase the wavelength to keep the velocity constant.
the correct answer is b increases its wavelength
Space vehicles traveling through Earth's radiation belts can intercept a significant number of electrons. The resulting charge buildup can damage electronic components and disrupt operations. Suppose a spherical metallic satellite 1.7 m in diameter accumulates 3.1 µC of charge in one orbital revolution. (a) Find the resulting surface charge density. (b) Calculate the magnitude of the electric field just outside the surface of the satellite, due to the surface charge.
Answer:
(a) σ = 3.41*10⁻7C/m^2
(b) E = 38,530.1 N/C
Explanation:
(a) In order to calculate the resulting surface charge density, you use the following formula:
[tex]\sigma=\frac{Q}{S}[/tex] (1)
σ: surface charge density
Q: charge of the satellite = 3.1 µC = 3.1*10^-6C
S: surface area of the satellite
The satellite has a spherical form, then, the area of the surface is given by:
[tex]S=4\pi r^2[/tex] (2)
r: radius of the satellite = d/2 = 1.7m/2 = 0.85m
You replace the equation (2) into the equation (1) and solve for the surface charge density:
[tex]\sigma=\frac{3.1*10^{-6}C}{4\pi (0.85m)^2}=3.41*10^{-7}\frac{C}{m^2}[/tex]
The surface charge density acquired by the satellite on one orbit is 3.41*10⁻7C/m^2
(b) The electric field just outside the surface is calculate d by using the following formula:
[tex]E=k\frac{Q}{R^2}[/tex] (3)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
R: radius of the satellite = 0.85m
[tex]E=(8.98*10^9Nm^2/C^2)\frac{3.1*10^{-6}C}{(0.85m)^2}=38530.1\frac{N}{C}[/tex]
The magnitude of the electric field just outside the sphere is 38,530.1 N/C
At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the left side of the pivot an adult pushes straight down on the teeter-totter with a force of 151 N. Part A In which direction does the teeter-totter rotate if the adult applies the force at a distance of 3.0 m from the pivot?Part B
In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.5 m from the pivot?
(clockwise/counterclockwise)
Part C
In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.0 m from the pivot?
(clockwise/counterclockwise)
Answer:
By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.
here weight of the child =21kgx9.8m/s2 = 205.8N
the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.
torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.
net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.
b) Ta = 2.5x151 = 377.5N-m
Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.
c)Ta = 2x151 = 302N-m
Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.
Imagine a metal rod 0.4 m long with a mass of 2 kg. You attach the rod at one end by a lightweight 3.0-m-long cord and twirl the rod around your head. Which expression below gives the best estimate of the moment of inertia of the rod-cord system?
a) I=mr^2, where m = 2kg and r = 3.2m
b) I=1/3mr^2, where m = 2kg and r = 3.4
c) I=1/12mr^2, where m = 2kg and r = 1.7m
d) I=mr^2, where m= 2kg and r = 3.4m
Answer:
Explanation:
Given that
a metal rod 0.4 m long
mass of 2 kg
lightweight 3.0-m-long cord
a) I=mr^2, where m = 2kg and r = 3.2m
b) I=1/3mr^2, where m = 2kg and r = 3.4
c) I=1/12mr^2, where m = 2kg and r = 1.7m
d) I=mr^2, where m= 2kg and r = 3.4m
[tex]I=\int\limits^{3.4}_3 \lambda dx \,x^2\\\\=\lambda\frac{x^3}{3} |^{3.4}_{3}[/tex]
[tex]I=\frac{M}{L}[\frac{3.4^3}{3}-\frac{3^3}{3} ] \\\\I=\frac{2}{(0.4\times3)}[\frac{3.4^3}{3}-\frac{3^3}{3} ]\\\\I=\frac{2}{1.2}(39.304-9)\\\\I=1.67\times12.304\\\\I=20.51[/tex]
[tex]a)I_1=mr^2\\\\I_1=2\times3.2^2\\\\I_1=20.48kgm^2\\\\b)I_2=\frac{1}{3} mr^2\\\\=\frac{1}{3} \times2\times3.4^2\\\\I_2=7.71kgm^2\\\\c)I_3=\frac{1}{12} mr^2\\\\I_3=\frac{1}{12} \times2\times1.7^2\\\\I_3=0.461kg/m^2\\\\d)I_4=mr^2\\\\I_4=2\times3.462\\\\I_4=23.12kg/m^2[/tex]
What is a possible state for an object in the absence of a net force?
There is only one possible state: constant uniform motion. That means constant speed in a straight line.
(If the constant speed happens to be zero, this description also covers the case where the object isn't moving. That special case is called "at rest".)
Answer:
at restzero accelerationconstant speedHope this helps
How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of electrons from an initial point where the electric potential is 9.20 V to a point where the electric potential is -6.90 V
Answer:
-1486 KJ
Explanation:
The work done by an electric field on a charged body is:
W = ΔV * q
where ΔV = change in voltage
q = total charge
The total charge of Avogadro's number of electrons is:
6.0221409 * 10^(23) * -1.6023 * 10^(-19) = -9.65 * 10^(4)
The change in voltage, ΔV, is:
9.20 - (6.90) = 15.4
Therefore, the work done is:
W = -9.65 * 10^(4) * 15.4 = -1.486 * 10^6 J = -1486 KJ
The negative sign means that the motion of the electrons is opposite the electrostatic force.
A box on a ramp is connected by a rope to a winch. The winch is turned so that the box moves down the ramp at a constant speed. The box experiences kinetic friction with the ramp. Which forces on the box do zero work as the box moves down the ramp?
a. Weight (gravitational force)
b. Normal force
c. Kinetic friction force
d. Tension force
e. None
Answer:
Option B:
The normal force
Explanation:
The normal force does no work as the box slides down the ramp.
Work can only be done when the force succeeds in moving the object in the direction of the force.
All the other forces involved have a component that is moving the box in their direction.
However, the normal force does not, as it points downwards into the ramp. Since the normal force is pointing into the ramp, and the box is sliding down the ramp, we can say that no work is being done by the normal force because the box is not moving in its direction (which would have been the box moving into the ramp)
An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and 0.1250 T, respectively. The particle passes out of the electric field, but the magnetic field continues, and the particle makes a semicircle of diameter 25.05 cm.
Part A. What is the particle's charge-to-mass ratio?
Part B. Can you identify the particle?
a. can't identify
b. proton
c. electron
d. neutron
Answer:
Explanation:
Given that
The electric fields of strengths E = 187,500 V/m and
and The magnetic fields of strengths B = 0.1250 T
The diameter d is 25.05 cm which is converted to 0.2505m
The radius is (d/2)
= 0.2505m / 2 = 0.12525m
The given formula to find the magnetic force is [tex]F_{ma}=BqV---(i)[/tex]
The given formula to find the electric force is [tex]F_{el}=qE---(ii)[/tex]
The velocity of electric field and magnetic field is said to be perpendicular
Electric field is equal to magnectic field
Equate equation (i) and equation (ii)
[tex]Bqv=qE\\\\v=\frac{E}{B}[/tex]
[tex]v=\frac{187500}{0.125} \\\\v=15\times10^5m/s[/tex]
It is said that the particles moves in semi circle, so we are going to consider using centripetal force
[tex]F_{ce}=\frac{mv^2}{r}---(iii)[/tex]
magnectic field is equal to centripetal force
Lets equate equation (i) and (iii)
[tex]Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg[/tex]
Therefore, the particle's charge-to-mass ratio is [tex]958.1\times10^5C/kg[/tex]
b)
To identify the particle
Then 1/ 958.1 × 10⁵ C/kg
The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg
Therefore the particle is proton.
Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above the reference point. If the mass of the car is 1000-kg, and the car traveled a distance of 400 m, estimate the magnitude of the frictional force between the car and the track.
Answer:
The magnitude of the frictional force between the car and the track is 367.763 N.
Explanation:
The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:
[tex]U_{g,1} = U_{g,2} + W_{dis}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]W_{dis}[/tex] - Dissipated work due to friction, measured in joules.
Gravitational potential energy is described by the following formula:
[tex]U = m \cdot g \cdot y[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]y[/tex] - Height with respect to reference point, measured in meters.
In addition, dissipated work due to friction is:
[tex]W_{dis} = f \cdot \Delta s[/tex]
Where:
[tex]f[/tex] - Friction force, measured in newtons.
[tex]\Delta s[/tex] - Travelled distance, measured in meters.
Now, the energy equation is expanded and frictional force is cleared:
[tex]m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s[/tex]
[tex]f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}[/tex]
If [tex]m = 1000\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 40\,m[/tex], [tex]y_{2} = 25\,m[/tex] and [tex]\Delta s = 400\,m[/tex], then:
[tex]f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}[/tex]
[tex]f = 367.763\,N[/tex]
The magnitude of the frictional force between the car and the track is 367.763 N.
A lens collects light and focuses it into a small spot. This increases the ________ of the light wave. A lens collects light and focuses it into a small spot. This increases the ________ of the light wave. wavelength energy intensity frequency
Answer:
This increases the intensity of the light wave.
3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charge on the paper of the same magnitude, find the weight of the paper in newtons. Remember to convert the distance to meters and show your work here.
Answer:
[tex]\mathbf{1.51\times10^{-15}N}[/tex]
Explanation:
The computation of the weight of the paper in newtons is shown below:
On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.
Therefore the paper charge is
[tex]q_{paper}=-4.1\times10^{-15}C[/tex]
Now the distance from the charge is
[tex]r=1cm=0.01m[/tex]
Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.
[tex]mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}[/tex]
[tex]\Rightarrow W=mg[/tex]
[tex]=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}[/tex]
[tex]=\mathbf{1.51\times10^{-15}N}[/tex]
A charge of 4.5 × 10-9 C is located 3.2 m from a charge of -2.8 × 10-9 C. Find the
electrostatic force exerted by one charge on the other.
Answer:
[tex]F = -1.107*10^{-8} N\\|F| = 1.107*10^{-8} N[/tex]
Explanation:
[tex]q_1 = 4.5 * 10^{-9} C[/tex]
[tex]q_2 = -2.8 * 10^{-9} C[/tex]
The distance separating the two charges, r = 3.2 m
According to Coulomb's law of electrostatic attraction, the electrostatic force between the two charges can be given by the formula:
[tex]F = \frac{kq_{1} q_{2} }{r^2}[/tex]
Where [tex]k = 9.0 * 10^9 Nm^2/C^2[/tex]
[tex]F = \frac{9*10^9 * 4.5*10^{-9} * (-2.8*10^{-9}}{3.2^2} \\F = \frac{-113.4*10^{-9}}{10.24}\\F = -11.07 *10^{-9}\\F = -1.107*10^{-8}N[/tex]
Answer:
F = 1.1074 × [tex]10^{-8}[/tex] N
Explanation:
An electrostatic force is either a force or attraction or repulsion between two charges. It can be determined by:
F = [tex]\frac{kq_{1}q_{2} }{r^{2} }[/tex]
where: F is the force, k is a constant, [tex]q_{1}[/tex] is the first charge, [tex]q_{2}[/tex] is the second charge and r the distance between the charges.
Given that: k = 9 × [tex]10^{9}[/tex] N[tex]m^{2}[/tex][tex]C^{-2}[/tex], [tex]q_{1}[/tex] = 4.5 × [tex]10^{-9}[/tex]C, [tex]q_{2}[/tex] = -2.8 × [tex]10^{-9}[/tex]C and r = 3.2 m.
Then,
F = [tex]\frac{9*10^{9}*4.5*10^{-9}*2.8*10^{-9} }{3.2^{2} }[/tex]
= [tex]\frac{1.134*10^{-7} }{10.24}[/tex]
= 1.1074 × [tex]10^{-8}[/tex]
The electrostatic force exerted is 1.1074 × [tex]10^{-8}[/tex] N, and it is a force of attraction.
A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 15.0 m: (a) the initially stationary spelunker is accelerated to a speed of 2.40 m/s; (b) he is then lifted at the constant speed of 2.40 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 85.0 kg rescue by the force lifting him during each stage
Answer:
A) 12752.55 J
B) 12507.75 J
C) 12262.95 J
Explanation:
We are given;
Mass;m = 85 kg
Vertical distance; d = 15 m
From change in kinetic energy, the work done by the applied force to pull the spelunker is given by;
Change in kinetic energy = Wa + Wg
Where Wg = -mgd
A) In the first stage;the the work done is given by;
Wa = mgd + ½m(v_f)² - ½m(v_i)²
Since initially stationary, v_i = 0
So, we have;
Wa = mgd + ½m(v_f)²
v_f = 2.4 m/s
So,
Wa = (85 × 9.81 × 15) + ((1/2) × 85 × 2.4²)
Wa = 12752.55 J
B) for the second stage, there is a constant speed of 2.4 m/s
So, v_f = v_i
Thus; Wa = mgd
Wa = (85 × 9.81 × 15)
Wa = 12507.75 J
C) he finally decelerated to zero.
So v_f = 0 while v_i is 2.4 m/s
Thus;
Wa = mgd - ½m(v_i)²
Wa = (85 × 9.81 × 15) - (½ × 85 × 2.4²)
Wa = 12507.75 - 244.8
Wa = 12262.95 J
A light source simultaneously emits light of two wavelengths, 480 nm and 560 nm, respectively. The source is used in a double-slit interference experiment where the slit spacing is a 0.040 mm, and the distance between double slits and the screen is 1.2 m. What is the separation between the second-order bright fringes of the two wavelengths as they appear on the screen
Answer:
0.48 cm
Explanation:
given data
wavelength = 480 nm
wavelength = 560 nm
slit spacing = 0.040 mm
distance between double slits and the screen = 1.2 m
solution
we know that (1 nm= [tex]10^{-9}[/tex] m)
we wil take here equation of equations of interference that is
ym = R × (m λ)/d ..........................1
here m = 2 R i.e distance of screen and slit
so put here value and we get
separation between the second-order bright fringes = 0.48 cm
A skydiver of mass 87 kg falls straight down. If air resistance exerts 355 N of force on him as he falls, what is the net force on the skydiver?
Answer:
497.6 N
Explanation:
From the question,
The net force on the skydiver = weight of the skydiver- the resistive force of air
F' = W-F...................... Equation 1
Where W = weight of the skydiver, F = resistive force of air.
But,
W = mg................ Equation 2
Where m = mass of the skydiver, g = acceleration due to gravity.
Substitute equation 2 into equation 1
F' = mg-F............ Equation 3
Given: m = 87 kg, F = 355 N, g = 9.8 m/s²
Substitute these values into equation 3
F' = 87(9.8)-355
F' = 852.6-355
F' = 487.6 N
Answer:
498 N down
Explanation:
In 1898, the world land speed record was set by Gaston Chasseloup-Laubat driving a car named Jeantaud. His speed was 39.24 mph (63.15 km/h), much lower than the limit on our interstate highways today. Repeat the calculations of Example 2.7 (assume the car accelerates for 6 miles to get up to speed, is then timed for a one-mile distance, and accelerates for another 6 miles to come to a stop) for the Jeantaud car. (Assume the car moves in the +x direction.)
Find the acceleration for the first 6 miles.
Answer:
the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]
Explanation:
Given that:
the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.
So; the acceleration for the first 6 miles can be calculated by using the formula:
v₂² = v₁² + 2a (Δx)
Making acceleration a the subject of the formula in the above expression ; we have:
v₂² - v₁² = 2a (Δx)
[tex]a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}[/tex]
[tex]a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}[/tex]
[tex]a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}[/tex]
[tex]a =0.0159 \ m/s^2[/tex]
Thus;
Assume the car moves in the +x direction;
the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]
A power of 100 W is delivered to a certain resistor when the applied voltage is 100 V. Find the resistance. Suppose that the voltage is reduced by 10% to 90 V. By what percentage is the power reduced
Answer:
The resistance is 100 ohmsThe power has reduced 19%Explanation:
The formula for power is given as
[tex]P=\frac{V^2}{R}[/tex]
Given data
power= 100 W
voltage= 100 V
Substituting into the power formula the resistance is
[tex]100=\frac{100^2}{R} \\\\100*R=10000\\\\R= \frac{10000}{100} \\\\R= 100 ohms[/tex]
The resistance is 100 ohms
Now the power has been reduced by 10% (from 100 to 90V)
the power is
[tex]P= \frac{90^2}{100} \\\\P= \frac{8100}{100} \\\\P= 81 watts[/tex]
The power has reduced 19% from 100 watt to 81 watt
1. The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of positions measured with the best Earth-based optical telescopes. If you cannot measure an angle smaller than this, what is the maximum distance at which a star can be located and still have a measurable parallax
Answer:
Explanation:
each grid corresponding 0.1s⁻¹.
0.2grid unit = 0.2×0.1 =0.02s⁻¹
distance of the star from telescope
d = 1/p
d= 1/0.02= 50 par sec
1par sec = 3.26 light year
1 light year = 9.5×10¹²km
3.26ly=3.084×10¹³km
d= 50×3.084×10¹³=1.55×10¹⁵km
Imagine that you want to make sure the battery for your string of lights will last as long as possible. A battery will last longer if it powers a circuit with low current. How could you hook up a battery and 2 light bulbs so the least amount of current flows through the battery
Answer:
Hooking up the bulb to the battery in a series arrangement will draw the least amount of current.
Explanation:
In this case now, the bulb will serve as the load on the battery (resistance).
For the current to last longer, the least amount of energy must be drawn.
The least amount of energy will be drawn when the arrangement provides the maximum resistance possible.
Let us take the resistance of each bulb as 'R'
If we arrange the bulbs in series, then, the total resistance will be
Rt = R + R = 2R
at a EMF of V from the battery, current I through the battery will be
I = V/2R
If we arrange the bulbs in parallel, then , the total resistance will be
1/Rt = 1/R + 1/R
1/Rt = 2/R
therefore
Rt = R/2
at an EMF of V from the battery, the current I that will be drawn through the battery will be
I = 2V/R
we see that arranging the bulbs in parallel draws 4 times the current compared to arranging the bulb in series
From the above, we see that arranging the bulbs in series provides the maximum resistance, which means a lesser amount of current is drawn from the battery
During a particular time interval, the displacement of an object is equal to zero. Must the distance traveled by this object also equal to zero during this time interval? Group of answer choices
Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.
Explanation:
Ok, let's write the definitions:
Displacement: The displacement is equal to the difference between the final position and the initial position.
Distance traveled: Total distance that you moved.
So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:
The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.
But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.
So no, we can have a displacement equal to zero, but a distance traveled different than zero.
You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in terms of some unknown units r and v. Rank the wires in order of decreasing electron current. Rank from most to least electron current. To rank items as equivalent, overlap them.
a) radius=3r, drift speed=1v
b) radius=4r, drift speed=0.5v
c) radius=1r, drift speed=5v
d) radius=2r, drift speed=2.5v
Answer:
d > a > b > c
Explanation:
Given that
a) radius = 3r and drift speed = 1v
b) radius = 4r and drift speed = 0.5v
c) radius = 1r and drift speed = 5v
d) radius = 2r and drift speed = 2.5v
Based on the above information, the ranking of the wires for reducing the electron current is
As we can see that the radius i.e to be less and the drift speed that is highest so it should be rank one
And, According to that, other options are ranked
Therefore, the ranking would be d > a > b > c
High voltage power is often carried in wire bundles made up of individual strands. In your initial post to the discussion, discuss the forces on the strands of wire due to the current flowing through them. What would happen if the force acted opposite of the known behavior
Answer:
More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.
Explanation:
A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor. The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.
If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.
The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.
A circuit contains two elements, but it is not known if they are L, R, or C. The current in this circuit when connected to a 120-V 60 Hz source is 5.3 A and lags the voltage by 65∘.
Part A. What are the two elements?
Part B. What are their values?
Express your answer using two significant figures
Answer:
the two elements are resistor R and inductor L
answers in two significant figures
R = 9.6Ω
L = 54mH
Explanation:
One end of a uniform well-lagged metal bar of length 0.80 m and cross-sectional area of
4.0 x 10-m” is kept in steam at 100 °C while the other end is in melting ice in a well-lagged
container. The ice melts at a steady rate of 5,5 x 10 * kgs and the thermal conductivity of
the material of the bar is 401 Wm''K! Calculate the specific latent heat of fusion of ice.
(3 marks)
Complete question is;
One end of a uniform well-lagged metal bar of length 0.80 m and cross-sectional area of 4.0 x 10-3m2 is kept in steam at 100 ˚C while the other end is in melting ice in a well-lagged container. The ice melts at a steady rate of 5.5 x 10- 4 kgs-1 and the thermal conductivity of the material of the bar is 401 Wm-1K-1. Calculate the specific latent heat of fusion of ice
Answer:
Specific Latent heat of fusion;
L_f = 13.6 × 10^(5) J/kg
Explanation:
We are given;
Length of bar; L = 0.8 m
Area;A = 4 × 10^(-3) m²
Temperature;ΔT= 100°C = 100 + 273 = 373 K
Rate of melting;m/t = 5.5 × 10^(-4) kg/s
Thermal conductivity;k = 401 W/m·K
Latent heat of fusion has a formula;
ΔQ/Δt = (m/t)•L_f
So, L_f = (ΔQ/Δt)/(m/t) - - - (1)
We also know that ;
ΔQ/Δt = (ΔT × k × A)/L
Plugging in the relevant values, we have;
ΔQ/Δt = (373 × 401 × 4 × 10^(-3))/0.8
ΔQ/Δt = 747.865 J/S
Plugging this value for ΔQ/Δt in equation 1 gives;
L_f = 747.865/(5.5 × 10^(-4))
L_f = 13.6 × 10^(5) J/kg
A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-the loop which has a radius 5 m. The acceleration of gravity is 9.8 m/s^2.
Required:
How large is the normal force on it if its mass is 5g?
Answer:
N₁ = 393.96 N and N = 197.96 N
Explanation:
In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop
Lowest point, we write Newton's second law n for the y-axis
N -W = m a
where the acceleration is ccentripeta
a = v² / r
N = W + m v² / r
N = mg + mv² / r
we can use energy to find the speed at the bottom of the circle
starting point. Highest point where the ball is released
Em₀ = U = m g h
lowest point. Stop curl down
[tex]Em_{f}[/tex] = K = ½ m v²
Emo = Em_{f}
m g h = ½ m v²
v² = 2 gh
we substitute
N = m (g + 2gh / r)
N = mg (1 + 2h / r)
let's calculate
N₁ = 5 9.8 (1 + 2 17.6 / 5)
N₁ = 393.96 N
headed up
we repeat the calculation in the longest part of the loop
-N -W = - m v₂² / r
N = m v₂² / r - W
N = m (v₂²/r - g)
we seek speed with the conservation of energy
Em₀ = U = m g h
final point. Top of circle with height 2r
[tex]Em_{f}[/tex] = K + U = ½ m v₂² + mg (2r)
Em₀ = Em_{f}
mgh = ½ m v₂² + 2mgr
v₂² = 2 g (h-2r)
we substitute
N = m (2g (h-2r) / r - g)
N = mg (2 (h-r) / r 1) = mg (2h/r -2 -1)
N = mg (2h/r - 3)
N = 5 9.8 (2 17.6 / 5 -3)
N = 197.96 N
Directed down
An astronaut is in an all-metal chamber outside the space station when a solar storm results in the deposit of a large positive charge on the station. Which statement is correct?
a. The astronaut must abandon the chamber immediately to avoid being electrocuted.
b. The astronaut will be safe only if she is wearing a spacesuit made of non-conducting materials.
c. The astronaut does not need to worry: the charge will remain on the outside surface.
d. The astronaut must abandon the chamber if the electric field on the outside surface becomes greater than the breakdown field of air.
d. The astronaut must abandon the chamber immediately because the electric field inside the chamber is non-uniform.
Answer:
c. The astronaut does not need to worry: the charge will remain on the outside surface.
Explanation:
The astronaut need not worry because according to Gauss's law of electrostatic, a hollow charged surface will have a net zero charge on the inside. This is the case of a Gauss surface, and all the charges stay on the surface of the metal chamber. This same principle explains why passengers are safe from electrostatic charges, in an enclosed aircraft, high up in the atmosphere; all the charges stay on the surface of the aircraft.
A wave travels at a speed of 242 m/s. If the distance between crests is 0.11
m, what is the frequency of the wave? Use |
A. 0.00045 Hz
B. 27 Hz
C. 2200 Hz
D. 190 Hz
Answer:
f = 2200 Hz
Explanation:
It is given that,
Speed of a wave is 242 m/s
The distance between crests is 0.11 m
We need to find the frequency of the wave. The distance between crests is called wavelength of a wave. So,
[tex]v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{242}{0.11}\\\\f=2200\ Hz[/tex]
So, the frequency of the wave is 2200 Hz.
Answer:2200 hz
Explanation:
A 432 kg merry-go-round in the shape of a horizontal disk with a radius of 2.3 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.1 rad/s in 2.1 s
Answer:
τ = 1679.68Nm
Explanation:
In order to calculate the required torque you first take into account the following formula:
[tex]\tau=I\alpha[/tex] (1)
τ: torque
I: moment of inertia of the merry-go-round
α: angular acceleration
Next, you use the following formulas for the calculation of the angular acceleration and the moment of inertia:
[tex]\omega=\omega_o+\alpha t[/tex] (2)
[tex]I=\frac{1}{2}MR^2[/tex] (3) (it is considered that the merry-go-round is a disk)
w: final angular speed = 3.1 rad/s
wo: initial angular speed = 0 rad/s
M: mass of the merry-go-round = 432 kg
R: radius of the merry-go-round = 2.3m
You solve the equation (2) for α. Furthermore you calculate the moment of inertia:
[tex]\alpha=\frac{\omega}{t}=\frac{3.1rad/s}{2.1s}=1.47\frac{rad}{s^2}\\\\I=\frac{1}{2}(432kg)(2.3)^2=1142.64kg\frac{m}{s}[/tex]
Finally, you replace the values of the moment of inertia and angular acceleration in the equation (1):
[tex]\tau=(1142.64kgm/s)(1.47rad/s^2)=1679.68Nm[/tex]
The required torque is 1679.68Nm
a What CE describes the way energy is stored in a sandwich
An electron of mass 9.11 x 10^-31 kg has an initial speed of 4.00 x 10^5 m/s. It travels in a straight line, and its speed increases to 6.60 x10^5 m/s in a distance of 5.40 cm. Assume its acceleration is constant.
Required:
a. Determine the magnitude of the force exerted on the electron.
b. Compare this force (F) with the weight of the electron (Fg), which we ignored.
Answer:
a. F = 2.32*10^-18 N
b. The force F is 2.59*10^11 times the weight of the electron
Explanation:
a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:
[tex]v^2=v_o^2+2ax[/tex] (1)
v: final speed of the electron = 6.60*10^5 m/s
vo: initial speed of the electron = 4.00*10^5 m/s
a: acceleration of the electron = ?
x: distance traveled by the electron = 5.40cm = 0.054m
you solve the equation (2) for a and replace the values of the parameters:
[tex]a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}[/tex]
Next, you use the second Newton law to calculate the force:
[tex]F=ma[/tex]
m: mass of the electron = 9.11*10^-31kg
[tex]F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N[/tex]
The magnitude of the force exerted on the electron is 2.32*10^-18 N
b. The weight of the electron is given by:
[tex]F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N[/tex]
The quotient between the weight of the electron and the force F is:
[tex]\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}[/tex]
The force F is 2.59*10^11 times the weight of the electron