Answer:
The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.
Explanation:
Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:
[tex]F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}[/tex]
and since the electric field E in between parallel plates separated a distance d and under a potential difference [tex]\Delta V[/tex], is given by:
[tex]E=\frac{\Delta\,V}{d}[/tex]
then :
[tex]a=\frac{q\,\Delta V}{m\,d}[/tex]
We want to find when the particle reaches velocity zero via kinematics:
[tex]v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a[/tex]
We replace this time (t) in the kinematic equation for the particle displacement:
[tex]\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}[/tex]
Replacing the values with the information given, converting the distance d into meters (0.01 m), using [tex]\Delta V=100\,V[/tex], and the electron's kinetic energy:
[tex]\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J[/tex]
we get:
[tex]\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters[/tex]Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:
0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]
A wire of 0.50m length is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.98T. The vurrent in the wire is 2.0A in the direction shown. What is the mass of the wire if the current and the magnetic field are sufficient to remove the tension in the supporting leads?
Answer:
0.1 kg or 100 g
Explanation:
The length of the wire = 0.5 m
the field magnitude = 0.98 T
the current through the wire = 2.0 A
magnetic force due to a wire carrying current is
F = [tex]IlB[/tex]
where
F is the force
[tex]I[/tex] is the current = 2 A
[tex]l[/tex] is the length of the wire
B is the magnetic field strength
Substituting, we have
F = 2 x 0.5 x 0.98 = 0.98 N
This force balances the weight of the mass
weight = mg
where m is the mass of the wire
g is acceleration due to gravity = 9.81 m/s^2
therefore, weight = m x 9.81 = 9.81m
equating this weight with the force, we have
0.98 = 9.81m
m = 0.98/9.81 = 0.099 kg ≅ 0.1 kg or 100 g
Answer:
100 g
Explanation:
What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is the radius of the Earth
Answer:
g' = g/9 = 1.09 m/s²
Explanation:
The magnitude of free fall acceleration at the surface of earth is given by the following formula:
g = GM/R² ----- equation 1
where,
g = free fall acceleration
G = Universal Gravitational Constant
M = Mass of Earth
R = Distance between the center of earth and the object
So, in our case,
R = R + 2 R = 3 R
Therefore,
g' = GM/(3R)²
g' = (1/9) GM/R²
using equation 1:
g' = g/9
g' = (9.8 m/s)/9
g' = 1.09 m/s²
Answer:
The magnitude of the free-fall acceleration [tex]g_h = 1.09m/s^2[/tex]Explanation:
Surface of earth,
[tex]g = \frac{GM}{R^2}\\\\g = 9.8m/s^2[/tex]
free fall acceleration at height h,
[tex]g_h = \frac{GM}{(R+h)^2}[/tex]
where
G = gravitational constant
R = Radius of earth
M = mass of earth
therefore,
[tex]\frac{g_h}{g} = \frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}}\\\\ \frac{g_h}{g} = \frac{R^2}{(R+h)^2}\\\\g_h = g\frac{R^2}{(R+h)^2}[/tex]
Where height h = 2R
[tex]g_h = 9.8\frac{R^2}{(R+2R)^2}\\\\g_h = 9.8\frac{R^2}{(3R)^2}\\\\g_h = 9.8\frac{R^2}{(9R^2}\\\\g_h = 1.09m/s^2[/tex]
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In a photoelectric-effect experiment it is observed that no current flows unless the wavelength is less than 540 nmnm . Part A What is the work function of this material
Answer:
Φ = 36.84 × 10^(-20) J
Explanation:
In the photoelectric effect, the energy of the incoming photon is usually used in part to extract the photoelectron from the material (work function) and then the rest is converted into kinetic energy of the photoelectron which is given by the formula;
K_max = hf - Φ
where;
hf represents the energy of the incoming photon
h is the Planck's constant
f is the light frequency
Φ is the work function of the material
K_max is the maximum kinetic energy of the photoelectrons.
From the question, we are told that no current flows unless the wavelength is less than 540 nm. This means that when the wavelength has this value, the maximum kinetic energy of the photoelectrons is zero i.e K_max = 0. Thus the energy of the incoming photons is just enough to extract the photoelectrons from the material.
Thus,
hf - Φ = 0
hf = Φ - - - (1)
We are given the wavelength as;
λ = 540 nm = 540 × 10^(-9) m
Now, let's find the frequency of the light by using the relationship between frequency and wavelength. The equation is;
f = c/λ
Where c is speed of light = 3 × 10^(8) m/s
f = (3 × 10^(8))/(540 × 10^(-9))
f = 5.56 × 10^(14) Hz
Thus, from equation 1,we can now find the work function;
Φ = hf
h is Planck's constant and has a value of 6.626 × 10^(-34) J.s
Thus;
Φ = 6.626 × 10^(-34) × 5.56 × 10^(14)
Φ = 36.84 × 10^(-20) J
White light containing wavelengths from 410 nm to 750 nm falls on a grating with 7800 slits/cm. Part APart complete How wide is the first-order spectrum on a screen 3.20 m away
Answer:
1.227 m
Explanation:
Given that
Minimum wavelength is 410 nm
Maximum wavelength is 750 nm
Grating is 7800 slits/cm
Distance is 3.2 m
To solve this question, we would use the formula
sin θ = λ/d
sin θ = (410*10^-9) / (0.01/7800)
Sin θ = 410*10^-9 / 1.282*10^-6
Sin θ = 0.32 and θ = 18.67 degrees
For the second wavelength = 750 nm
sin θ = [(0.32x750)/410]
sin θ = (240 / 410)
sin θ = 0.5853 or
θ = 35.8 degrees
And finally, the width of spectrum would be
3.2[tan 35.8 - tan 18.67]
3.2 * 0.3833
= 1.227 m
An 18g bullet is shot vertically into a 10kg block. The block lifts upward 9mm. The bullet penetrates the block in a time interval of 0.001s. Assume the force on the bullet is constant during penetration. The initial kinetic energy of the bullet is closest to:
Answer:
The initial kinetic energy of the bullet is closest to 491.87 J
Explanation:
Given;
mass of bullet, m₁ = 18g = 0.018kg
mass of block, m₂ = 10kg
height moved by the block, h = 9 mm = 0.009 m
time taken for the bullet to travel through the block, t = 0.001s
let the initial velocity of the bullet = v₁
let the final velocity of the bullet = v₂
Apply the principle of conservation of linear momentum;
initial momentum = final momentum
0.018v₁ = v₂(0.018 + 10)
0.018v₁ = 10.018v₂ -----equation (1)
Apply the law of conservation of energy when the bullet lifts the block through 9mm
mgh = ¹/₂mv₂²
gh = ¹/₂v₂²
v₂² = 2gh
v₂ = √2gh
v₂ = √(2 x 9.8 x 0.009)
v₂ = 0.42 m/s
Substitute in v₂ in equation 1, to determine the initial velocity of the bullet;
0.018v₁ = 10.018v₂
0.018v₁ = 10.018(0.42)
0.018v₁ = 4.208
v₁ = 4.208 / 0.018
v₁ = 233.78 m/s
Now, determine the initial kinetic energy of the bullet;
K.E₁ = ¹/₂m₁v₁²
K.E₁ = ¹/₂(0.018)(233.78)²
K.E₁ = 491.87 J
Therefore, the initial kinetic energy of the bullet is closest to 491.87 J
Determine the next possible thickness of the film (in nm) that will provide the proper destructive interference. The index of refraction of the glass is 1.58 and the index of refraction of the film material is 1.48.
Answer:
I know the answer
Explanation:
We want to choose the film thickness such that destructive interference occurs between the light reflected from the air-film interface (call it wave 1) and from the film-lens interface (call it wave 2). For destructive interference to occur, the phase difference between the two waves must be an odd multiple of half-wavelengths.
You can think of the phases of the two waves as second hands on a clock; as the light travels, the hands tick-tock around the clock. Consider the clocks on the two waves in question. As both waves travel to the air-film interface, their clocks both tick-tock the same time-no phase difference. When wave 1 is reflected from the air-film boundary, its clock is set forward 30 seconds; i.e., if the hand was pointing toward 12, it's now pointing toward 6. It's set forward because the index of refraction of air is smaller than that of the film.
Now wave 1 pauses while wave two goes into and out of the film. The clock on wave 2 continues to tick as it travels in the film-tick, tock, tick, tock.... Clock 2 is set forward 30 seconds when it hits the film-lens interface because the index of refraction of the film is smaller than that of the lens. Then as it travels back through the film, its clock still continues ticking. When wave 2 gets back to the air-film interface, the two waves continue side by side, both their clocks ticking; there is no change in phase as they continue on their merry way.
So, to recap, since both clocks were shifted forward at the two different interfaces, there was no net phase shift due to reflection. There was also no phase shift as the waves travelled into and out from the air-film interface. The only phase shift occured as clock 2 ticked inside the film.
Call the thickness of the film t. Then the total distance travelled by wave 2 inside the film is 2t, if we assume the light entered pretty much normal to the interface. This total distance should equal to half the wavelength of the light in the film (for the minimum condition; it could also be 3/2, 5/2, etc., but that wouldn't be the minimum thickness) since the hand of the clock makes one revolution for each distance of one wavelength the wave travels (right?).
A golfer hits a 42 g ball, which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s. Determine the height the ball will rise after the bounce. Show all your work.
Answer:
12.2 m
Explanation:
Given:
v₀ = 15.6 m/s
v = 0 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy
Δy = 12.2 m
[tex] \LARGE{ \boxed{ \rm{ \green{Answer:}}}}[/tex]
Given,
The initial speed is 15.6 m/s The mass of the ball is 42g = 0.042kgFinding the initial kinetic energy,
[tex]\large{ \boxed{ \rm{K.E. = \frac{1}{2}m {v}^{2}}}}[/tex]
⇛ KE = (1/2)mv²
⇛ KE = (1/2)(0.042)(15.6)²
⇛ KE = 5.11 J
|| ⚡By conservation of energy, the potential energy at the highest point will also be 5.11 J, since there is no kinetic energy at the highest point because the ball is not moving (we neglect energy lost due to air resistance, heat, sound, etc.) ⚡||
So, we have:
[tex] \large{ \boxed{ \rm{P.E. = mgh}}}[/tex]
⇛ h = PE/(mg)
⇛ h = 5.11 J /(0.042 × 9.8)
⇛ h = 12.41 m
✏The ball will rise upto a height of 12.41 m
━━━━━━━━━━━━━━━━━━━━
What is the impedance of an AC series circuit that is constructed of a 10.0-W resistor along with 12.0 W inductive reactance and 7.0 W capacitive reactance
Answer:
11.2 Ω
Explanation:
The impedance of a circuit is given by;
Z= √R^2 +(XL-XC)^2
Since
Resistance R= 10 Ω
Inductive reactance XL= 12 Ω
Capacitive reactance XC= 7 Ω
Z= √10^2 + (12-7)^2
Z= √100 + 25
Z= √125
Z= 11.2 Ω
Now the friends are ready to tackle a homework problem. A pulse is sent traveling along a rope under a tension of 29 N whose mass per unit length abruptly changes, from 19 kg/m to 45 kg/m. The length of the rope is 2.5 m for the first section and 2.8 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, Δt, between the two pulses detected at the origin? s
Answer:
The time difference is 2.97 sec.
Explanation:
Given that,
Tension = 29 N
Mass per unit length [tex]\mu_{1}=19\ kg/m[/tex]
Mass per unit length [tex]\mu_{2}=45\ kg/m[/tex]
Length of first section = 2.5 m
Length of second section = 2.8 m
We need to total distance of first pulse
Using formula for distance
[tex]d=2.5+2.5[/tex]
[tex]d_{1}=5.0\ m[/tex]
We need to total distance of second pulse
Using formula for distance
[tex]d=2.8+2.8[/tex]
[tex]d_{2}=5.6\ m[/tex]
We need to calculate the speed of pulse in the first string
Using formula of speed
[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]
Put the value into the formula
[tex]v_{1}=\sqrt{\dfrac{29}{19}}[/tex]
[tex]v_{1}=1.24\ m/s[/tex]
We need to calculate the speed of pulse in the second string
Using formula of speed
[tex]v_{2}=\sqrt{\dfrac{T}}{\mu_{2}}}[/tex]
Put the value into the formula
[tex]v_{2}=\sqrt{\dfrac{29}{45}}[/tex]
[tex]v_{2}=0.80\ m/s[/tex]
We need to calculate the time for first pulse
Using formula of time
[tex]t_{1}=\dfrac{d_{1}}{v_{1}}[/tex]
Put the value into the formula
[tex]t_{1}=\dfrac{5.0}{1.24}[/tex]
[tex]t_{1}=4.03\ sec[/tex]
We need to calculate the time for second pulse
Using formula of time
[tex]t_{2}=\dfrac{d_{1}}{v_{1}}[/tex]
Put the value into the formula
[tex]t_{2}=\dfrac{5.6}{0.80}[/tex]
[tex]t_{2}=7\ sec[/tex]
We need to calculate the time difference
Using formula of time difference
[tex]\Delta t=t_{2}-t_{1}[/tex]
Put the value into the formula
[tex]\Delta t=7-4.03[/tex]
[tex]\Delta t=2.97\ sec[/tex]
Hence, The time difference is 2.97 sec.
A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the magnitude of the magnetic field at the center of the loop (in T).
Answer:
The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
Explanation:
Given;
number of turns of the flat circular loop, N = 18 turns
radius of the loop, R = 15.0 cm = 0.15 m
current through the wire, I = 0.51 A
The magnetic field through the center of the loop is given by;
[tex]B = \frac{N\mu_o I}{2R}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]
Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 W/m2 at the floor of the facility.
A) Find the average radiation pressure (in pascals) on a totally absorbing section of the floor.B) Find the average radiation pressure (in atmospheres) on a totally absorbing section of the floor.C) Find the average radiation pressure (in pascals) on a totally reflecting section of the floor.D) Find the average radiation pressure (in atmospheres) on a totally reflecting section of the floor.
Answer:
a) 8.33 x 10^-6 Pa
b) 8.23 x 10^-11 atm
c) 1.67 x 10^-5 Pa
d) 1.65 x 10^-10 atm
Explanation:
Intensity of the light [tex]I[/tex] = 2500 W/m^2
speed of light [tex]c[/tex] = 3 x 10^8 m/s
a) we know that the pressure for for a totally absorbing surface is given as
[tex]P_{abs}[/tex] = [tex]I/c[/tex] = 2500/(3 x 10^8) = 8.33 x 10^-6 Pa
b) 1 atm = 101325 Pa
[tex]P_{abs}[/tex] = (8.33 x 10^-6)/101325 = 8.23 x 10^-11 atm
c) for a totally reflecting surface
[tex]P_{ref}[/tex] = [tex]2I/c[/tex] = twice the value for totally absorbing
[tex]P_{ref}[/tex] = 2 x 8.33 x 10^-6 = 1.67 x 10^-5 Pa
d) 1 atm = 101325 Pa
[tex]P_{ref}[/tex] = 2 x 8.23 x 10^-11 = 1.65 x 10^-10 atm
You need to repair a broken fence in your yard. The hole in your fence is
around 3 meters in length and for whatever reason, the store you go to
has oddly specific width 20cm wood. Each plank of wood costs $16.20,
how much will it cost to repair your fence? (Hint: 1 meter = 100 cm) *
Answer:
cost = $ 243.00
Explanation:
This exercise must assume that it uses a complete table for each piece, we can use a direct ratio of proportions, if 1 table is 0.20 m wide, how many tables will be 3.00 m
#_tables = 3 m (1 / 0.20 m)
#_tables = 15 tables
Let's use another direct ratio, or rule of three, for cost. If a board costs $ 16.20, how much do 15 boards cost?
Cost = 15 (16.20 / 1)
cost = $ 243.00
Convert 7,348 grams to kilograms
an ideal gas is confined to a container with adjustable volume. the number of moles, n, and temperature, t, are constant. by what factor will the volume change if pressure increase by a factor of 5.1
Answer:
The volume will decrease by a factor of 10/51.
Explanation:
Hello,
In this case, since both moles and temperature remain constant, we can use the Boyle's law that relates the volume and pressure as an inversely proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
Thus, since the pressure increases by a factor of 5.1 (statement), we have:
[tex]P_2=5.1P_1[/tex]
Thus, the final volume is:
[tex]V_2=\frac{P_1V_1}{P_2} =\frac{P_1V_1}{5.1P_1}\\\\V_2=\frac{10}{51}V_1[/tex]
It means that the volume will decrease by a factor of 10/51.
Regards.
Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto the diameter of a circle. When this is done, the analog along the diameter of the acceleration of the particle executing simple harmonic motion is
Answer:
the analog along the diameter of the acceleration of the particle executing simple harmonic motion is the projection along the diameter of the centripetal acceleration of the particle in the circle
If a ray of light traveling in the liquid has an angle of incidence at the interface of 33.0 ∘, what angle does the refracted ray in the air make with the normal?
Answer:
29°
Explanation:
because the refracted ray angle is small than angle of incidence
An airplane propeller is rotating at 2200 rpm . You may want to review (Pages 255 - 259) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Rotation of a compact disc.
A. How many seconds does it take for the propeller to turn through 49.0?
t = 4.41x10^-3 S
B. Compute the propeller's angular velocity in rad/s
w = 194 rad/s
Answer:
a) 3.7 x 10^-3 s
b) 230.41 rad/s
Explanation:
The angular speed N = 2200 rpm (revolution per minute)
==> 2200/60 revolutions per sec = 36.67 rps
The total angle turned in one second = 36.67 x 360° = 13201.2°
if it takes 1 sec to revolve 13201.2°
then it will take t sec to rotate 49.0°
time t = 49/13201.2 = 3.7 x 10^-3 s
conversion to rad/s = 2πN/60 = (2 x 3.142 x 2200)/60 = 230.41 rad/s
A/An ____________________ is a small, flexible tube with a light and lens on the end that is used for examination. Question 96 options:
Answer:
"Endoscope" is the correct answer.
Explanation:
A surgical tool sometimes used visually to view the internal of either a body cavity or maybe even an empty organ like the lung, bladder, as well as stomach. There seems to be a solid or elastic tube filled with optics, a source of fiber-optic light, and sometimes even a sample, epidurals, suction tool, and perhaps other equipment for sample analysis or recovery.A simple arrangement by means of which e.m.f,s. are compared is known
Answer:
A simple arrangement by means of which e.m.f,s. are compared is known as?
(a)Voltmeter
(b)Potentiometer
(c)Ammeter
(d)None of the above
Explanation:
How does the direction of current flow in the coil affect the orientation of the magnetic field produced by the electromagnet
Answer:
The magnetic field produced by an electric current is always oriented perpendicular to the direction of the current flow. And.Direction of magnetic field is governed by the 'right hand thumb rule, The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of Force . Similar to the situation with electric field lines, the greater the number of lines (or the closer they are together) in an area the stronger the magnetic field.
________ is a thermodynamic function that increases with the number of energetically equivalent ways to arrange components of a system to achieve a particular state.
Answer:
entropy
Explanation:
A concrete slab shown in Figure 5 is being lifted by using three cables connected to the slab at points A, B and C. The slab is in the xy plane. The vertical force required to lift this slab is 60 kN (F 60 kN). Find the tensions in cables DA, DB and DC (show all your workings that you do to find these)
Answer:
Fad = 28.8 kN
Fbd = 16.4 kN
Fcd = 28.1 kN
Explanation:
First, find the length of each cable.
AD = √((2 m)² + (0.5 m)² + (2.5 m)²)
AD = √10.5 m
AD ≈ 3.24 m
BD = √((1.5 m)² + (1 m)² + (2.5 m)²)
BD = √9.5 m
BD ≈ 3.08 m
CD = √((1 m)² + (1 m)² + (2.5 m)²)
CD = √8.25 m
CD ≈ 2.87 m
Next, use similar triangles to find the x, y, and z components of each tension force.
Fadx = 2/3.24 Fad = 0.617 Fad
Fady = 0.5/3.24 Fad = 0.154 Fad
Fadz = 2.5/3.24 Fad = 0.772 Fad
Fbdx = 1.5/3.08 Fbd = 0.487 Fbd
Fbdy = 1/3.08 Fbd = 0.324 Fbd
Fbdz = 2.5 / 3.08 Fbd = 0.811 Fbd
Fcdx = 1/2.87 Fcd = 0.348 Fcd
Fcdy = 1/2.87 Fcd = 0.348 Fcd
Fcdz = 2.5/2.87 Fcd = 0.870 Fcd
Now sum the forces in the x, y, and z directions:
∑Fx = ma
-0.617 Fad + 0.487 Fbd + 0.348 Fcd = 0
∑Fy = ma
-0.154 Fad − 0.324 Fbd + 0.348 Fcd = 0
∑Fz = ma
60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0
To solve this system of equations algebraically, start by subtracting the first two equations, eliminating Fcd.
-0.463 Fad + 0.811 Fbd = 0
0.811 Fbd = 0.463 Fad
Fbd = 0.571 Fad
Substitute into either of the first two equations:
-0.617 Fad + 0.487 (0.571 Fad) + 0.348 Fcd = 0
-0.617 Fad + 0.278 Fad + 0.348 Fcd = 0
-0.339 Fad + 0.348 Fcd = 0
0.348 Fcd = 0.339 Fad
Fcd = 0.975 Fad
Now substituting into the third equation:
60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0
60 kN − 0.772 Fad − 0.811 (0.571 Fad) − 0.870 (0.975 Fad) = 0
60 kN − 0.772 Fad − 0.463 Fad − 0.849 Fad = 0
60 kN − 2.083 Fad = 0
Fad = 28.8 kN
Solving for the other two tension forces:
Fbd = 0.571 Fad = 16.4 kN
Fcd = 0.975 Fad = 28.1 kN
Answer:
Tensions of:
DA = 28.81 KN
DB = 16.45 KN
DC = 28.07 KN
Explanation:
see attached
A 1000 kg car experiences a net force of 9500 N while slowing down from 30 m/s to 16 m/s. How far does it travel while slowing down?
Answer:
33.89 m
Explanation:
We must first obtain the acceleration of the car from;
F=ma
Where
F= force= 9500 N
m= mass of the car= 1000kg
a= acceleration
a= F/m= 9500/1000
a= 9.5 m/s^2
From;
V^2=u^2 + 2as
Where;
V= final velocity
u= initial velocity
s= distance covered
a= acceleration
s= v^2 -u^2/2a
s= (30)^2 -(16)^2/2×9.5
s= 900 - 256/19
s= 644/19
s= 33.89 m
The distance is 33.89 m
The first step is to calculate the acceleration
F= ma
force= 9500N
mass= 1000 kg
9500= 1000 × a
a= 9500/1000
= 9.5 m/s
v²= u² + 2as
30²= 16² + 2(9.5)(s)
900= 256 + 19s
900-256= 19s
644= 19s
s= 644/19
s= 33.89 m
Hence the distance traveled by the car is 33.89 m
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"Light traveling in a medium with a refractive index 1.11 is incident on a plate of another medium with index of refraction 1.66. At what angle of incidence is the reflected light fully polarized?"
Answer:
56°
Explanation:
Brewsters angle can be simply derived from
n1sin theta1= n2sintheta2= n2costheta1
because the reflected light will be 100% polarized if it is reflected at an angle 90o to the refracted light. Hence, Brewsters angle is
Tan theta= n2/n1
1.66/1.11= 1.495
Theta = 56°
Explanation:
A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire
Answer:
The direction of the force is towards the East.
Explanation:
Using the right hand rule, the force on the current carrying conductor is east.
In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.
In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.
To protect her new two-wheeler, Iroda Bike
buys a length of chain. She finds that its
linear density is 0.65 lb/ft.
If she wants to keep its weight below 1.4 lb,
what length of chain is she allowed?
Answer in units of ft.
Answer:
2.2 ft
Explanation:
0.65 lb / 1 ft = 1.4 lb / x
x ≈ 2.2 ft
An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength of 880 nm. What is the potential difference though which this electron was accelerated
Answer:
3x10⁴v
Explanation:
Using
Wavelength= h/ √(2m.Ke)
880nm = 6.6E-34/√ 2.9.1E-31 x me
Ke= 6.6E-34/880nm x 18.2E -31.
5.6E-27/18.2E-31
= 3 x 10⁴ Volts
A long bar slides on two contact points and is in motion with velocity ν. A steady, uniform, magnetic field B is present. The induced current through resistor R is:
Answer:
The induced current in the resistor is I = BLv/R
Explanation:
The induced emf ε in the long bar of length, L in a magnetic field of strength, B moving with a velocity, v is given by
ε = BLv.
Now, the current I in the resistor is given by
I = ε/R where ε = induced emf in circuit and R = resistance of resistor.
So, the current I = ε/R.
substituting the value of ε the induced emf, we have
I = ε/R
I = BLv/R
So, the induced current through the resistor is given by I = BLv/R
The isotope (_90^234)Th has a half-life of 24days and decays to (_91^234)Pa. How long does it take for 90% of a sample of (_90^234)Th to decay to (_91^234)Pa?
Answer:
79.7 days
Explanation:
Half-life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is the amount of time,
and T is the half life.
If 90% decays, then 10% is left.
A = A₀ (½)^(t / T)
0.1 A₀ = A₀ (½)^(t / 24)
0.1 = ½^(t / 24)
ln(0.1) = (t / 24) ln(0.5)
t ≈ 79.7 days
Proposed Exercises: Strength and Acceleration in Circular Movement In the situation illustrated below, a 7kg sphere is connected to a rope so that it can rotate in a vertical plane around an O axis perpendicular to the plane of the figure. When the sphere is in position A, it has a speed of 3m/s. Determine for this position the modulus of tension on the string and the rate at which the tangential velocity is increased.
Answer:
81 N
7.1 m/s²
Explanation:
Draw a free body diagram of the sphere. There are two forces:
Weight force mg pulling straight down,
and tension force T pulling up along the rope.
Sum of forces in the centripetal direction:
∑F = ma
T − mg sin 45° = m v² / r
T = m (g sin 45° + v² / r)
T = (7 kg) (10 m/s² sin 45° + (3 m/s)² / 2 m)
T = 81 N
Sum of forces in the tangential direction:
mg cos 45° = ma
a = g cos 45°
a = (10 m/s²) cos 45°
a = 7.1 m/s²
