Answer:
0.99m
Explanation:
Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:
[tex]v=\lambda f=\lambda\frac{1}{T}=(110m)\frac{1}{8.3s}=13.25\frac{m}{s}[/tex]
the relative velocity is:
[tex]v'=13.25m/s-5m/s=8.25\frac{m}{s}[/tex]
This velocity is used to know which is the distance traveled by the boat after 20 seconds:
[tex]x'=v't=(8.25m/s)(20s)=165m[/tex]
Next, you use the general for of a wave:
[tex]f(x,t)=Acos(kx-\omega t)=Acos(\frac{2\pi}{\lambda}x-\omega t)[/tex]
you take the amplitude as 2.0/2 = 1.0m.
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{8.3s}=0.75\frac{rad}{s}[/tex]
by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:
[tex]f(165,20)=1.0m\ cos(\frac{2\pi}{110m}(165)-(0.75\frac{rad}{s})(20s))\\\\f(165,20)=0.99m[/tex]
During an earthquake, _______ travels through the Earth's interior as _______ waves.
Answer:
During an earthquake, seismic waves travels through the Earth's interior as body or p waves.
Explanation:
If neither of the bold words look familiar from your lesson feel free to ignore this answer
What is the major difference between herbal and conventional medicines
What is an independent variable?
A. A variable that is intentionally changed during an experiment
B. A variable that depends on the experimental variable
C. A variable that is not used in an experiment
D. A variable that is unknown during the experiment
Answer:
The answer is A
Explanation:
Independent variables don't have to depend on other factors of the experiment because they're independent
Answer:
A.
Explanation:
Independent variables don't have to depend on other factors of the experiment because they're independent.
A particular coil has 100 turns and a diameter of 6.0 m. When it's time for a measurement, a 4.5 A current is turned on. The large diameter of the coil means that the field in the water flowing directly above the center of the coil is approximately equal to the field in the center of the coil. The field is directed downward and the water is flowing east. The water is flowing above the center of the coil at 1.5 m/s .
What is the magnitude of the field at the center of the coil?
Answer:
The magnetic field at the center of the coil = 5.23 * 10 ^ -5 T
Explanation:
Information from the question:
Number of turns of the coil = 100 turns
The diameter of the coil = 6 m
The radius of the coil = diameter / 2 = 3 m
The coil current = 2.5 A
Formula : The Magnetic field at the center of the coil =
k * number of turns * current / 2 * radius
Therefore, The Magnetic field at the center of the coil=
(4 * [tex]\pi[/tex] * 10 ^ -7 * 100 * 2.5 ) / (2 * 3)
The Magnetic field at the center of the coil = 5.23 * 10 ^ -5 T
While watching a movie a spaceship explodes and there is a loud bang and flash of light. What is wrong with this scene? Explain how you know using evidence and scientific reasoning from the lesson.
What happens if you move a magnet near a could of wire
Answer:
The wire would stick to the magnet????????????????????????
Explanation:
You are designing an optical fiber scope for directing light into a confined area. You want to keep light within the fiber. Based on the specifications, you know that the greatest angle that the light will make with the horizontal is no greater than 25⁰. Assuming you will be using the scope in the body which has the same index of refraction of water (n = 1.33). What is the minimum index of refraction n2 required for the design to be functional?
Answer:
Explanation:
For entry of light into tube of unknown refractive index
sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube
r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.
sin ( 90 - 25 ) / sin( 90 - C) = μ
sin65 / cos C = μ
sinC = 1.33 / μ , where 1.33 is the refractive index of body liquid.
From these equations
sin65 / cos C = 1.33 / sinC
TanC = 1.33 / sin65
TanC = 1.33 / .9063
TanC = 1.4675
C= 56°
sinC = 1.33 / μ
μ = 1.33 / sinC
= 1.33 / sin56
= 1.33 / .829
μ = 1.6 Ans
A cobalt-60 source with activity 2.60×10-4 Ci is embedded in a tumor that has
mas 0.20 kg. The source emits gamma photons with average energy 1.25 MeV.
Half the photons are absorbed in the tumor, and half escape.
i. What energy is delivered to the tumor per second? [4 marks]
ii. What absorbed dose, in rad, is delivered per second? [2 marks]
iii. What equivalent dose, in rem, is delivered per second if the RBE for
these gamma rays is 0.70? [2 marks]
Page 6 of 7
iv. What exposure time is required for an equivalent dose of 200 rem? [2
marks]
B. A laser with power output of 2.0 mW at a wavelength of 400 nm is projected
onto a Calcium metal. The binding energy is 2.31 eV.
i. How many electrons per second are ejected? [6 marks]
ii. What power is carried away by the electrons? [4 marks]
C. A hypodermic needle of diameter 1.19 mm and length 50 mm is used to
withdraw blood from a patient? How long would it take for 500 ml of blood to be
taken? Assume a blood viscosity of 0.0027 Pa.s and a pressure in the vein of
1,900 Pa. [10 marks]
D. A person with lymphoma receives a dose of 35 gray in the form of gamma
radiation during a course of radiotherapy. Most of this dose is absorbed in 18
grams of cancerous lymphatic tissue.
i. How much energy is absorbed by the cancerous tissue? [2 marks]
ii. If this treatment consists of five 15-minute sessions per week over the
course of 5 weeks and just one percent of the gamma photons in the
gamma ray beam are absorbed, what is the power of the gamma ray
beam? [4 marks]
iii. If the gamma ray beam consists of just 0.5 percent of the photons
emitted by the gamma source, each of which has an energy of 0.03
MeV, what is the activity, in Curies, of the gamma ray source? [4 marks]
E. A water heater that is connected across the terminals of a 15.0 V power supply
is able to heat 250 ml of water from room temperature of 25°C to boiling point
in 45.0 secs. What is the resistance of the heater? The density of water is 1,000
kg/m2 and the specific heat capacity of water is 4,200 J/kg/°C. [10 marks]
Answer:
A i. E = 9.62 × 10⁻⁷ J/s
ii. The absorbed dose is 4.81 × 10⁻⁶ Gy
iii. The equivalent dose is 3.37 × 10⁻⁴ rem/s
iv. t = 593471.81 seconds
B. i. 4.025 × 10¹⁵/s
ii. 0.512 mW
C. 7218092.2 seconds
D. i. 6.3 × 10⁻¹ J
ii. 1.4 × 10⁻² W
iii. 1.57 × 10³ Curie
E. 0.129 Ω
Explanation:
The given parameters are;
Mass of tumor = 0.20 kg
Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci
Photon energy = 1.25 MeV
(i) The energy, E, delivered to the tumor is given by the relation;
[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]
[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]
E = 9.62 × 10⁻⁷ J/s
(ii) The equation for absorbed dose is given as follows;
Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg
Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy
1 Gray = 100 rad
4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s
(iii) Equivalent dose, H, is given by the relation;
H = D × Radiation factor, [tex]w_R[/tex]
∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s
(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;
[tex]\dot{H} = \dfrac{H}{t}[/tex]
Therefore;
[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]
∴ t = 6.9 days
B. The number of electrons ejected is given by the relation;
[tex]N = \frac{P}{E} = \frac{P \times \lambda}{hc}[/tex]
[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]
(ii) The power carried by the electron
The energy carried away by the electrons is given by the relation;
[tex]KE_e = hv - \Phi[/tex]
[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]
[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]
Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW
C. The given parameters are;
d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m
l = 50 mm = 5 × 10⁻³ m
V = 500 ml = 5 × 10⁻⁴ m³
η = 0.0027 Pa
p = 1,900 Pa.
[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]
[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]
[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]
t = 7218092.2 seconds
D) i. Energy absorbed is given by the relation;
E = m×D
Where:
D = 35 Gray = 35 J/kg
m = 18 g = 18 × 10⁻³ kg
∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J
ii. Total time for treatment = 15 × 5 = 75 minutes
Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J
Power = Energy(in Joules)/Time (in seconds)
∴ Power = 63/(75×60) = 1.4 × 10⁻² W
iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;
[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]
1 MeV = 1.60218 × 10⁻¹³ J
0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon
Therefore, the number of disintegration per second = 0.28 J/s ÷ 4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second
1 Curie = 3.7 × 10¹⁰ disintegrations per second
Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie
= 1.57 × 10³ Curie
E. The parameters given are;
Density of water = 1000 kg/m³
Volume of water = 250 ml = 0.00025 m³
Initial temperature, T₁, = 25°C
Final temperature, T₂, = 100°C
Change in temperature, ΔT = 100 - 25 = 75°
Specific heat capacity of the water = 4200 J/kg/°C
Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg
∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J
Time to heat the water = 45.0 sec
Therefore, power = Energy/time = 78750/45 = 1750 W
The formula for electrical power = I²R =VI = V²/R
Therefore, where V = 15.0 V, we have;
15²/R = 1750
R = 15²/1750 = 0.129 Ω.
The resistance of the heater = 0.129 Ω.
Find the frequency of the 4th harmonic waves on a violin string that is 48.0cm long with a mass of 0.300 grams
and is under a tension of 4.00N.
Answer:
The frequency of the 4th harmonic of the string is 481.13 Hz.
Explanation:
When a stretch string fixed at both ends is set into vibration, it produces its lowest sound of possible note called the fundamental frequency. Under certain conditions on the string, higher frequencies called harmonics or overtones can be produced.
The frequency of the forth harmonic is the third overtone of the string and can be determined by:
f = [tex]\frac{2}{L}[/tex][tex]\sqrt{\frac{T}{m} }[/tex]
Given that; L = 48.0 cm = 0.48 m,
m = 0.3 g = 0.0003 Kg,
T = 4.0 N,
f = [tex]\frac{2}{0.48}[/tex][tex]\sqrt{\frac{4}{0.0003} }[/tex]
f = 4.1667 × 115.4701
= 481.1252
f = 481.13 Hz
The frequency of the 4th harmonic of the string is 481.13 Hz.
A car speeds up from 18.54 m/s to
29.52 m/s in 13.84 s.
The acceleration of the car is:
Answer:
.7934[tex]m/s^{2}[/tex]
Explanation:
Acceleration = change in velocity / change in time
A = 10.98[tex]m/s[/tex] / 13.84[tex]s[/tex]
A = .7934[tex]m/s^{2}[/tex]
Answer:0.8 m/s^2
Explanation:
initial velocity(u)=18.54m/s
Final velocity(v)=29.52m/s
Time(t)=13.84 sec
Acceleration =(v-u)/t
acceleration =(29.52-18.54)/13.84
Acceleration =10.98/13.34
Acceleration=0.8 m/s^2
Which of the following characteristics of Earth's crust is likely to contribute to geological events?
A. can be broken into many plates
B. has has uniform thickness throughout
C. cysts on top of the solid rock of the lower mantle
D. is composed of a single layer that surrounds Earth
Answer:A
Explanation:
The crust Of The earth's has plates boundaries,when the layers that forms to boundaries glide on each other, vibrations occur which are known as earthquakes
Answer:
A) Can be Broken Into Many Plates
Derive the equation relating the total charge Q that flows through a search coil (Conceptual Example 29.3) to the magnetic-field magnitude B. The search coil has N turns, each with area A, and the flux through the coil is decreased from its initial maximum value to zero in a time Δt. The resistance of the coil is R, and the total charge is Q=IΔt, where I is the average current induced by the change in flux.
Answer:
Q= NBA/R
Explanation:
Check attachment for derivation
The equation relating the total charge, magnitude, turns, time will be "[tex]\frac{NBA}{R}[/tex]".
Magnetic fieldAccording to the question,
Resistance = R
Total charge = Q
Current = I
Number of turns = N
Time = Δt
and,
Q = IΔt ...(equation 1)
We know the flux,
→ [tex]\Phi[/tex] = NBA
Emf induced,
ε = [tex]\frac{- \Delta \Phi}{\Delta t}[/tex]
Δ[tex]\Phi[/tex] = [tex]\Phi_2 - \Phi_1[/tex]
then,
ε = [tex]\frac{NBA}{\Delta t}[/tex]
As we know, Voltage (V) = iR
then, ε = [tex]\frac{NBA}{\Delta t}[/tex] = iR
i = [tex]\frac{NBA}{R \Delta t}[/tex]
Hence, by applying the values in "equation 1"
→ Q = iΔt
= [tex]\frac{NBA}{R \Delta t}[/tex] × Δt
= [tex]\frac{NBA}{R}[/tex]
Thus the response above is correct.
Find out more information magnetic field here:
https://brainly.com/question/14411049
Modified Newtonian dynamics(MoND)proposes that, for small accelerations, Newton’s second law, F = ma, approaches the form F = ma2/a0, where a0 is a constant.
(a) (10 points) Show how such a modified version of Newton’s second law can lead to flat rotation curves, without the need for dark matter.
(b) (10 points) Alternatively, propose a new law of gravitation to replace F = GMm/r2 at distances greater than some characteristic scale r0 so that again, you can explain the observed flat rotation curved of galaxies without dark matter.
Answer:
Explanation:
The two pictures attached here shows the solution to the two questions from the problem. thank you and I hope it helps you
One of your classmates, Kevin, is trying to calculate the acceleration due to gravity at the top of Mt. Everest. Looking at an equation sheet, he sees that the acceleration due to gravity is g = G M r 2. For G, he plugs in the gravitational constant. For M, he plugs in the mass of the Earth. For r, Kevin plugs in the elevation (the height above sea level) of Mt. Everest. Will Kevin arrive at the right answer for g at the top of Mt Everest?
Answer:
no.
Explanation:
No because for M he put the mass of the earth instead of the mass of the object.
Kevin will not arrive at the right answer for g if he calculates the height from sea level, it must be from the center of the earth.
Gravitational acceleration:The force of gravity on an object of mass m is given by:
F = GMm/r²
where G is the gravitational constant
M is the mass of the earth
r is the distance from the center of the earth
This force is equal to the weight of the object given by:
mg = GMm/r²
so,
g = GM/r²
But here r is the distance of the object from the center of the earth not from the sea level.
So, Kevin will not arrive at the right answer for g if he calculates the height from sea level.
Learn more about gravitational acceleration:
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Why does current flow in a coil when a magnet is pushed in and out of the coil ?
Answer:
So the induced current opposes the motion that induced it (from Lenz's Law). When we pull the magnet out, the left hand end of the coil becomes a south pole (to try and hold the magnet back). Therefore the induced current must be flowing clockwise.
hope this helps u...
In order to get going fast, eagles will use a technique called stooping, in which they dive nearly straight down and tuck in their wings to reduce their surface area. While stooping, a 6- kg golden eagle can reach speeds of up to 53 m/s . While golden eagles are not very vocal, they sometimes make a weak, high-pitched sound. Suppose that while traveling at maximum speed, a golden eagle heads directly towards a pigeon while emitting a sound at 1.1 kHz. The emitted sound has a sound intensity level of 30 dB when heard at a distance of 5 m .A) Model this stooping golden eagle as an object moving at terminal velocity. The eagle’s drag coefficient is 0.5 and the density of air is 1.2 kg/m 3 . What is the effective cross-sectional area of the eagle’s body while stooping?B) What is the doppler-shifted frequency that the pigeon will hear coming from the eagle?C) Consider the moment when the pigeon is 5 m away from the eagle. At the pigeon’s position, what is the intensity (in W/m^2 ) of the sound the eagle makes?D) The golden eagle slams into the 250- g pigeon, which is initially moving at 10 m/s in the opposite direction (toward the eagle). The eagle grabs the pigeon in its talons, and they move off together in a perfectly inelastic collision. How fast do they move after the collision?
Answer:
Check the explanation
Explanation:
Part A
F = CA
this drag force balances the weight = 6X 9.8
so
6X9.8 = 0.5 X A X0.5 X 1.2 X 532
A= 0.069 m2
Part B
here the sorce is moving and the observer is at rest
so f= f(- 1 - 1
f = 1.1X10 343 343 – 53
f' = 1.3 KHz
Part C:
given the intensity = 30 dB
we know that I dB = 10 log (I(W/m2))
so we get I (W/m2) = 1000
Part D : The catch
Given that U1 = 53 M1 = 6 kg
U2 =-10 M2=0.25
V1=V2
now conserving momentum
6 X 53 -0.25 X10 =(6+0.25)V
V= 50.48 m/sec
The universal law of gravitation states that the force of attraction between two objects depends on which quantities?
the masses of the objects and their densities
the distance between the objects and their shapes
the densities of the objects and their shapes
the masses of the objects and the distance between them
Save and Exit
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Subm
Kandretum
Answer:depends on the masses of the objects and the distance between them
Explanation:
According to Newton's law of universal gravitation,the force of attraction between two objects depends on the masses of the objects and the distance between them
Two wires, both with current out of the page, are next to one another. The wire on the left has a current of 1 A and the wire on the right has a current of 2 A. We can say that:
A. The left wire attracts the right wire and exerts twice the force as the right wire does.
B. The left wire attracts the right wire and exerts half the force as the right wire does.
C. The left wire attracts the right wire and exerts as much force as the right wire does.
D. The left wire repels the right wire and exerts twice the force as the right wire does.
E. The left wire repels the right wire and exerts half the force as the right wire does.
F. The left wire repels the right wire and exerts as much force as the right wire does.
Answer:
C. The left wire attracts the right wire and exerts as much force as the right wire does.
Explanation:
To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:
[tex]B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\[/tex]
Next, you use the formula for the magnetic force produced by the wires:
[tex]\vec{F_B}=I\vec{L}\ X \vec{B}[/tex]
if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:
[tex]\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}[/tex]
Hence, due to this result you have that:
C. The left wire attracts the right wire and exerts as much force as the right wire does.
In a shipping company distribution center, an open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. You can ignore friction between the cart and the floor. A 15 kg package slides down a chute that makes an angle of 27 degrees below the horizontal. The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds. The package comes to a stop in the cart after 4 seconds. What is:a) the speed of the package just before it lands in the cart
Answer:
Explanation:
The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds . During .75 second duration . package undergoes free fall due to which additional vertical velocity is added
velocity added = a x t
= 9.8 x .75
= 7.35 m /s
Total vertical velocity
= 3 sin27 + 7.35
= 8.71 m /s
Horizontal component = 3 cos 27
= 2.67 m /s
If v be the resultant velocity of these components
v² = 2.67² + 8.71²
v² = 7.13 + 75.86
v = 9.11 m /s .
Two charged particles are accelerated through a uniform electric field and zero magnetic field, then enter a region with zero electric field and a uniform magnetic field. The particles start at rest from the same position (but at different times; they do not interact with each other). They have identical charges, but different masses. Particle 2 has a cyclotron radius 1.5 times as large as that of particle 1. Find ratio m2/m1
Answer:
Explanation:
In magnetic field , charged particle will have circular path . Let the radius of their circular path be r₁ and r₂ . Let their velocity at the time of entering magnetic field be v₁ and v₂ .
The velocity with which they will come out of electric field can be measured from following equation
Eq = 1/2 m v² , E is electric field , q is charge on the particle , m is mass and v is velocity .
v² = 2Eq / m
radius of circular path can be measured by the following expression
m v² / r = Bqv
2Eq / r = Bqv
r = 2Eq / Bqv
= 2E / Bv
r² = 4E² / B²v²
= 4E²m / B²x 2Eq
since E , B and q are constant
r² = K . m
r₂² / r₁² = m₂ / m₁
1.5²
m₂ / m₁ = 1.5²
= 2.25
Air is matter which backs best support the statement
Answer: A. Balloons can be filled with air.
C. Air has mass.
Explanation:
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Balloons are able to be filled with air and air has mass.
which one of the following statements is true? A.in an elastic collision,only momentum is conserved B. in any collision,both momentum & kinetic energy are conserved C.in an inelastic collision,both momentum & kinetic energy are conserved D.in an elastic collision,only kinetic energy is conserved
Answer:
option C is correct
................
Answer:
C- in an inelastic collision, both momentum & kinetic energy are conserved
Explanation:
Took the test
An electric dipole consists of a positive and a negative charge of equal magnitude. Consider an electric dipole with each charge having a magnitude of 1 × 10−6 C. The negative charge is located at (3 cm, 0) and the positive charge is located at (−3 cm, 0). Calculate the electric field from each charge at the points A through E, described below. Use symmetry as much as possible! Using the scale 1 cm = 105 N/C, draw the vector to represent the magnitude and direction of the electric field from each charge. (When entering angle values, enter a number greater than or equal to 0° and less than 360° measured counterclockwise from the +x-axis.) • A = (−13 cm, 0) • B = (−3 cm, 10 cm) • C = (0, 10 cm ) • D = (3 cm, 10 cm) • E = (13 cm, 0) For the negative charge:
Answer:
Explanation:
To find the electric field you use the equation for an electrostatic electric field:
[tex]E=k\frac{q_1q_2}{r^2}[/tex]
r: distance in which E is calculated, from each charge
In the of a dipole you have two contributions to E:
[tex]\vec{E}=\vec{E_1}+\vec{E_2}[/tex]
where E1 is the electric field generated by the first charge and E2 by the second one.
A. (-13 cm, 0):
First you calculate the vectors E1 and E2:
[tex]E_1=(8.98*10^9)\frac{(1*10^{-6}C)}{(-0.13-0.03)^2}\hat{j}\\\\E_1=350781.25N/C\\\\E_2=-(8.98*10^9)\frac{(1*10^{-6}C)}{(-0.13+0.03)^2}\hat{j}\\\\E_2=-989000N/C[/tex]
Then, you sum both contributions:
[tex]\vec{E}=-547218.75N/C\hat{j}[/tex]
B. (-3cm, 10cm):
[tex]r_1=\sqrt{(0.06)^2+(0.1)^2}=0.116m\\\\\theta=tan^{-1}(\frac{0.06}{0.1})=30.96\°\\\\r_2=0.1m\\\\E_1=(8.98*10^9Nm^2/C)\frac{(1.6*10^{-6}C)}{(0.116m)^2}[cos(30.96\°)\hat{i}+sin(30.96\°)\hat{j}]\\\\E_1=[-915646\hat{i}-549306.42\hat{j}]N/C\\\\\theta=(90-30.96)+180=239.04\°\\\\[/tex]
the last angle is calculated again because the vector direction is measured from the +x axis.
and for the second vector:
[tex]E_2=(8.98Nm^2/C)\frac{1.6*10^{-6}C}{(0.1m)^2}\hat{j}\\\\E_2=1436800N/C\hat{j}[/tex]
the total E is:
[tex]\vec{E}=[-915646\hat{i}+887493.58\hat{j}]N/C[/tex]
1. A tennis ball is dropped from a second story window. It is in free fall, accelerating downward at a rate of 9.8 m/s2. At the exact same time, another person throws a tennis ball out the adjacent window with a horizontal velocity of 30 m/s. Assuming no air resistance, which ball hits the ground first? Why?
Answer:
Both balls will hit the ground at the same time, because gravity is constant.Explanation:
Notice that both balls are being thrown at the same initial height.
It's important to know that these movements which depends of gravity (a constant acceleratio) they would fall at the same rythm, because the gravity is a constant.
Remember that gravity is an acceleration, which it's defined as the change of the velocity, so if both balls change their vertical velocity at the same rate, then they will fall at the same time, because they have the same initial height.
Additionally, when you throw a ball horizontally, it will bend down due to gravity, and this falling movement is the same as if you throw vertically as a free falling movement.
Therefore, both balls will hit the ground at the same time, because gravity is constant.
(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached an altitude of 500m and is rising at a velocity of 125 m/s. Gravity then takes over. The height of the rocket as a function of time is h(t)=-9.8/2 t^2+125t+500,t>0. Using your function file from HW2A: Generate a plot of height (vertical axis) vs. time (horizontal axis) from 0 to 30 seconds. Include proper axis labels. Find the maximum height and the time at which it occurs: Analytically, showing your steps and equations. This part should be done entirely in the write-up: no coding Using the data cursor on the plot. Using the MAX function on your data from part (a) Using FMINSEARCH on your m file Comment on the differences between the methods. How closely does each method match the "true" (analytical) value? Find the time when the rocket hits the ground: Analytically, showing your equations. This part should be done entirely in the write-up: no coding Using the data cursor on the plot. Using FZERO on your m file Comment on the differences between the methods in each of part (B) and (C). How closely does each method match the "true" (analytical) value? Use a quantitative comparison to make your argument.
Answer:
Explanation:
Given that,
h(t) = -9.8t² / 2 + 125t + 500
for t > 0.
At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.
We want to find the maximum height reached by rocket
Using mathematics maxima and minima
let find the turning point when dh/dt = 0
dh/dt = -9.8t + 125
dh / dt = 0 = -9.8t + 125
9.8t = 125
t = 125 / 9.8
t = 12.76s
Let find the turning point to know if this time t = 12.76 is maximum or minimum point
Let find d²h / dt²
d²h / dt² = -9.8
Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.
Then, the maximum height reached is
h = -9.8t² / 2 + 125t + 500
h = -9.8(12.76)² / 2 + 125(12.76) + 500
h = -797.80 + 1595 + 500
h = 1297.2 m
The maximum height reached is 1297.2 m
From the attachment, the maximum height is 1297.2m at t = 12.76sec.
Comment, the result are the same for both the analysis aspect and the graphical aspect.
This is a measure of quantity of matter
Answer:
Mass
Explanation:
Mass is the measure of amount of matter contained within any substance and hence mass determines the weight. Unit of mass is kilogram as per ISI system of units.
Mass is measured through a balance. The more is the mass of an object, the more the balance tilts towards the object side.
Weight is equal to product of mass and the gravitational constant i.e 9.8m/s^2
If A = (6i-8j) units, B = (-8i-3j) units, and C = (26i-19j) units, determine a and b
such that aA + bB + C = 0
Answer:
Explanation:
given equation
aA + bB + C = 0
Putting the given values
a(6i-8j) +b (-8i-3j) +(26i-19j) = 0
i ( 6a - 8b ) - j ( 8a + 3 b ) = - 26 i + 19 j
comparing the coefficients of i and j
6a - 8b = -26
8a + 3b = -19.
multiplying first equation by 4 and second equation by 3
24a - 32 b = - 104
24a + 9b = -57
9b + 32b = -57 + 104
41 b = 47
b = 1.41
6 a - 8 x 1.41 = -26
6a = -14.72
a = - 2.45
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram-meter^3 and the density of silicon in other units: 2.33 gram-centimeter^3. You decide to convert the density of silicon into units of kilogram-meter^3 to perform the comparison.
By which combination of conversion factors will you multiply 2.33 gram-centimeter^3 to perform the unit conversion?
Answer:
Explanation:
To convert gram / centimeter³ to kg / m³
gram / centimeter³
= 10⁻³ kg / centimeter³
= 10⁻³ / (10⁻²)³ kg / m³
= 10⁻³ / 10⁻⁶ kg / m³
= 10⁻³⁺⁶ kg / m³
= 10³ kg / m³
So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³
2.33 gram / cm³
= 2.33 x 10³ kg / m³ .
Suppose you wanted to use a non-reflecting layer for radar waves to make an aircraft invisible. What would the thickness of the layer be to avoid reflecting 2 cm radar waves. (You can neglect changes of wavelength in the layer for this problem.) Would there be any problems as the aircraft turn
Answer:
the thickness of the film for destructive interference is 1 cm
Explanation:
We can assume that the radar wave penetrates the layer and is reflected in the inner part of it, giving rise to an interference phenomenon of the two reflected rays, we must be careful that the ray has a phase change when
* the wave passes from the air to the film with a higher refractive index
* the wavelength inside the film changes by the refractive index
λ = λ₀ / n
so the ratio for destructive interference is
2 n t = m λ
t = m λ / 2n
indicate that the wavelength λ = 2 cm, suppose that the interference occurs for m = 1, therefore it is thickness
t = 1 2/2 n
t = 1 / n
where n is the index of refraction of the anti-reflective layer. As they tell us not to take into account the change in wavelength when penetrating the film n = 1
t = 1 cm
So the thickness of the film for destructive interference is 1 cm
Answer:
the thickness of the film for destructive interference is 1 cm
Explanation:
What spectacles are required for reading purposes by a person whose near point is 2.0m
Answer:Convex lens spectacles is required for reading purpose..
Explanation:
I don't say you have to mark my ans as brainliest but if it has really helped you please don't forget to thank me...
