Answer:
Materials that allow electricity to pass through them are called conductors.
Copper wire is a good conductor. Materials that do not allow electricity to pass through them are called insulators.
Explanation:
Can I have brainliest?
Is mercury (the planet) rocky or gaseous(meaning relating to or having the characteristics of a gas.)
Answer:
Mercury is rocky
Explanation:
Answer:
Rocky
Explanation:
It has no atmosphere so it cannot hold gas.
Which term defines the distance from crest to crest
Answer:
The horizontal distance between two adjacent crests or troughs is known as the wavelength.
Answer: Wavelength
Explanation:
From crest to crest, it is one full wavelength
Dogs keep themselves cool by panting, rapidly breathing air in and out. Panting results in evaporation from moist tissues of the airway and lungs, which cools the animal. Measurements show that, on a 35∘C day with a relative humidity of 50%, a 12 kg dog loses 1.0 g of water per minute if it is panting vigorously. What rate of heat loss, in watts, does this achieve?
The rate of heat loss, in watts, does this achieve is 37.66 W
Evaporation:It leads in cooling since water absorbs heat equivalent to mass times latent heat of evaporation to get converted into vapor .
So,
latent heat of evaporation of water = 2260 x 10³ J / kg or 2260 J / g
Now
in the evaporation of 1 g of water , heat lost = 2260 J
And,
heat lost per minute = 2260 J
So,
heat lost per second = 2260 / 60
= 37.66 J /s
= 37.66 W
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The uniform slender bar of mass m and length l is released from rest in the vertical position and pivots on its square end about the corner at O. (a) If the bar is observed to slip when 30 , find the coefficient of static friction s between the bar and the corner. (b)If the end of the bar is notched so that it cannot slip, find the angle at which contact between the bar and the corner ceases.
Answer:
A) 0.188
B) 53.1 ⁰
Explanation:
taking moment about 0
∑ Mo = Lo∝
mg 1/2 sin∅ = 1/3 m L^2∝
note ∝ = w[tex]\frac{dw}{d}[/tex]∅
forces acting along t-direction ( ASSUMED t direction)
∑ Ft = Ma(t) = mr∝
mg sin ∅ - F = m* 1/2 * 3g/2l sin∅
therefore F = mg/4 sin∅
forces acting along n - direction ( ASSUMED n direction)
∑ Fn = ma(n) = mr([tex]w^{2}[/tex])
= mg cos∅ - N = m*1/2*3g/1 ( 1 - cos∅ )
hence N = mg/2 ( 5cos∅ -3 )
A ) Angle given = 30⁰c find coefficient of static friction
∪ = F/N
= [tex]\frac{\frac{mg}{4}sin30 }{\frac{mg}{2}(5cos30 -3) }[/tex] = 0.188
B) when there is no slip
N = O
= 5 cos ∅ -3 =0
therefore cos ∅ = 3/5 hence ∅ = 53.1⁰
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.
Match these items.
1 . pls help
asteroids
between Mars and Jupiter
2 .
fission
ice, dust, frozen gases
3 .
energy
sun's atmosphere
4 .
fusion
ability to do work
5 .
corona
splitting atoms
6 .
comets
the combining of atomic nuclei to form one nucleus
Answer:
Here's your answer :
Asteroids - Between mars and JupiterFission - splitting atomsEnergy - Sun's atmosphereFusion - The combining of atomic nuclei to form one nucleusCorona - Ability to do workComets - Ice, dust, frozen gaseshope it helps!
An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is true about this situation?
Answer:
Answer: the true statement form the given statements is “the athletes is not doing any work because he does not move weight”
Explanation:
The athlete isn’t doing any work because he doesn’t move the weight is the correct statement.
What is Work? Work is the energy transferred to or from an object via the application of force along a displacement.Work = Force x Displacement.How to solve this Problem?The weight of an object given is 50kgsThe time of holding an object given is 10 secondsWe need to justify the statements
Here ,
There is no displacement that means displacement is zero.If displacement is zero then work done will also be zeroHence there is no work done by the athlete
Therefore ,The athlete isn’t doing any work because he doesn’t move the weight is the correct statement
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In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give your answer in units of Amps. An Amp is 1 Coulomb of charge flowing through a cross-sectional area of the wire per second - that's a lot of charge per second and will warm up a typical wire quite a bit! Most devices have circuits with larger resistors - kLaTeX: \OmegaΩ (103 LaTeX: \OmegaΩ) and MLaTeX: \OmegaΩ (106 LaTeX: \OmegaΩ) are common.
Answer:
The current pass the [tex]R_2[/tex] is [tex]I = 0.25 A[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The voltage is [tex]V = 3V[/tex]
The first resistance is [tex]R_1 = 7 \Omega[/tex]
The second resistance is [tex]R_2 = 5 \Omega[/tex]
Since the resistors are connected in series their equivalent resistance is
[tex]R_{eq} = R_1 +R_2[/tex]
Substituting values
[tex]R_{eq} = 7 + 5[/tex]
[tex]R_{eq} = 12 \Omega[/tex]
Since the resistance are connected in serie the current passing through the circuit is the same current passing through [tex]R_2[/tex] which is mathematically evaluated as
[tex]I = \frac{V}{R_{eq}}[/tex]
Substituting values
[tex]I = \frac{3}{12}[/tex]
[tex]I = 0.25 A[/tex]
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.
plzzz help will mark the brainliest
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
The throwing back by a wall or barrier of a sound wave without absorbing
it. *
1 point
Answer:Reflection
Explanation:
The throwing back of a sound wave without absorbing it is called reflection
In acoustic reflection of sound is termed as echo i.e. sound arrived at the listener after a particular delay depending upon the position of barrier to the observer.
The reflection of sound is used in many devices like megaphone, trumpets, etc. It is also used in auditorium such that the ceiling of the auditorium is curved for multiple reflections of sound so that sound can be reached at every corner of the auditorium.
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.
Which of the following BEST summarizes the relationship between groups and culture and critical thinking?
Answer:
Groups and culture helps in influencing our values,ethics and beliefs. This influence should always be questioned through the process of thinking critically.
This best summarizes the relationship between groups and culture and critical thinking.
An astronaut is being tested in a centrifuge. The centrifuge has a radius of 11.0 m and, in starting, rotates according to θ = 0.260t2, where t is in seconds and θ is in radians. When t = 2.40 s, what are the magnitudes of the astronaut's (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?
Answer:
a) 1.248 rad/s
b) 13.728 m/s
c) 0.52 rad/s^2
d) 17.132m/s^2
Explanation:
You have that the angles described by a astronaut is given by:
[tex]\theta=0.260t^2[/tex]
(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:
[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t[/tex]
Then, you evaluate for t=2.40 s:
[tex]\omega=0.52(2.40)=1.248\frac{rad}{s}[/tex]
(b) The linear velocity is calculated by using the following formula:
[tex]v=\omega r[/tex]
r: radius if the trajectory of the astronaut = 11.0m
You replace r and w and obtain:
[tex]v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}[/tex]
(c) The tangential acceleration is:
[tex]a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}[/tex]
(d) The radial acceleration is:
[tex]a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}[/tex]
Q1. What is the frequency of rotation of 1000 loop coil of area 20cm2 in a magnetic field of 5T to
generate an emf that has a maximum value of 15.7V?
Answer:
Explanation:
Emf e generated in a coil with no of turn n and area A rotating in a magnetic field B with angular speed of ω is given by the expression
e = e₀ sinωt
where e₀ = nωAB which is the maximum emf generated
Putting the given values
15.7 = 1000xω x 20 x 10⁻² x 5
ω = .0157
frequency of rotation
= ω / 2π
= .0157 / 2 x 3.14
= .0025 /s
9 rotation / hour .
19
Which gas is the most abundant greenhouse gas?
A.
ozone
B.
chlorofluorocarbon
C.
carbon dioxide
OD.
methane
E.
water vapor
Reset
Next
Carbon dioxide is the most abundant greenhouse gas in the atmosphere.
Answer:C
Explanation:
Carbon dioxide is the most abundance greenhouse gas in The atmosphere.
I need some help!!!!!!!!!
Answer:
The Object will immediately begin moving toward the left
Explanation:
Because the force of thirteen is greater than ten and applied to the opposite side
(10 points) A spring with a 7-kg mass and a damping constant 12 can be held stretched 1 meters beyond its natural length by a force of 4 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t of the form c1eαt+c2eβt where α= (the larger of the two) β=
Answer:
......................
Dual Nature of Light
Assignment
Active
Explaining the Nature of Light
Why do scientists believe that light is made of streams of
particles?
The voltage in a circuit is given by the equation V= IR.in this equation v is the voltage Iis correct and R is the resistance which answer shows this equation solved for current?
Answer:I=V/R
Explanation:
V=IR
Divide both sides by R
V/R=IR/R
V/R=I
I=V/R
The voltage in a circuit is given by the equation V= IR, in this equation v is the voltage Iis correct and R is the resistance the solution for the current is given as follows,
V= IR
I = V/R
What is resistance?Resistance is the obstruction of electrons in an electrically conducting material. The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.
V=IR
As for the given problem if we have to solve for the current from the equation V=IR
V=IR
I = V/R
Let us suppose a 60-volt battery connected in a closed circuit with a resistor of 15 ohms then we have o find out the amount of current flowing in the circuit,
Voltage = 60V
Resistance = 15 Ohm
Current =?
By using Ohm,s Law,
V=IR
I = V/R
By substituting the respective values,
I = 60/15
I = 4 Ampere
Hence, we solved for the current from the equation V=IR.
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I need help plz help me out 10 points!!!!!!!
Answer:
The answer is diffraction
Explanation:
Answer:
The answer is diffraction
Explanation:
I did the test! HOPE THIS HELPS!
A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming to rest. what was the deceleration experienced by the survivor? Use g = 9.8 m/s2 Calculate to one decimal.
Answer:
The deceleration is [tex]a = - 76.27 m/s^2[/tex]
Explanation:
From the question we are told that
The height above firefighter safety net is [tex]H = 14 \ m[/tex]
The length by which the net is stretched is [tex]s = 1.8 \ m[/tex]
From the law of energy conservation
[tex]KE_T + PE_T = KE_B + PE_B[/tex]
Where [tex]KE_T[/tex] is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )
and [tex]PE_T[/tex] is the potential energy of the before jumping which is mathematically represented at
[tex]PE_T = mg H[/tex]
and [tex]KE_B[/tex] is the kinetic energy of the person just before landing on the safety net which is mathematically represented at
[tex]KE_B = \frac{1}{2} m v^2[/tex]
and [tex]PE_B[/tex] is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )
So the above equation becomes
[tex]mgH = \frac{1}{2} m v^2[/tex]
=> [tex]v = \sqrt{2 gH }[/tex]
substituting values
[tex]v = 16.57 m/s[/tex]
Applying the equation o motion
[tex]v_f = v + 2 a s[/tex]
Now the final velocity is zero because the person comes to rest
So
[tex]0 = 16.57 + 2 * a * 1.8[/tex]
[tex]a = - \frac{16.57^2 }{2 * 1.8}[/tex]
[tex]a = - 76.27 m/s^2[/tex]
Which of the following is not true about of the use of MRI in medicine?
1) It produces no negative side-effects on the human body
2) It produces high resolution images of soft tissues
3) It is very cheap
4) It requires very strong magnetic fields
Answer:
3) False. It is expensive since it requires sophisticated equipment and very low temperatures
Explanation:
Nuclear magnetic resonance imaging measurements consist of magnetic resonance imaging to analyze tissues by the transition of the unpaired electron at carbon 13, giving information on the structure and composition of tissues. This information is processed in computers and transformed into images.
So the physical measurement is the MRN
Now we can analyze the statements in the problem
1) True by itself a magnetic measurement is non-invasive
2) True. Measuring carbon transitions has information about the soft tissue of the body
3) False. It is expensive since it requires sophisticated equipment and very low temperatures
4) Right. The applied magnetic field is high to be able to induce carbon transaction
A disk with a rotational inertia of 2.5 kg-m2 and a radius 1.1 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center. A force of 7.7 N is applied tangentially to the rim. The angular acceleration of the disk is _____ rad/s2. Round your answer to the nearest tenth.
Answer:
3.4 rad/sec^2
Explanation:
rotational inertia = 2.5 kg-m^2 radius = 1.1 m force = 7.7 N
t = rotational inertia * angular acceleration equation 1
also t = force * radius
therefore to calculate angular acceleration equation 1 becomes
f * r = inertia * angular acceleration hence
angular acceleration = f * r / inertia = [tex]\frac{7.7 * 1.1 }{2.5}[/tex] 8.47 / 2.5 = 3.388 ≈ 3.4 rad/sec^2
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance ddd is
|F|=K|QQ′|d2|F|=K|QQ′|d2,
where K=14πϵ0K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2)ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space.
Consider two point charges located on the x axis: one charge, q1q1q_1 = -15.0 nCnC , is located at x1x1x_1 = -1.660 mm ; the second charge, q2q2q_2 = 34.5 nCnC , is at the origin (x=0.0000)(x=0.0000).
What is the net force exerted by these two charges on a third charge q3q3q_3 = 47.0 nCnC placed between q1q1q_1 and q2q2q_2 at x3x3x_3 = -1.240 mm ?
Your answer may be positive or negative, depending on the direction of the force.
Answer:
Explanation:
Force between two charges of q₁ and q₂ at distance d is given by the expression
F = k q₁ q₂ / d₂
Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm
k = 1/ 4π x 8.85 x 10⁻¹²
putting the values in the expression
F = 1/ 4π x 8.85 x 10⁻¹² x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 9 x 10⁹ x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 35969.4 x 10⁻³ N .
force between charge q₂ = 34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .
putting the values in the expression
F = 1/ 4π x 8.85 x 10⁻¹² x 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 9 x 10⁹ x - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 82729.6 x 10⁻³ N
Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)
Total force = 118699 x 10⁻³
= 118.7 N.
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.
The figure shows a crane whose weight is 12.5 kN and center of gravity in G. (a) If the crane needs to suspend the 2.5kN drum, determine the reactions on the wheel in A and B when the boom is in the position shown.(b) Considering the same situation illustrated, what is the value of the maximum weight that the crane can suspend without tipping over?
Answer:
(a) Ra = 9.25 kN; Rb = 5.75 kN
(b) 26.7 kN
Explanation:
(a) Draw a free-body diagram of the crane. There are four forces:
Reaction Ra pushing up at A,
Reaction Rb pushing up at B,
Weight force 12.5 kN pulling down at G,
and weight force 2.5 kN pulling down at F.
Sum of moments about B in the counterclockwise direction:
∑τ = Iα
-Ra (0.66 m + 0.42 m + 2.52 m) + 12.5 kN (2.52 m + 0.42 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0
-Ra (3.6 m) + 12.5 kN (2.94 m) − 2.5 kN (1.38 m) = 0
Ra = 9.25 kN
Sum of moments about A in the counterclockwise direction:
∑τ = Iα
Rb (0.66 m + 0.42 m + 2.52 m) − 12.5 kN (0.66 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° + 0.66 m + 0.42 m) = 0
Rb (3.6 m) − 12.5 kN (0.66 m) − 2.5 kN (4.98 m) = 0
Rb = 5.75 kN
Alternatively, you can use sum of the forces in the y direction as your second equation.
∑F = ma
Ra + Rb − 12.5 kN − 2.5 kN = 0
Ra + Rb = 15 kN
9.25 kN + Rb = 15 kN
Rb = 5.75 kN
However, you must be careful. If you make a mistake in the first equation, it will carry over to this equation.
(b) At the maximum weight, Ra = 0.
Sum of the moments about B in the counterclockwise direction:
∑τ = Iα
12.5 kN (2.52 m + 0.42 m) − F ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0
12.5 kN (2.94 m) − F (1.38 m) = 0
F = 26.7 kN
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
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
