Answer:
Minority group - People who are singled out for unequal treatment and who regard themselves as objects of collective discrimination.
Ethnicity - A shared identity based on cultural elements such as heritage, language, and religion.
Majority group - A group's subset that consists of more than half of the entire group's members.
Race - A socially constructed group of people who share physical characteristics that are considered significant by a society and that are used to distinguish them from other groups.
Explanation:
To better understand the above-mentioned definitions we would use some examples:
Elder people can be considered as a minority group in the United States because of their reduced status as a consequence of popular discrimination and prejudice against them. In contrast to them, young men consisting of more than half the population in the age category is an example of the majority group.
Groups of people like Hispanic Americans, Jews, Italian America, Irish are some examples of people belonging to different ethnic backgrounds residing in the U.S.
African Americans who are distinguished based on their skin color is an example of race.
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.
What happens if you move a magnet near a could of wire
Answer:
The wire would stick to the magnet????????????????????????
Explanation:
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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(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.
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...
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
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.
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.
A pipe branches symmetrically into two legs of length L, and the whole system rotates with angular speed ω around its axis of symmetry. Each branch is inclined at angle α to the axis of rotation. Liquid enters the pipe steadily, with zero angular momentum, at volume flow rate Q. The pipe diameter, D, is much smaller than L. Obtain an expression for the external torque required to turn the pipe. What additional torque would be required to impart angular acceleration ω_ ?
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
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!
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
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]
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
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]
plzzz help will mark the brainliest
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
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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 forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-degrees above the forward direction .Find the magnitude and direction(relative to forward direction of the resultant force that these forces exert on the body)
Answer:
F = (913.14 , 274.87 )
|F| = 953.61 direction 16.71°
Explanation:
To calculate the resultant force you take into account both x and y component of the implied forces:
[tex]\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N[/tex]
Thus, the net force over the body is:
[tex]F=(913.14N)\hat{i}+(274.87N)\hat{j}[/tex]
Next, you calculate the magnitude of the force:
[tex]F=\sqrt{(913.14N)+(274.87N)^2}=953.61N[/tex]
and the direction is:
[tex]\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°[/tex]
Question 7
Review
Which particles are not affected by the strong force?
A.
hadrons
B.
protons
C.
neutrons
D
electrons
Submit A
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Answer:
Electron
Explanation:
Because electron are not hadrons so electron are not affected by strong force
Particles that can not be affected by strong forces are electrons.
What are electrons?Electrons are the rotating material around the nucleus of an atomic element in orbit.
Atoms have electrostatic energy between their electrons. This force is not broken by a force as strong as nuclear power.
What are strong forces?Strong force is a fundamental interaction of nature that acts between subatomic particles of matter.
There are four basic forces in nature:
Gravity: the gravitational force used between any heavy objects. It has an infinite range.Electrical energy: energy used between electrically charged objects. It can be either attraction or repulsion.Nuclear power: is the magnetic field that responds to the binding of protons and neutrons within the nucleus of an atom. It only works for very short distances.Weak nuclear power: a force that causes nuclear decay. It only works for very short distances.Therefore, particles that are unaffected by strong force are electrons.
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Which of these parameters is directly related to sound frequency?
Answer:Velocity
Explanation:
Velocity is directly proportional to the frequency of a wave.
Velocity=frequency x wavelength
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 current in the wires of a circuit is 60 milliamps. If the resistance of the circuit were doubled (with no change in voltage), then it’s new current would be _____ milliamps
Answer:30
Explanation:
Current=60 milliamps
Current=(voltage)/(resistance)
60=(voltage)/(resistance)
Doubling the resistance means multiplying both sides by 1/2
60x1/2=(voltage)/(resistance) x 1/2
30=(voltage)/2(resistance)
Therefore the resistance would be 30 milliamp if we double the resistance
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.
Distributions of electric charges in a cell play a role in moving ions into and out of a cell. In this situation, the motion of the ion is affected by two forces: the electric force due to the non-uniform charge distribution in the cell membrane, and the resistive force (viscosity) due to colliding with the fluid molecules. In order to begin our analysis of this, let's consider a toy model in which the ion is moving in response to electric forces alone.
Charges in a cell membrane are distributed along the opposite sides of the membrane approximately uniformly. This leads to an (on the average) constant electric field inside the membrane. A simple model that gives this kind of field is two large parallel plates close together. The field between the plates is approximately constant pointing from the negative to the parallel plate. This results in a charge feeling a constant force anywhere between the plates (sort of like flat-earth gravity turned sideways). Outside of the plates the electric fields from the two plates cancel and there is no force.
2. The electric field between the plates (inside the membrane) is about 107 N/C and the thickness of the membrane is about 7 nm. Estimate:
2.1 The electric force on the ion when it is in the center of the channel.
F = N
Explain your reasoning.
2.2 The acceleration of the ion when it is in the center of the channel.
a = nm/s2
Explain your reasoning.
2.3 The magnitude of the change in the ion's potential energy as it crosses from one side of the plates to the other.
U = J
Explain your reasoning.
2.4 The kinetic energy the ion would gain as it crosses from one side of the plates to the other.
KE = J
Explain your reasoning.
Could you explain 2.3!
Answer:
An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element. Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.Explanation:
An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.
What is atom?Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.
Each atom is made up of a nucleus and one or more electrons that are linked to it. One or more protons and a significant number of neutrons make up the nucleus. Only the most prevalent type of hydrogen is neutron-free.
Atoms that are neutral or ionized make up every solid, liquid, gas, and form of plasma. Atoms are incredibly tiny, measuring typically 100 picometers across. The nucleus of an atom contains more than 99.94% of its mass.
Therefore, An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.
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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.
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.
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
A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its motion? What is the coefficient of kinetic friction?
A block is given an initial speed of 3m/s up a 25° incline. Coefficient of friction
Answer:
a) μ = 0.475 , b) μ = 0.433
Explanation:
a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
y Axis
N - [tex]W_{y}[/tex] = 0
let's use trigonometry for the components the weight
sin 27 = Wₓ / W
Wₓ = W sin 27
cos 27 = W_{y} / W
W_{y} = W cos 27
N = W cos 27
W sin 27 - μ W cos 27 = m a
mg sin 27 - μ mg cos 27 = m a
μ = (g sin 27 - a) / (g cos 27)
very = tan 27 - a / g sec 27
μ = 0.510 - 0.0344
μ = 0.475
b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result
μ = tan 25 - 0.3 / 9.8 sec 25
μ = 0.466 -0.03378
μ = 0.433
Pendulum clock. Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.
Answer:
The time period for this pendulum is 1.68 seconds
Explanation:
Solution
Given that:
The length of the pendulum is measured from the axis of rotation to the center of mass of the bob of the pendulum
Now,
In this case, the length becomes:
L= 80 - 15+5
L = 70 cm
The time period = T = 2π √L/g
T = 2* 3.14 *√0.7/9.8
= 1.68 seconds
Note: Kindly find an attached work to the part of the solution of the given question
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
