Answer:
We can remove ions and water molecules
Explanation:
We can remove ions and water molecules because both of them have an electric field . So this will help in repelling or attractive the above mentioned particles. Water is polar so it has an electric field due to net dipole moment.
Which of the following choices is not an example of climate?
0000
San Diego has mild, warm temperatures and sea breezes year-round.
Anchorage has short, cool summers and long, snowy winters.
It will be 78° on Friday in Clovis.
Florida is tropical, with a significant rainy season.
Answer:
Florida is tropical, with a significant rainy seson
What is an internal resistance?
Explanation:
some thing inside a resistor
find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4
The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:
Q(t) = Aeσ[tex]T^{4}[/tex]
where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.
To determine the rate of energy radiated by the man in the given question;
[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ
But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.
So that;
[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]
= 3.8556 x [tex]10^{-8}[/tex]
= 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex]
Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
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A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects
Answer:
P = -1 D
Explanation:
For this exercise we must use the equation of the constructor
/ f = 1 / p + 1 / q
where f is the focal length, p and q is the distance to the object and the image, respectively
The far view point is at p =∞ and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image
[tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]
f = 1 m
P = 1/f
P = -1 D
define nortons theorem
Answer:
In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.
the number of significant figures in the measurement 4.300×10^5 km are
Answer:
6
Explanation
Any numbers in scientific notation are considered significant. For example, 4.300 x 10-4 has 4 significant figures.
Answer From Gauth Math
cyclist always bends when moving the direction opposite to the wind. Give reasons
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B
Explanation:
Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]
The sum of the two vectors is
[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]
[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]
The difference between the two vectors can be written as
[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]
[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]
Determine usando ecuación de Bernoulli la Presión P1 necesaria para mantener la condición mostrada dentro del sistema mostrado en la figura, sabiendo que el aceite tiene un s.g =0.45 y el valor de d=90mm.
Answer:
PlROCA
Explanation:
Which of the units of the following physical quantities are derived
Answer:
where is the attachment
Explanation:
I only need help with e (bottom of the page).
Explanation:
The box is accelerating along the y-axis at a rate of [tex]+2.5\:\text{m/s}^2[/tex] as well as along the x-axis at a rate of [tex]+5.1\:\text{m/s}^2.[/tex] So the magnitude of the box's total acceleration is given by
[tex]a_T = \sqrt{a_x^2 + a_y^2}[/tex]
[tex]\:\:\:\:= \sqrt{(5.1\:\text{m/s}^2)^2 + (2.5\:\text{m/s}^2)^2}[/tex]
[tex]\:\:\:\:=5.7\:\text{m/s}^2[/tex]
The direction of the acceleration [tex]\theta[/tex] with respect to the horizontal direction is given by
[tex]\theta = \tan^{-1}\!\left(\dfrac{a_y}{a_x}\right) = \tan^{-1}\!\left(\dfrac{2.5\:\text{m/s}^2}{5.1\:\text{m/s}^2}\right)[/tex]
[tex]\:\:\:\:= 26.1°[/tex]
Una persona de 76 kg está siendo retirada de un edificio en llamas mientras se muestra en la figura. Calcule la tensión
en las dos cuerdas si la persona está momentáneamente inmovil.
Ayuda por favor.
Answer:
T1 = 736.6 N, T2 = 193.5 N
Explanation:
W = 76 N
The tension is T1 and T2.
By use of Lami's theorem
[tex]\frac{T_1}{Sin100}=\frac{T_2}{Sin165}=\frac{W}{Sin 95}\\\\So, \\\\T_1 = \frac{76\times 9.8\times Sin 100}{Sin 95} = 736.6 N \\And\\T_2 = \frac{76\times 9.8\times Sin 165}{Sin 95} = 193.5 N \\[/tex]
The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the gravitational force between them ? ans. 2.668× 10-9N
[tex]F = 5.93×10^{13}\:\text{N}[/tex]
Explanation:
Given:
[tex]m_1= 2×10^{16}\:\text{kg}[/tex]
[tex]m_2= 4×10^{22}\:\text{kg}[/tex]
[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]
Using Newton's universal law of gravitation, we can write
[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]
[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]
[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]
15 . A scientist who studies the whole environment as a working unit .
Botanist
Chemist
Ecologist
Entomologist
Answer:
Ecologist.
Your answer is Ecologist.
(Ecologist) is a scientist who studies the whole environment as a working unit.
Is it true that as we gain mass the force of gravity on us decreases
Answer:
No. As we gain mass the force of gravity on us does not decrease
Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 655 nm and a peak electric field magnitude of 1.5 V/m. 0.002984 W/m2 (b) an electromagnetic wave with an angular frequency of 6.5 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T. 1.19366E-6 W/m2
The intensity of the electromagnetic wave in terms of the electric field is 0.00298 W/m² and the intensity of the electromagnetic wave in terms of the magnetic field is 1.193x10⁻⁶ W/m².
The intensity of the electromagnetic wave is related to the electric field as well as to the magnetic field.
a) Intensity of the electromagnetic wave for the electromagnetic field.
The intensity of the electromagnetic wave (I) in terms of the electromagnetic field is given by:
[tex] I = \frac{E^{2}*c*\epsilon_{0}}{2} [/tex] (1)
Where:
c: is the speed of light = 3.00*10⁸ m/s
E: is the magnitude of the electric field = 1.5 V/m
ε₀: is the permittivity of free space = 8.85*10⁻¹² C²/Nm²
Hence, the intensity of the electromagnetic wave (eq 1) is:
[tex] I = \frac{(1.5 V/m)^{2}*3.00 \cdot 10^{8} m/s*8.85 \cdot 10^{-12} C^{2}/(N*m^{2})}{2} = 0.00298 W/m^{2} [/tex]
b) Intensity of the electromagnetic wave for the magnetic field
We can calculate the intensity of the electromagnetic wave (I) in terms of the magnetic field with the following equation:
[tex] I = \frac{cB^{2}}{2\mu_{0}} [/tex] (2)
Where:
B: is the magnitude of the magnetic field = 10⁻¹⁰ T
μ₀: is the vacuum permeability = 4π*10⁻⁷ m*T/A
Therefore, the intensity of the electromagnetic wave (eq 2) is:
[tex] I = \frac{3.00 \cdot 10^{8} m/s*(1\cdot 10^{-10} m*T/A)^{2}}{2*4\pi \cdot 10^{-7} T/A} = 1.193 \cdot 10^{-6} W/m^{2} [/tex]
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A force of 1000N is used to kick a football of mass 0.8kg find the velocity with which the ball moves if it takes 0.8 sec to be kicked.
The velocity of the ball is 100m/s
The first step is to write out the parameters;
The force used to kick the ball is 1000N
The mass of the ball is 0.8 kg
Time is 0.8 seconds
Therefore the velocity can be calculated as follows
F= Mv-mu/t
1000= 0.8(v) - 0.8(0)/0.8
1000= 0.8v- 0.8/0.8
Cross multiply both sides
1000(0.8) = 0.8v
800= 0.8v
divide both sides by the coefficient of v which is 8
800/0.8= 0.8v/0.8
v= 1000m/s
Hence the velocity is 1000m/s
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The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?
Answer:
[tex]v_1=87.40m/s[/tex]
Explanation:
From the question we are told that:
Mass of arrow [tex]m=52g[/tex]
Mass of rock [tex]m_r=1.50kg[/tex]
Height [tex]h=0.47m[/tex]
Generally the equation for Velocity is mathematically given by
[tex]v = \sqrt{(2gh)}[/tex]
[tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]
[tex]v= 3.035m/s[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_1v_1=m_2v_2[/tex]
[tex]0.052kg * v = 1.5 * 3.03m/s[/tex]
[tex]v_1=87.40m/s[/tex]
a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10.2 m
Answer:
Explanation:
Here's the info we have:
initial velocity is 20 m/s;
final velocity is our unknown;
displacement is -10.2 m; and
acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation
v² = v₀² + 2aΔx and filling in accordingly to solve for v:
[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex] Rounding to the correct number of sig fig's to simplify:
[tex]v=\sqrt{400+2.0*10^2}[/tex] to get
v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.
So either 20 m/s or 24 m/s
A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.
a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.
(A) The pressure will be greater than 10% overpressure limit.
(B) The final pressure will be "582.915 kPa".
Given:
Volume,
[tex]V = 0.0021 \ m^3[/tex]Initial pressure,
[tex]P_o= 495 \ kPa[/tex]Initial temperature,
[tex]T_o = -14^{\circ} C[/tex][tex]= 259 \ K[/tex]
Final temperature,
[tex]T = 32^{\circ} C[/tex](B)
Number of moles,
→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]
then,
The final absolute pressure,
→ [tex]P = \frac{nRT}{V}[/tex]
[tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]
[tex]=(\frac{T}{T_o} )P_o[/tex]
[tex]= (\frac{305}{259} )\times 495[/tex]
[tex]= 582.915 \ kPa[/tex]
Thus the above approach is correct.
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Determine the density in kg \cm of solid whose Made is 1080 and whose dimension in cm are length=3 ,width=4,and height=3
Answer:
d = 30kg/cm³
Explanation:
d = m/v
d = 1080kg/(3cm*4cm*3cm)
d = 30kg/cm³
A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?
Answer:
Charge of the particle is 1 coulomb.
Explanation:
Force, F:
[tex]{ \bf{F=BeV}}[/tex]
F is magnetic force.
B is the magnetic flux density.
e is the charge of the particle.
V is the velocity
[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]
The source of sound moves away from the listener.The listener has the impression that the source is lower in pitch. Why?
When the source is moving away from the observer the velocity of the source is added to the speed of light. This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.
A block weighing 400 kg rest on a horizontal surface and supports on top of it another block of weight 100 kg placed on the top of it as shown. The block W2 is attached to a vertical wall by a string 6 m long. Ifthe coefficient of friction between all surfaces is 0.25 and the system is in equilibrium find the magnitude of the horizontal force P applied to the lower block.
The horizontal force applied to the lower block is approximately 1,420.85 Newtons
The known parameters are;
The mass of the block, m₁ = 400 kg, weight, W₁ = 3,924 N
The mass of the block resting on the first block, m₂ = 100 kg, weight, W₂ = 981 N
The length of the string attached to the block, W₂, l = 6 m
The horizontal distance from the point of attachment of the second block to the block W₂, x = 5 m
The coefficient of friction between the surfaces, μ = 0.25
Let T represent the tension in the string
The upward force on W₂ due to the string = T × sin(θ)
The normal force of W₁ on W₂, N₂ = W₂ - T × sin(θ)
The tension in the string, T = N₂ × μ × cos(θ)
∴ T = (W₂ - T × sin(θ)) × μ × cos(θ)
sin(θ) = √(6² - 5²)/6
cos(θ) = 5/6
∴ T = (981 - T × √(6² - 5²)/6) × 0.25 × 5/6
Solving, we get;
T ≈ 183.27 N
The normal reaction on W₂, N₂ = T/(μ × cos(θ))
∴ N₂ = 183.27/(0.25 × 5/6) = 879.7
N₂ ≈ 879.7 N
The friction force, [tex]F_{f2}[/tex] = N₂ × μ
∴ [tex]F_{f2}[/tex] = 879.7 N × 0.25 = 219.925 N
The total normal reaction on the ground, [tex]\mathbf{N_T}[/tex] = W₁ + N₂
[tex]N_T[/tex] = 3,924 N + 879.7 N = 4,803.7 N
The friction force, on the ground [tex]\mathbf{F_T}[/tex] = [tex]\mathbf{N_T}[/tex] × μ
∴ [tex]F_T[/tex] = 4,803.7 N × 0.25 = 1,200.925 N
The horizontal force applied to the lower block, P = [tex]\mathbf{F_T}[/tex] + [tex]\mathbf{F_{f2}}[/tex]
Therefore;
P = 1,200.925 N + 219.925 N = 1,420.85 N
The horizontal force applied to the lower block, P ≈ 1,420.85 N
The graph below shows a cycle of a heat engine. Add the following labels to the graph. Some labels are used more than once.
Labels: Isobaric process; W= 0J; Work done on the system; Work done by the system.
I will give brainliest!
P.S. AL2006 if you see this please help!
I'm not very good at this material. I'll try it, but if I were you, I wouldn't bet money on these answers.
"Isobaric" means constant pressure. So those are the horizontal lines, where every point on the line is at the same pressure. Those are the processes 1>2 and 3>4 .
I'm going around and around in my mind with the other labels, and I can't decide. So I'm afraid I can't answer any more of them ... they might be wrong.
Answer:
1 -> 2 & 3 -> 4: Isobaric process
4 -> 1: Work done BY the system
2 -> 3: Work done ON the system
W(total): W = 0J
Explanation:
The two horizontal lines (1 -> 2 & 3 -> 4) are "Isobaric" since isobaric processes take place at constant pressure. I believe 4 -> 1 is "Work done BY the system" since pressure increases when there is an increase of thermal energy, in other words, the system is absorbing heat. This is why the volume increases from 1 -> 2 after the system has absorbed heat in 4 -> 1. Following the directions of the arrows, 2 -> 3 would be "Work done ON the system" since pressure is DECREASING, meaning temperature is also exiting the system. That's why the next step (3 -> 4) shows a decrease in volume. This model depicts a process that has a W(total) of 0 J because this is a cycle.
I hope this helps :))
When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity
The cells lie odjacent to the sieve tubes
Answer:
Almost always adjacent to nucleus containing companion cells, which have been produced as sister cells with the sieve elements from the same mother cell.A Ball A and a Ball B collide elastically. The initial momentum of Ball A is -2.00kgm/s and the initial momentum of Ball B is -5.00kgm/s. Ball A has a mass of 4.00kg and is traveling at 2.50 m/s after the collision. What is the velocity of ball B if it has a mass of 6.50kg?
The velocity of B after the collision is obtained as -2.6 m/s.
What is the principle of conservation of momentum?Now we now that the principle of conservation of momentum states that the momentum before collision is equal to the momentum after collision.
Thus;
(-2.00kgm/s) + ( -5.00kgm/s) = ( 4.00kg * 2.50 m/s) + ( 6.50kg * v)
-7 = 10 + 6.5v
-7 - 10 = 6.5v
v = -7 - 10 /6.5
v = -2.6 m/s
Hence, the velocity of B after the collision is obtained as -2.6 m/s.
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You are working on a project to make a more efficient engine. Your team is investigating the possibility of making electrically controlled valves that open and close the input and exhaust openings for an internal combustion engine. Determine the stability of the valve by calculating the force on each of its sides and the net force on the valve.
The valve is made of a thin but strong rectangular piece of non-magnetic material that has a current-carrying wire along its edges. The rectangle is 0.35 cm x 1.83 cm. The valve is placed in a uniform magnetic field of 0.15 T such that the field lies in the plane of the valve and is parallel to the short sides of the rectangle. The region with the magnetic field is slightly larger than the valve. When a switch is closed, a 1.7 A current enters the short side of the rectangle on one side and leaves on the opposite short side of the rectangle. At the suggestion of a colleague, who is hoping to ensure different currents along the sides of the valve, resistors have been included along the wire on each of the short sides of the valve. The value of the resistor on one side is twice that on the other side.
Answer:
The answer is "0.00466 N".
Explanation:
[tex]F=(B \times i) L\\\\[/tex]
therefore the smaller side is parallel to magnetic field
[tex]\therefore \\\\F= B i L\ \sin\ 'o'=0 \ N[/tex]
calculating the force on the layer side:
[tex]\to F=0.15 \times 1.7 \times 0.0183 \times \sin 90^{\circ}=0.00466\ N\\\\[/tex]
Therefore [tex]F_o[/tex] the net force on the rectangular loop [tex]= 0.00466 \ N[/tex]
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.
We have that the spring constant is mathematically given as
[tex]k=2.37*10^{11}N/m[/tex]
Generally, the equation for angular velocity is mathematically given by
[tex]\omega=\sqrt{k}{m}[/tex]
Where
k=spring constant
And
[tex]\omega =\frac{2\pi}{T}[/tex]
Therefore
[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]
Hence giving spring constant k
[tex]k=m((\frac{2 \pi}{T})^2[/tex]
Generally
Mass of earth [tex]m=5.97*10^{24}[/tex]
Period for on complete resolution of Earth around the Sun
[tex]T=365 days[/tex]
[tex]T=365*24*3600[/tex]
Therefore
[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]
[tex]k=2.37*10^{11}N/m[/tex]
In conclusion
The effective spring constant of this simple harmonic motion is
[tex]k=2.37*10^{11}N/m[/tex]
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