Answer:
a
When [tex]r \ge R[/tex]
[tex]B = \frac{ \mu_o * I}{ 2 \pi r }[/tex]
b
When [tex]r< R[/tex]
[tex]B = [\frac{\mu_o * I }{ 2 \pi R^2} ]* r[/tex]
Explanation:
From the question we are told that
The radius is R
The current is I
The distance from the center
Ampere's law is mathematically represented as
[tex]B[2 \pi r] = \mu_o * \frac{I r^2 }{R^2 }[/tex]
[tex]B = \frac{ \mu_o}{2 \pi } * \frac{r}{R^2}[/tex]
When [tex]r \ge R[/tex]
=> [tex]B = \frac{ \mu_o * I}{ 2 \pi r }[/tex]
But when [tex]r< R[/tex]
[tex]B = [\frac{\mu_o * I }{ 2 \pi R^2} ]* r[/tex]
What is the root mean square speed of an As4 particle as it is sublimed? (Assume at high temperatures arsenic acts like an ideal gas; the Boltzmann constant, kB, can be approximated as 1.4 x 10-23 J∙K-1) and the kinetic energy of the gas is equal to 3/2kB T
Answer: [tex]v_{rms} =[/tex] 273m/s
Explanation: Root Mean Square Speed of an atom or molecule is the speed of a particle in a gas. It is the average speed a particle in a gas can have.
It can be calculated:
[tex]\frac{1}{2}mv^{2} = \frac{3}{2} k_{B}T[/tex]
[tex]v^{2} = \frac{3k_{B}T}{m}[/tex]
[tex]v = \sqrt{\frac{3k_{B}T}{m}}[/tex]
m is mass of one atom or molecule in kg.
An atom of Arsenic sublimes at 614°C. Converting to Kelvin:
T = 614 + 273 = 887K
Molecular mass of As4 is approximately 0.3kg.
[tex]m = \frac{0.3}{6.10^{23}}[/tex]
[tex]m=5.10^{-25}[/tex]kg
Calculating Root mean square speed :
[tex]v = \sqrt{\frac{3*1.4.10^{-23}*887}{5.10^{-25}}[/tex]
[tex]v = \sqrt{745.08.10^{2}[/tex]
v = 273m/s
The root mean square speed of As4 is approximately 273m/s.
A fire engine approaches a wall at 5 m/s while the siren emits a tone of 500 Hz frequency. At the time, the speed of sound in air is 340 m/s. How many beats per second do the people on the fire engine hear
Answer:
The values is [tex]f_b =14.9 \ beats/s[/tex]
Explanation:
From the question we are told that
The speed of the fire engine is [tex]v = 5\ m/s[/tex]
The frequency of the tone is [tex]f = 500 \ Hz[/tex]
The speed of sound in air is [tex]v_s = 340 \ m/s[/tex]
The beat frequency is mathematically represented as
[tex]f_b = f_a - f[/tex]
Where [tex]f_a[/tex] is the frequency of sound heard by the people in the fire engine and is is mathematically evaluated as
[tex]f_a = [\frac{v_s + v }{v_s -v} ]* f[/tex]
substituting values
[tex]f_a = [\frac{340 + 5 }{340 -5} ]* 500[/tex]
[tex]f_a = 514.9 \ Hz[/tex]
Thus
[tex]f_b =514.9 - 500[/tex]
[tex]f_b =14.9 \ beats/s[/tex]
Is the statement "An object always moves in the direction of the net force acting on it" true or false
Answer:
False
Explanation:
This is because according to newtons second law which says the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. So take for example a net a net force in opposite direction will cause an object to slow down.
velocity vector here is not the same as acceleration vector
A sprinter is running a 100m sprint race. The race begins, and stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and passes the 59m [N] mark at 7.98s, What was the sprinter's average velocity between the two time marks? Show all your work.
Answer:
velocity = 7.7558 m/s
Explanation:
s = u × t
s - distance u - velocity t - time
59 - 12 = (7.98 - 1.92) *u
47 = ( 6.06) u
u = 7.7558 m/s
When a sprinter is running a 100m sprint race. The race begins, and the stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and passes the 59m [N] mark at 7.98s, then the sprinter's average velocity between the two-time marks would be
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. The velocity of an object depends on the magnitude as well as the direction of the object.
the mathematical expression for velocity is given by
velocity = total displacement /time
As given in the problem a sprinter is running a 100m sprint race. The race begins, and the stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and passes the 59m [N] mark at 7.98s
The total displacement covered by the sprinter
S = 59-12
S= 47 m
The total time is taken by the sprinter
T= 7.98 -1.92
= 6.06 seconds
Velocity = S/T
=47/6.06
=7.75 m/s
Thus. the average velocity of the sprinter would be 7.75 m/s
Learn more about Velocity from here
brainly.com/question/18084516
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A parallel plate capacitor consists of two square parallel plates separated by a distance d. the potential across the plates, while keeping everything else constant, what happens to the energy stored in the capacitor?
A) There will be % of the energy stored
B) There will be % of the energy stored
C) The energy stored will remain constant
D) The energy stored will double
E) the energy stored will quadruple
Answer:
E) the energy stored will quadrupled
Explanation:
The correct question is
A parallel plate capacitor consists of two square parallel plates separated by a distance d. If i double the potential across the plates, while keeping everything else constant, what happens to the energy stored in the capacitor?
A) There will be 1/4 of the energy stored
B) There will be 1/2 of the energy stored
C) The energy stored will remain constant
D) The energy stored will double
E) the energy stored will quadruple
The initial energy stored in the capacitor [tex]E_{i}[/tex] = [tex]\frac{1}{2}CV^{2}[/tex]
where C is the capacitance
V is the potential difference
If I double this voltage, while holding every other parameters constant, the new energy stored will be
[tex]E_{n}[/tex] = [tex]\frac{1}{2}C(2V)^{2}[/tex] = [tex]\frac{4}{2}CV^{2}[/tex]
[tex]E_{n}[/tex] = [tex]2CV^{2}[/tex]
dividing new energy stored by the initial energy stored, we have
[tex]E_{n}/E_{i}[/tex] = [tex]2CV^{2}[/tex] ÷ [tex]\frac{1}{2}CV^{2}[/tex] = 4
the energy stored will be quadrupled.
A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the smaller piston to obtain a force of 1600 N at the larger piston
Answer:
The force is [tex]F_1 = 400.8 \ N[/tex]
Explanation:
From the question we are told that
The first diameter is [tex]d_1 = 4.0 \ cm = 0.04 \ m[/tex]
The second diameter is [tex]d_2 = 8.0 \ cm = 0.08 \ m[/tex]
Generally the first area is
[tex]A_1 = \pi * \frac{d^2_1 }{4}[/tex]
=> [tex]A_1 = 3.142 * \frac{0.04^2}{4}[/tex]
=> [tex]A_1 = 0.00126 \ m^2[/tex]
The second area is
[tex]A_2 = \pi * \frac{d^2_2 }{4}[/tex]
[tex]A_2 = 3.142 * \frac{0.08^2}{4}[/tex]
[tex]A_2 = 0.00503 \ m^2[/tex]
For a hydraulic press the pressure at both end must be equal .
Generally pressure is mathematically represented as
[tex]P = \frac{F}{A}[/tex]
=>
[tex]\frac{F_1}{A_1 } = \frac{F_2}{A_2 }[/tex]
=> [tex]F_1 = \frac{1600}{0.00503} * 0.00126[/tex]
=> [tex]F_1 = 400.8 \ N[/tex]
Using the Bohr model, find the ionization energy of the ground He ion.
Answer:
54.4ev
Explanation:
Using
E = - me⁴Z²/8Eo²h²n²
Where n= 1
Z= 1
h=6.6E-31Js
me= 9.1x 10-31kg
Eo=8.85E-12m-3 kg-1 s4 A2
So by substituting
E= -(9.1x 10-31kg)⁴x 1²/8(8.85E-12m-3 kg-1 s4 A2)² x (6.6E-31Js)² x 1²
So E= 54.4ev
Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 5.00 m from two double slits 0.500 mm apart illuminated by 500-nm light.
Answer:
1.25cm
Explanation:
Using
Minimum, as dsinစ = (m+1/2) lambda
Third dark fringe m= 2
dsinစ = (2+1/2)lambda
d(y/L)= (5/2) lambda
Y= 5/2* lambda *L/d
So substituting
=[ (500E-9m)(5m)/0.5E-3] 5/2
=0.0125m
= 1.25cm
Explanation:
Fill in the nuclide symbol for the missing particle in the following nuclear equation.
_______ → + 0 e1 + 13C6
Answer:
Nitrogen 13
Explanation:
We need fill in the blank the nuclide symbol for the missing particle in the following nuclear equation.
[tex]X\rightarrow ^0e_1+^{13}C_6[/tex]
In this reaction, the emission of positron takes place. The atomic number in the LHS and the RHS should be same. So,
[tex]^{13}_7 N \rightarrow ^0e_1+^{13}C_6[/tex]
So, the LHS have nitrogen 13, a radioactive isotope.
In a Young experiment two slits are speared by 6(um), the third dark fringe is formed at an angle 5.6°. The distance between slits and viewing screen is 2 (m).
A- What is the frequency of light used for this experiment?
B- What is the distance between second bright fringe and central fringe?
Answer:
A) f = 1.28x10¹⁵ Hz
B) y = 0.20 m
Explanation:
A) The frequency of light can be found as follows:
[tex] f = \frac{c}{\lambda} [/tex]
Where:
c: is the speed of light = 3.0x10⁸ m/s
λ: is the wavelength
The wavelength can be calculated using the following equation:
[tex] sin(\theta) = \frac{\lambda(m - 1/2)}{d} [/tex]
Where:
m = 3, d = 6 μm, θ = 5.6°
[tex] \lambda = \frac{d*sin(\theta)}{m - 1/2} = \frac{6 \cdot 10^{-6} m*sin(5.6)}{3 - 1/2} = 2.34 \cdot 10^{-7} m [/tex]
Now, the frequency is:
[tex] f = \frac{c}{\lambda} = \frac{3.0 \cdot 10^{8} m/s}{2.34 \cdot 10^{-7} m} = 1.28 \cdot 10^{15} Hz [/tex]
Hence, the frequency of light used for this experiment is 1.28x10¹⁵ Hz.
B) The distance between the second bright fringe and central fringe (y) is:
[tex] tan(\theta) = \frac{y}{D} [/tex]
[tex] y = D*tan(\theta) = 2 m*tan(5.6) = 0.20 m [/tex]
Therefore, the distance between the second bright fringe and central fringe is 0.20 m.
I hope it helps you!
Describe two homeostasis mechanisms
Answer:
The interpretation including its given subject is listed in the subsection below on explanatory.
Explanation:
Homeostasis should be any mechanism of self-regulation from which an individual seeks to preserve equilibrium when adapting to requirements better suited towards its existence.
Two mechanisms are given below:
Blood Pressure: Pressure is regulated by either a homeostatic system through which the blood becomes circulated throughout the individual's body.Body Temperature: The body requires multiple forms of heat regulatory oversight such as:Endothermic is something a living creature can hold one's own internal body temperature. Ectothermic is really where the external atmosphere absorbs temperature.What is the mass of erath
Calculate the force that must be applied to an object weighing 25 pounds for it to be in equilibrium, if it is on a plane inclined at 45 °
Answer:
17.7 lb
Explanation:
Assuming the force is parallel to the incline, draw a free body diagram of the object. There are 3 forces:
Normal force N pushing perpendicular to the incline,
Weight force mg pulling down,
and applied force F pushing parallel to the incline.
Sum the forces in the parallel direction:
∑F = ma
F − mg sin 45° = 0
F = mg sin 45°
F = (25 lb) sin 45°
F = 17.7 lb
Convert 37 mg to Dg (Dekagram)
0.37 dg
Explanation:1 dg ....... 100 mg
x dg .........37 mg
x = 1×37/100 = 37/100 = 0.37 dg
A 8700-kgkg boxcar traveling at 14 m/sm/s strikes a second boxcar at rest. The two stick together and move off with a speed of 5.0 m/sm/s.What is the mass of thesecond car?
Answer:
m₂ = 15660 kg
Explanation:
Given that,
Before collision,
Mass of box car 1, m₁ = 8700 kg
Speed of box car 1, u₁ = 14 m/s
Speed of box car 2, u₂ = 0 (at rest)
After collision,
The two stick together and move off with a speed of 5 m/s
Let m₂ is the mass of the second car. As cars stick together, it is a case of inelastic collision. Using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\8700\times 14+0=(8700+m_2)5\\\\121800=43500+5m_2\\\\121800-43500=5m_2\\\\m_2=\dfrac{78300}{5}\\\\m_2=15660\ kg[/tex]
So, the mass of the second car is 15660 kg.
Answer:
15000 kg
Explanation:
m1v1+m2v2=(m1m2)Vf
8700(15)=(8700+m)5.5
130500=8700+m(5.5)
130500/5.5=8700+m
130500-8700=m
m=15027.3 kg
ANSWER ASAP PLEASEEE Kendra wonders what would happen if the oceans did not exist on Earth. Based on her understanding of the oceans and the water cycle she predicts that Earth would become warmer, with less rain, more clouds, and that living things would not survive. Which part of her prediction is incorrect answers; 1.That living things would not survive 2.That Earth would get warmer 3.That there would be less rain 4.That there would be more clouds
Answer:
that there would be more clouds.
Explanation:
one of the cycles is precipitation. without water there would be no rain and clouds are what causes the rain to come down
Answer:
it's D or 4 that there would be more clouds
Explanation:
hoped it helped!!
Two sources emit beams of light of wavelength 550 nm. The light from source A has an intensity of 10 μW/m2, and the light from source B has an intensity of 20 μW/m2. This is all we know about the two beams. Which of the following statements about these beams are correct? A) Beam B carries twice as many photons per second as beam A. B) A photon in beam B has twice the energy of a photon in beam A. C) The frequency of the light in beam B is twice as great as the frequency of the light in beam A. D) A photon in beam B has the same energy as a photon in beam A. E) None of the above statements are true.
Answer:
A) Beam B carries twice as many photons per second as beam A.
Explanation:
If we have two waves with the same wavelength, then their intensity is proportional to their power, or the energy per unit time.
We also know that the amount of photon present in an electromagnetic beam is proportional to the energy of the beam, hence the amount of beam per second is proportional to the power.
With these two facts, we can say that the intensity is a measure of the amount of photon per second in an electromagnetic beam. So we can say that beam B carries twice as more power than beam A, or Beam B carries twice as many photons per second as beam A.
An ion has 109 neutrons, 87 protons and 84 electrons. What is the net charge in Coulombs on the ion?
Answer:
4.8E-19C
Explanation:
Charge is calculated by getting the net sum of protons and electrons
So
87protons - 84electrons= +3
So the charge of the ion is +3
But 1proton= 1.6*10^-19C
So charge will be 3 x 1.6E-19C
4.8x10^-19C
A block that weighs 200 N is initially at rest on a horizontal surface where the coefficient of static friction is 0.500 and the coefficient of kinetic friction is not yet known.
A) What is the gravitational force on this block?
B) What is the mass of the block?
C) What is the normal force on this block?
D) What is the smallest force needed to start moving the block?
E) If you were pushing the block with the force you found in C, what force would the block exert on you?
F) If you were pushing the block with the force you found in C, and the block began to accelerate at 1.96 meters per second squared, what is the net force on the block?
G) If the block accelerates at 1.96 meters per second squared, what is the coefficient of kinetic friction?
Answer:
A) W= 200N, B) m = 20.4 kg , C) N = 200 N , D) F = 100 N ,
E) F_reaction = 200 N , F) F _net = 40 N
Explanation:
This exercise can be solved using Newton's second law
A) The gravitational force is the weight of the block
Fg = W = m g
W = 200 N
B) m = W / g
m = 200 / 9.8
m = 20.4 kg
C) Normal is the reaction of the floor to the weight of the block
N-W = 0
N = W
N = 200 N
D) we write Newton's second law on the x-axis
F - fr = 0
F = fr
the friction force equation is
fr = μ_s N
fr = μ_s W
subtitute
F = 0.50 200
F = 100 N
E) As the forces in the natural are in pairs, by Newton's third law or law of action and reaction, the block responds with a force of equal magnitude, but opposite direction
F_reaction = 200 N
F and G) We write Newton's second law
F - fr = m a
fr = N = μ_k mg
F - μ_k m g = m a
μ_k = (F - ma) / mg
μ_k = (200 - 20.4 1.96) / 200
μ_k = 0.8
In general, the coefficient of kinetic friction is lower than the static one
the net force is
F_net = F -fr = F - μ_k W
F_net = 200 - 0.8 200
F _net = 40 N
Open the Faraday Law simulation and discover what you can about induction. Make a list of ways to cause induction.
Answer:
Current can be induced in a lot of ways. I'll try and list as much of these methods as I can here.
1. Moving a bar magnet relative to a wire coil
2. Moving a wire coil relative to a magnet.
3. Move a wire coil that has electricity flowing though it relative to a wire coil without electricity flowing through it.
4. Moving a current carrying circuit relative to a non-current carrying circuit
5. Rapidly opening and closing the switch of a current carrying circuit beside a non current carrying circuit.
6. Moving a bar magnet through the middle of a wire coil.
7. Moving a wire coil through a magnet
8. Moving a circuit relative to a magnetic field.
The main idea is to cause a change in the magnetic field, or to change the available area of the wire loop, or to change the angle between the field and the loop.
A point charge Q moves on the x-axis in the positive direction with a speed of A point P is on the y-axis at The magnetic field produced at the point P, as the charge moves through the origin, is equal to What is the charge Q?
The question is missing. Here is the complete question.
A point charge Q moves on the x-axis in the positive direction with a speed of 160 m/s. A point P is on the y-axis at y = + 20mm. The magnetic field produced at point P, as the charge moves through the origin is equal to -0.6μT k. What is the charge Q? [tex]\mu_{0}=4.\pi.10^{-7}[/tex]T.m/A)
Answer: Q = [tex]-0.015.10^{-3}[/tex]C
Explanation: Magnetic Field (B) is a vector field, i.e., has magnitude and direction, and describes the distribution of magnetic force in the space around. It happens when an electrical charge is in movement.
Its magnitude is determined by the formula:
[tex]B = \frac{\mu_{0}}{4.\pi}\frac{Qvsin(\theta)}{r^{2}}[/tex]
where
[tex]\mu_{0}[/tex] is the vacuum permeability constant;
r is the distance between charge and a point you want to know the magnetic field;
θ is the angle between velocity and distance r;
For this question, magnetic field has that magnitude when charge is passing through the origin. So, angle between velocity and distance is 90°.
Calculating to determine charge:
[tex]-0.6.10^{-6} = \frac{4.\pi.10^{-7}}{4.\pi}\frac{Q.160.sin(90)}{(2.10^{-2})^{2}}[/tex]
[tex]-0.6.10^{-6} = 10^{-7}\frac{Q.160.1}{(2.10^{-2})^{2}}[/tex]
[tex]-2.4.10^{-10} = Q.160.10^{-7}[/tex]
[tex]Q = \frac{-2.4.10^{-10}}{160.10^{-7}}[/tex]
[tex]Q = -0.015.10^{-3}[/tex]
Charge Q is [tex]-0.015.10^{-3}[/tex]C or [tex]-0.015[/tex]mC.
Q10. Explain the Retention time, in GC. What will happen to retention time, if you increase the detector temperature
Answer:
The detector temperature doesn't affect retention time
Explanation:
Retention time is one of the chromatographic parameters. Is defined as the time of a compound spends from injection to detection.
A solute in GC is added to the injector where is volatilized. When volatilized, it pass through a column until the detector.
The detector temperature doesn't affect retention time. To change retention time you must change injector temperature or column temperature. An increase in column or injector temperature results in a decrease in retention time.
a vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .
what is will its velocity be
a: 1s later
b: 5s later.
Answer:
1 second later the vehicle's velocity will be:
[tex]v(1)= 6\,\,\frac{m}{s} \\[/tex]
5 seconds later the vehicle's velocity will be:
[tex]v(5)=14\,\,\frac{m}{s}[/tex]
Explanation:
Recall the formula for the velocity of an object under constant accelerated motion (with acceleration "[tex]a[/tex]"):
[tex]v(t)=v_0+a\,t[/tex]
Therefore, in this case [tex]v_0=4\,\,\frac{m}{s}[/tex] and [tex]a=2\,\,\frac{m}{s^2}[/tex]
so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:
[tex]v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}[/tex]
During which stage of the water cycle does water from the ocean form clouds?
Answer:
Condensation
Explanation:
Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.
Answer:
This process is called condensation.
Explanation:
When a cloud becomes full of liquid water, it falls from the sky as rain or snow.
Show that (a)KE=1/2mv2
Answer:
[tex] \boxed{ \bold{ \huge{ \boxed{ \sf{see \: below}}}}}[/tex]
Explanation:
[tex] \underline{ \bold{ \sf{To \: prove \: that \: kinetic \: energy = \frac{1}{2} m {v}^{2} }}}[/tex]
Let us consider, a body of mass ' m ' is lying at rest ( initial velocity = 0 ) on a smooth surface. Let a constant force F displaces this body in its own direction by a displacement ' d '. Let 'v' be it's final velocity. The work done ' W ' by the force is given by :
[tex] \sf{W = FD}[/tex]
⇒[tex] \sf{W = m \: \times a \: \times s} \: \: \: \: \: \: \: \: \: ( \: ∴ \: f \: = \: ma \: ; \: s \: = d)[/tex]
⇒[tex] \sf{W = m \: \times \frac{v - u}{t} \times \frac{u + v}{2} \times t \: \: \: \: \: \: \: \: \: (∴ \: a = \frac{v - u}{t} and \: s = \frac{u + v}{2} \times t}[/tex]
⇒[tex] \sf{W = m \times \frac{ {v}^{2} - {u}^{2} }{2} }[/tex]
⇒[tex] \sf{W = \frac{1}{2} m {v}^{2} \: \: \: \: \: \: \: \: \: \: \: \: (since, \: initial \: velocity(u) = 0)}[/tex]
The work done becomes the kinetic energy of the body. Thus, the kinetic energy of a body of mass ' m : moving with the velocity equal to 'v ' is 1 / 2 mv²
∴ [tex] \sf{KE= \frac{1}{2} m {v}^{2} }[/tex]
[tex] \sf{ \underline{ \bold{ {proved}}}}[/tex]
Hope I helped!
Best regards!!
Calculate (in MeV) the total binding energy for 40Ar. Express your answer in mega-electron volts to four significant figures.
Answer:
299.4 MeV
Explanation:
For 40 Ar
Number of neutrons = 22
Number of protons= 18
Mass of each proton = 1.007277 amu
Mass of each neutron= 1.008665 amu
Total mass of protons= 18 × 1.007277 amu = 18.130986 amu
Total mass of neutrons = 22 × 1.008665 amu = 22.19063 amu
Total mass of protons and neutrons= 18.130986 + 22.19063 = 40.321616 amu
Mass defect = 40.321616 amu - 40 amu
Mass defect = 0.321616 amu
Since 931 is the conversion factor from amu to MeV
Binding energy = 0.321616 amu × 931 = 299.4 MeV
Binding energy = 299.4 MeV
2.18) The calorie is a unit of energy defined as the amount of energy needed to raise 1 g of water by 1oC. a) How many calories are required to bring a pot of water at 1oC to a boil
Answer:
Q = 80000 cal
Explanation:
La expresin para el calor es
Q = m [tex]c_{e}[/tex]ce ΔT
el calor especifico del agua es ce= 1 cal/gr ºC
la temperatura de ebullición del agua es Ef = 100ºC y partimos de la temperatura ambiente To= 20 OC
Q = m 1 ( 100 – 20)
q= 80 m
en el ejercicio no se da la masa de agua, pro podemos suponer, pote lleva 1 litros = 100 G
Q = 80 1000
Q = 80000 cal
A teapot with a surface area of 700 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic cell containing silver nitrate (Ag+ NO3-). If the cell is powered by a 12.0-V battery and has a resistance of 1.30 , how long does it take for a 0.133-mm layer of silver to build up on the teapot? (The density of silver is 10.5 multiplied by 103 kg/m3.)
The piston of a hydraulic elevator used for lifting trucks has a 0.3m radius. What pressure is required to lift a truck of 2500 kg mass? What force was applied to the small piston if it has a radius of 3cm?
Answer:
1. 88370.45 N/[tex]m^{2}[/tex]
2. 2500 N.
Explanation:
Pressure, P, is define as the force, F, per unit area, A, applied on/ by an object.
i.e P = [tex]\frac{F}{A}[/tex]
1. The area, A, of the piston of the hydraulic elevator can be determined by;
A = [tex]\pi[/tex][tex]r^{2}[/tex]
where r is the radius of the piston.
A = [tex]\frac{22}{7}[/tex] × [tex](0.3)^{2}[/tex]
= [tex]\frac{22}{7}[/tex] × 0.09
= 0.2829 [tex]m^{2}[/tex]
Pressure required to lift a truck of 2500 kg mass can be determined as;
P = [tex]\frac{F}{A}[/tex]
F = W = mg
= 2500 × 10
= 25 000 N
So that,
P = [tex]\frac{25000}{0.2829}[/tex]
= 88370.45 N/[tex]m^{2}[/tex]
The pressure required is 88370.45 N/[tex]m^{2}[/tex].
2. [tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]
Area of small piston = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] × [tex](0.03)^{2}[/tex]
= 0.02829 [tex]m^{2}[/tex]
So that,
[tex]\frac{25000}{0.2829}[/tex] = [tex]\frac{F_{2} }{0.02829}[/tex]
[tex]F_{2}[/tex] = 2500 N
The force applied to the small piston is 2500 N.
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 38.0 mph and half the distance at 62.0 mph. On her return trip, she drives half the time at 38.0 mph and half the time at 62.0 mph.A) What is Julie's average speed on the way to grandmother's house?B) What is her average speed in the return trip?
Answer:
A.46.7mph
B.50mph
Explanation:
We know that from the data given
Julie drove 50 miles at a speed of 38 mph, and another 50 miles for 62mph. Thus , for the first 50 miles, she drove for the following time
T1= distance/speed
====> 50/38= 1.3hrs
the next 50 miles, she drove for
T2= 50/62
= 0.8hrs
So average speed is
Totaldistance/ total time
100/2.1
= 46.7mph
B. Because the time she used driving at 35 mph is the same amount of time she used driving at 65 mph, the average speed is just the average of the two speeds given which is
(50+50)/2 = 50mph
Explanation:
Answer:
46.95 mph
50 mph
Explanation:
first, we start by finding the average speed. The average speed if given by the relation
Average Speed = Total distance / Total Time
from the question, we know that
total distance is 100 mi
total time is time of the first distance + time of the second distance
t = (50mi / 38) + (50mi / 62)
t = 1.32 + 0.81
t = 2.13 hrs
Then, the average speed going to grandmothers house will be
Speed = 100mi / 2.13 hrs
Speed = 46.95 mph
on her return trip
speed on the first distance * half the time + speed on the second distance * half the time
x = V1 * t/2 + V2 * t/2
100 = 38 * t/2 + 62 * t/2
100 = 100 * t/2
t/2 = 100/100
t/2 = 1
t = 2 hrs
The average speed then will be, = 100 mi / 2 hrs = 50mph