Answer:
No - It is connected in parallel instead of series
Explanation:
The answer is No, the reason behind the answer that the ammeter is connected in parallel which is wrong it must be connected in series.
What is Ammeter?An ammeter is a device for detecting electric current in amperes, either directly (DC) or alternating (AC). Due to the fact that a shunt running parallel to the meter carries the majority of the electricity at high current values, the digital multimeter can measure a wide range of current values. A circle with the capital A in it serves as the icon for an ammeter for circuit diagrams.
The moving coil of the electrodynamics ammeter rotates in the field created by the fixed coil. It measures alternating and direct current with 0.1 to 0.25 percentage points (by converting Ac power To DC power using a rectifier).
The ammeter is connected in series and, on the other hand, the voltmeter is connected in parallel.
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Three point charges are placed on the y-axis: a chargeqaty=a,a charge-2qat the origin, and a chargeqaty= -a.Such an arrangement is called an electricquadrupole.(a) Find the magnitude and direction of the electric field at points on thepositive x-axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for x >> a.Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole.
A tow truck pulls a car 5.00 km along a horizontal roadway using a cable having a tension of 850 N. (a.) How much work does the cable do on the car if it pulls horizontally? If it pulls at 35.0° above the horizontal? (b.) How much work does the cable do on the tow truck in both cases of part (a)? (c.) How much work does gravity do on the car in part (a)?
I think 1980is the answer because you add???
Identify:
In each case the forces are constant and the displacement is along a straight line, so
[tex]$$W=F s \cos \phi \text {. }$$[/tex]
Set-Up:
In part (a), when the cable pulls horizontally [tex]$\phi=0^{\circ}$[/tex] and when it pulls [tex]$35.0^{\circ}$[/tex] above the horizontal [tex]$\phi=35.0^{\circ}$[/tex]. In part (b), if the cable pulls horizontally[tex]$\phi=180^{\circ}$[/tex]. If the cable pulls on the car [tex]$35.0^{\circ}$[/tex] above the horizontal it pulls on the truck at below the horizontal and [tex]$\phi=145.0^{\circ}$[/tex]. For the gravity force [tex]$\phi=90^{\circ}$[/tex], since the force is vertical and the displacement is horizontal.
Execute:
(a) When the cable is horizontal, [tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 0^{\circ}=4.25 \times 10^{6} \mathrm{~J}$[/tex].
When the cable is[tex]$35.0^{\circ}$[/tex] above the horizontal,[tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 35.0^{\circ}=3.48 \times 10^{6} \mathrm{~J}$[/tex].
(b)[tex]$\cos 180^{\circ}=-\cos 0^{\circ}$[/tex] and [tex]$\cos 145.0^{\circ}=-\cos 35.0^{\circ}$[/tex],
So the answers are [tex]$-4.26 \times 10^{6} \mathrm{~J}$[/tex] and [tex]$-3.48 \times 10^{6} \mathrm{~J}$[/tex].
(c) Since [tex]$\cos \phi=\cos 90^{\circ}=0, W=0$[/tex] in both cases.
Evaluate: If the car and truck are taken together as the system, the tension in the cable does no net wnetwork
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A car of mass 200 kg, moving with a forward acceleration of 3 m/s-ıs acted upon by
constant resistive force of 500 N. Calculate the force exerted from the engine to
maintain this forward acceleration.
Answer:
600 N
Explanation:
The force exerted from the engine to maintain this forward acceleration is 600 N.
What is force?Force is defined in physics as: the push or pull on an object with mass that causes it to change velocity. Force is an external agent that can change the state of rest or motion of a body.
The term "force" has a specific meaning in science. At this level, it is perfectly acceptable to refer to a force as a push or a pull.
A force is not something that an object possesses or possesses. Another object applies a force to another.
We know that,
F = ma
Here, it is given that
Acceleration = 3 m/s
Mass = 200Kg.
So, by keeping the value
F = 200 x 3
F = 600 N
Thus, the force is 600 N.
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Una turbina de vapor
recibe vapor con un flujo másico de 30 kg/s a 6205 kPa, 811 K, con una velocidad a la
entrara de 10 m/s. El vapor a la entrada tiene una energía interna específica de 3150.3
kJ/kg y un volumen específico de 0.05789 m3
/kg. El vapor sale de la turbina a 9.859 kPa,
318.8 K. El vapor sale a 200 m/s con una energía interna específica de 2211.8 kJ/kg y
un volumen específico de 13.36 m3
/kg. Encuentre la potencia producida por la turbina
si ésta pierde calor a una tasa de 30 kW.
Este problema está describiendo una turbina de vapor a la que entra vapor a 30 kg/s, 6.205 kPa y 811 K con una velocidad de 10 m/s y sale a 9.859 kPa, 318.8 K y con una velocidad de 200 m/s. Adicionalmente, tanto el volumen específico como la energía interna son dados para ambas corrientes.
Con lo anterior, resulta posible escribir un balance de energía para esta turbina, despreciando todo efecto por energía potencial ya que no hay diferencia significativa entre la altura de la entrada (1) y la salida (2), pues están practicamente al mismo nivel:
[tex]mh_1+\frac{1}{2} mv^2_1=mh_2+\frac{1}{2} mv^2_2+Q_2+W_2[/tex]
Aquí vemos que la incógnita es [tex]W_2[/tex] como la potencia que produce la turbina. Ahora, el primer cáculo a realizar es el de las entalpías de las corrientes de entrada y salida, dada la energía interna, presión y volumen específico:
[tex]h_1=3150.3\frac{kJ}{kg}+6205kPa*0.05789\frac{m^3}{kg} =3509.51\frac{kJ}{kg}\\\\h_2=2211.8\frac{kJ}{kg}+9.859kPa*13.36\frac{m^3}{kg} =2342.72\frac{kJ}{kg}[/tex]
Ahora, podemos reacomodar el balance de energía con el fin de resolver [tex]W_2[/tex]:
[tex]W_2=m(h_1-h_2)+\frac{1}{2} m(v^2_1-v^2_2)-Q_2[/tex]
Finalmente, reemplazamos los valores para obtener:
[tex]W_2=10\frac{kg}{s} (3509.51-2342.72)\frac{kJ}{kg} +\frac{1}{2} *10\frac{kg}{s} [(10\frac{m}{s}) ^2-(200\frac{m}{s} )^2]*\frac{1kJ}{1000J} -30\frac{kJ}{s}\\\\W_2=11438.4 kJ/s=11438.4kW[/tex]
Es de precisar que la energía cinética como 1/2 m*v² resulta en Joules, por lo que hay que convertir a kilojoules para tener unidades consistentes de kilowatts al final.
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Which of the following statements are true?
(a) An object can move even when no force acts on it.
(b) If an object isn't moving, no external forces act on it.
(c) If a single force acts on an object, the object accelerates.
(d) If an object accelerates, a force is acting on it.
(e) If an object isn't accelerating, no external force is acting on it.
(f) If the net force acting on an object is in the positive x-direction, the object moves only in the positive x-direction.
I can't understand how A is true.
Answer: a, c, d
Explanation: a is true because the object will continue to move even without any force because of inertia (so yh thats why a is true). c is true because an object can accelerate if a single force acts on it. to accelerate (not move), it needs a force to act on it
Would appreciate brainly <3
Explanation:
inertia ...
you push or throw something, and you apply some force to it at that moment, but then it moves and keeps moving even long after you have no more connection to it, and no more force is applied to it.
please consider : we are only talking about moving. not about acceleration.
so, yes, (a) is true.
(c) is true.
(d) is true.
2.(01.01 LC)
Which of the following is true for gravitational force? (3 points)
Decreases with increase in mass
Increases with increase in mass
Increases with increase in distance
Decreases with decrease in distance
Answer:
Increases with increase in mass
Explanation:
gravity is proportional to mass and inversely proportional to the square of the distance between them
F = GMm/d²
A 5 kg bowling ball travelling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1 m/s, what will be the speed of larger ball after the collision? Hint: Use the conservation of momentum equation to solve this problem.
Answer:
1.5 m/s
Explanation:
Conservation of momentum means the momentum of the system before the collision is the same as after.
The before, after momentum of each ball is ...
5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)
10 kg ball: (10 kg)(0 m/s), (10 kg)(v)
The sum of the "before" products is the same as the sum of the "after" products:
(5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v
(10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides
v = (15 kg·m/s)/(10 kg) = 1.5 m/s
The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.
PLEASE HELP ME GET THIS RIGHT
Explanation:
I'm not sure, but I would go for the more than A since its orbital speed is at its fastest and the sweep occurs in about the same period of days.
Which strategy should you use if your research question is too broad for the scope of your project? (1 O narrow the focus of research question o add another research question o use the very first source you find for your project O change the scope of your project
Answer:
"Narrow the focus of research question"
Explanation:
O Narrow the focus of research question
This is good! You can still use your question, but focus in on something so you have a proper research project.
O Add another research question
Would adding another question to an already broad question help? No.
O Use the very first source you find for your project
If your question is too broad, you should not use whatever you see first as it may be incorrect or does not answer the question
O Change the scope of your project
You could, but if you have a set scope for your project (a) you might not be able to change it (b) you don't need to restart
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
Another box of samples is hoisted up by the same rope. If the rope is shaken with the same frequency as before, and the wavelength is found to be 7.9 m , what is the mass of this box of samples
The motion of the rope which is perpendicular to the direction of the
propagation of the wave is a transverse wave motion.
The mass of the box is approximately 9.93 kgReasons:
The given function for the wave speed is presented as follows;
[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]
Where;
[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]
Taking the mass of the rope as, m = 2.00 kg
The length of the rope, L = 80.0 m
The mass hanging on the rope, M = 20.0 kg
We have;
T = 20.0 kg × 9.81 m/s² = 196.2 N
[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]
Therefore;
Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;
v = f × λ
Therefore;
v = 7.9 Hz × 7.9 m = 62.41 m/s
Which gives;
[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]
T = 62.41² × 0.025 = 97.3752025
[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]
Where;
g = The acceleration due to gravity which is approximately 9.81 m/s²
[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]
Therefore;
The mass of the box, m ≈ 9.93 kg
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The parameters obtained from a similar question online are;
[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]
Length of the rope, L = 80.0 m
Mass of the rope, m = 2.0 kg
Frequency of a point on the rope, f = 20 Hz
What is the momentum of a 3 kg bowling ball moving at 3 m/s?
.
O 1 kg. m/s
O 3 kg. m/s
O 6 kg. m/s
O 9 kg • m/s
Explanation:
p = mvp denotes momentumm denotes massv denotes velocity→ p = 3 kg × 3 m/s
→ p = 9 kg.m/s
Option D is correct.
A very bouncy mushroom can be modeled as mass of 30g(the mushroom cap) on top of a spring(the mushroom stalk) when spring constant 20 N/m. A bird of mass 50g lands on the mushroom gently so that its velocity is zero when it lands. Questions are in the image.
The motion of the bouncy mushroom can be described as a simple
harmonic motion, SHM.
a) The equilibrium height of the mushroom is 0.024525 m below its initial heightb) The frequency of resulting oscillation is 0.5 Hzc) The maximum compression of the mushroom 0.03924 md) The equation that describes the oscillation of the mushroom as a function of time is; [tex]\underline{x(t) = 0.014715 \cdot cos(0.5\cdot t)}[/tex]Reasons:
The given parameters are;
Mass of the mushroom cap, m = 30 g = 0.03 kg
Mass of the bird = 50 g = 0.05 kg
The spring constant, K = 20 N/m
a) The equilibrium height of the mass spring system, is given as follows;
F = -K·x
[tex]x = \dfrac{F}{K}[/tex]
The applied force, F = The weight of the bird
∴ F = (0.05 kg) × 9.81 m/s² = 0.4905 N
[tex]x = \dfrac{0.05 \, kg \times 9.81 \ m/s^2}{20 \, N/m} = 0.024525 \, m[/tex]
The equilibrium height of the mushroom is 0.024525 m below its initial height.
b) The frequency of oscillation of a spring, ω, is given as follows;
[tex]\omega = \sqrt{\dfrac{K}{m} }[/tex]
Therefore;
[tex]\omega = \sqrt{\dfrac{20 \, N/m }{80 \, kg} } = \dfrac{1}{2} \, Hz[/tex]
The frequency of resulting oscillation, ω = [tex]\dfrac{1}{2} \, Hz[/tex] = 0.5 Hz
c) The applied force, F = The weight of the bird and the mushroom cap
F = (0.03 kg + 0.05 kg) × 9.81 m/s² = 0.7848 N
[tex]x = \dfrac{0.7848 \, N}{20 \, N/m } = 0.03924 \, m[/tex]
The maximum compression of the mushroom = 0.03924 m
d) The motion of the mushroom is a Simple Harmonic Motion, SHM.
The equation of a SHM as a function of time is; x(t) = A·cos(ω·t + Φ)
For the mushroom, we have;
The amplitude, A = 0.03924 m - 0.024525 m = 0.014715 m
Ф = The phase angle
When t = 0, cos(ω × 0 + Φ) = 1
cos(Φ) = 1
Ф = arcos(1) = 0
The equation is therefore;
x(t) = 0.014715·cos(0.5·t)
Equation of the oscillation of the mushroom is; [tex]\underline{x(t) = 0.014715 \cdot cos(0.5\cdot t)}[/tex]
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Which is NOT a function of the
cell wall?
A. Protects cell from bursting
B. Provides support for plant cells
C. Protects cell from harsh internal
environments
D. Absorbs sunlight to give energy to the cell
Answer:
The answer is D
Explanation:
The chloroplast absorbs sunlight for energy not the cell wall
PLEASE HELP BRAINLIEST TO THE RIGHT ANSWER!!!!!
A 7.8 kg object is suspended by a string from the ceiling of an elevator. The acceleration of gravity is 9.8 m/s^2.
A. Determine the tension in the string if it is accelerating upward at a rate of 1.5 m/s^2. Answer in units of N.
B. Determine the tension in the string if it is accelerating downward at a rate of
1.5 m/s^2
Gravitational pull downward on the object: 7.8 x 9.8 = 76.44N
A. 7.8 x 1.5 = 11.7N upward force
Tension = 76.44 + 11.7 = 88.14 N
Answer: 88.14 N
B. 76.44 - 11.7 = 64.74N
Answer: 64.74 N
Answer:
88.14 N
64.74 N
Explanation:
A large water tank is 3.70 m high and filled to the brim, the top of the tank open to the air. A small pipe with a faucet is attached to the side of the tank, 0.580 m above the ground. If the valve is opened, at what speed (in m/s) will water come out of the pipe
h =(3.7 - .58)m = 3.12m
Now put PE into KE and we have to use the formula:
√2gh (g = gravity and h = height) therefor:
√2 x 9.8 x 3.12
= 7.82m/s
I hope this helps!
A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When
sounded together, they produce 4 beats per second. On altering the temperature of the air in the pipes,
it is observed that the number of beats per second first diminishes to zero and then increases again to 4.
By how much has the temperature of the air in the pipe been altered?
The temperature of the air in the open orang pipe has been altered by 18.73° C
The frequency of an open orang pipe is estimated by using the formula:
[tex]\mathbf{f = \dfrac{v}{2L}}[/tex]
Then, the combination of the frequency of the tuning fork and the open orang pipe is:
[tex]\mathbf{254 - \dfrac{v}{2L} }[/tex]
These combinations of frequency produce 4 beats per sound.
i.e.
[tex]\mathbf{254 - \dfrac{v}{2L} =4}[/tex]
[tex]\mathbf{ \dfrac{v}{2L} = 254-4 }[/tex]
[tex]\mathbf{ \dfrac{v}{2L} = 250 ----(1)}[/tex]
When it is altered, the beats first diminish and increase again by 4.
i.e.
[tex]\mathbf{ \dfrac{v'}{2L} = 254+4 }[/tex]
[tex]\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }[/tex]
If we equate both equations (1) and (2) together, we have:
[tex]\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}[/tex]
However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.
Hence;
when the temperature of the pipe = unknown ???the temperature of the open orang pipe = 15∴
[tex]\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}[/tex]
By squaring both sides, we have:
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}[/tex]
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}[/tex]
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}[/tex]
[tex]\implies \mathbf{273 +T =306.726912 }[/tex]
T = 306.726912 - 273
T ≅ 33.73 ° C
∴
The change in temperature ΔT = 33.73° C - 15° C
The change in temperature ΔT = 18.73° C
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ПОМОГИТЕ ПЖЖППЖЖППЖЖПЖЖПЖПЖПЖ ООООООЧЕНЬ СРОЧНО!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! С ДАНО И РЕШЕНИЕМ!!!!!!
4. У сталевій коробці
масою 250 г
розплавляють 100 г
свинцю. Яка кількість
теплоти витратилася
на теплові процеси,
якщо початкова
температура тіл
становила 27 °С?
Answer:
thanks for the points dia
How did the study of the atom contribute to our understanding of the periodic table of the elements? (1 point) Atoms are representative of elements, so scientists scaled up O atomic char- derstand elemental characteristics, allowing sc Highlight e elements in a periodic table. The determination of electron charge led to an understanding of O how atoms interact with one another, which facilitated the organization of the periodic table. Elements are made of atoms, so understanding atoms provided O information about elements, which led to the organization of the periodic table. Experiments that identified characteristics of atoms provided O scientists with atomic weights and atomic numbers, which were used to organize the periodic table.
The study of the atom contribute d to our understanding of the periodic table of the elements by virtue of the following;
Elements are made of atoms, so understanding atoms provided information about elements, which led to the organization of the periodic tableExperiments that identified characteristics of atoms provided scientists with atomic weights and atomic numbers, which were used to organize the periodic table.As we know; the periodic table is an array of elements in order of their atomic number.
In which case, the periodic table is made up of 7 rows otherwise called Periods and 8 columns otherwise called Groups.
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A double-pane glass window is 60.0 cm x 90.0 cm and has 3.00-mm window panes. If the temperature difference between inside and outside is 24.0 K, how far apart should the panes be to have a heat loss of 4.09 W? Assume there is air in the gap.
The distance between the glass to have the given heat loss is 2.54 m.
The given parameters:
dimension of the window, = 60 cm by 90 cmtemperature, T = 24 Kheat lost, Q = 4.09 Wthermal conductivity of glass, k = 0.8 W/mKThe area of the glass window is calculated as follows;
[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]
The distance between the glass is calculated as follows;
[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]
Thus, the distance between the glass to have the given heat loss is 2.54 m.
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1. It’s fall and time for the corn maze and bonfire and you just can’t wait. On your way to the farm though a turkey flies out in front of you, so you slam on the brakes and go from from 30.0 m/s to 18.0 m/s. Luckily your date brought a stop watch and told you the whole thing took place in 10.5s. What is your acceleration and how far did you go?
Acceleration = (change in velocity ( final speed - starting speed))/ (time)
Acceleration = (18-30)/10.5
Acceleration = -12/10.5
Acceleration = -1.14 m/s^2
Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2
Distance = 252.2 meters
I tossed a ball straight up into the air and timed how long it took to return to the height I tossed it from. It took 4.2s. How fast did I throw the ball into the air?
Explanation:
[tex]v = u + at \\ 0 = u + ( - 10 { ms}^{ - 2}) \times 4.2s \\ u = 42 {ms}^{ - 1} [/tex]
A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Nm-1
. The mass is displaced 5.2 cm from its equilibrium position and then released to undergo simple
harmonic motion.
At what displacement from the equilibrium position is the potential energy equal to the kinetic energy.
The conservation of energy allows to find the result for the point where the kinetic and potential energy are equal is;
x = 0.026 m
Given parameters
The mass of the body m = 220 g = 0.220 kg The force constant k = 7.0 N / m The initial displacement or amplitude xo = 5.2 cm = 0.052 mTo find
The point where scientific and potential energy are equal.
The law of the conservation of mechanical energy is one of the most important in physics, stable that if there is no friction, the mechanical energy of the system is conserved. The mechanical energy is formed by the sum of the kinetic energy and the potential energies.
Em = K + U
Let's write the energy in two points.
Starting point. With maximum compression.
Em₀ = U = ½ k x²
Final point. Where the kinetic and potential energy are equal.
[tex]Em_f = K +U[/tex]
Since the mechanical energy is constant at this point K = U, therefore we can write the energy.
[tex]Em_f = 2U = 2 ( \frac{1}{2} \ k \ x_f^2 )[/tex]
Energy is conserved.
[tex]Em_o = Em_f \\\frac{1}{2} \ k x_o^2 = 2 ( \frac{1}{2} \ k x_f^2)[/tex]Emo = Emf
½ k x² = 2 (½ k xf²)
[tex]x_f = \frac{x_o}{2}[/tex]
let's calculate.
[tex]x_f = \frac{0.052}{2} \\x_f = 0.026 m[/tex]
In conclusion using the conservation of energy we can find the point where the kinetic and potential energy are equal is;
x = 0.026 m
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An electromagnet does not attract a piece of iron.Is it true ? Give reason
Answer:
False..
Explanation:
An electoMagnets attract iron due to the influence of their magnetic field upon the iron. ...
Light from the Sun takes about 8.0 min to reach Earth. How far away is the Sun? Answer in scientific notation
A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.
The magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].
The given parameters;
length of the solenoid, L = 91 cm = 0.91 mradius of the solenoid, r = 1.5 cm = 0.015 mnumber of turns of the solenoid, N = 1300 current in the solenoid, I = 3.6 AThe magnitude of the magnetic field inside the solenoid is calculated as;
[tex]B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\[/tex]
where;
[tex]\mu_o[/tex] is the permeability of frees space = 4π x 10⁻⁷ T.m/A
[tex]B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T[/tex]
Thus, the magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].
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An object weighs 573.0 N on planet Xyleneer. If the object's mass is 92.1 kg, what is the acceleration due to gravity on planet Xyleneer?
Answer:
a = 6.22 m//s²
Explanation:
F = ma
a = F/m
a = 573.0 / 92.1
a = 6.221498...
Answer:
[tex]\boxed {\boxed {\sf 6.22 \ m/s^2}}[/tex]
Explanation:
We are asked to find the acceleration due to gravity on another planet.
Weight is the measure of the force of gravity. Therefore, we can use the following version of the force formula:
[tex]F_g=mg[/tex]
In this formula, [tex]F_g[/tex] is the weight, m is the mass, and g is the acceleration due to gravity.
The object weights 573.0 Newtons (or 573.0 kg*m/s²) on the planet. The object has a mass of 92.1 kilograms.
[tex]F_g[/tex]= 573.0 kg* m/s²m= 92.1 kgSubstitute these values into the formula.
[tex]573.0 \ kg*m/s^2 = 92.1 \ kg * g[/tex]
We are solving for g, so we must isolate the variable. It is being multiplied by 92.1 kilograms. The inverse of multiplication is division, so divide both sides of the equation by 92.1 kg.
[tex]\frac {573.0 \ kg*m/s^2}{92.1 \ kg}= \frac{92.1 \ kg*a}{92.1 \ kg}[/tex]
[tex]\frac {573.0 \ kg*m/s^2}{92.1 \ kg}=a[/tex]
The units of kilograms cancel.
[tex]6.22149837 \ m/s^2=a[/tex]
The original measurements of weight and mass have 4 and 3 significant figures. Our answer must have the least number of sig figs, or 3. For the number we found, that is the hundredth place. The 1 in the thousandths place tells us to leave the 2 in the hundredth place.
[tex]6.22 \ m/s^2=a[/tex]
The acceleration due ot gravity on planet Xyleneer is approximately 6.22 meters per second squared.
70 POINT!!!
IF YOU DON'T KNOW THE ANSWER DO NOT PUT A COMMENT BELOW
THERE ARE TWO QUESTIONS YOU HAVE TO ANSWER
QUESTION 1: Identify any variables that are present as dependent variables, independent variables, and constants in your experimental group and your control group.
QUESTION 2: How does knowing the properties of matter help you separate the substances in mixtures?
Answer:
1) Dependent variable: Type of separation method used
Independent variables: Substance that separated out
Constants: Sand, pepper, and salt
2) you can use the properties of matter to decide on a method to separate out a particular substance based on its unique properties. For example, knowing pepper has a very small mass allows you to use the charge of the static electricity to separate out those particles. Additionally, knowing the solubility of salt allows you to add water to the mixture to be able to remove it from the sand, since the salt dissolves in the water which is poured away.
a wave travels at a constant speed. how does the wavelength change if the frequency is reduced by a factor of 3? assume the speed of the wave remains unchanged.
A. the wavelength decreases by a factor of 3
B. the wavelength does not change
C. the wavelength increases by a factor of 3
D. the wavelength increases by a factor of 9
Answer:
b
Explanation:
when the wavelength increase it doesnt affect the frequency of a wave.
Answer: The wavelength increases by a factor of 3
Explanation:
I will mark brainlist
A wave is disturbance that transfers energy and matter.
true
false
Answer:
False
Explanation:
A wave is a disturbance that transfers energy from one place to another without transferring matter.
Answer:
I'm pretty sure it true sorry if I'm wrong