In an electromagnet, what produces the magnetic field?
A. Electric charges moving through a conducting wire.
B. Electric charges moving through the metal core.
C. The metal core within a coil of wire.
D. The magnetic field of a permanent magnet.
Un proton penetra perpendiculares en un campo magnetico de 5 teslas con una velocidad de 2.10 m/s calcula
Answer:
The magnetic force acting on the proton is 1.68 x 10^-18 N.
Explanation:
magnetic field, B = 5 T
speed , v = 2.1 m/s
charge q = 1.6 x 10^-19 C
Angle, A = 90 degree
The magnetic force on the charge particle is given by
[tex]F = q v B sin A\\\\F = 1.6\times 10^{-19}\times 2.1\times 5\times sin 90\\\\F = 1.68\times 10^{-18} N[/tex]
A piano string of density 0.0050 kg/m is under a tension of 1,350 N. Find the velocity with which a wave travels on the string.
Answer:
519.62 m/s
Explanation:
Applying,
v = √(T/m').............. Equation 1
Where v = velocity of the wave, T = Tension on the string, m' = mass per unit length of the string
From the question,
Given: T = 1350 N, m' = 0.005 kg/m
Substitute these values into equation 1
v = √(1350/0.005)
v = √(270000)
v = 519.62 m/s
Un alambre de plástico, aislante y recto mide 10 cm de longitud y tiene una densidad de carga de +150 nC/m, distribuidos de manera uniforme por toda su longitud. Se encuentra sobre una mesa horizontal. A) Encuentre la magnitud y la dirección del campo eléctrico que produce este alambre en un punto que está 8 cm directamente arriba de su punto medio. B) Si el alambre ahora se dobla para formar un círculo que se coloca aplanado sobre la mesa, calcule la magnitud y la dirección del campo eléctrico que produce en un punto que se encuentra 6 cm directamente arriba de su centro.
Answer:
English only
Explanation:
When solving problems related to Electric Fields, care must be taken about symmetries. In our particular case when we take a look to at the drawings of the attached file, we realize:
1.-By symmetry each dx associated at a, has an opposite dx with point b as reference. The respective dE ( the charge is uniform ) is the same, as the charge of the wire is positive the force and the Field on a test charge (+) located at h will be upward, therefore the components dEx will cancel each other and the Electric Field becomes E = Ey = ∫ 2×dE× cosθ
The solutions:
A) Ey = 4623 N/C
B) Ey = 19.34 N/C
E = Ey = ∫ 2×dE× cosθ
Here cosθ = h/ d ⇒ cosθ = h/√h² + x² dE = K× dQ / d²
d² = h² + x²
k = 8.9 ×10⁹ Nm²C⁻² ; dQ = λ×dx λ = 150×10⁻⁹ C h = 0.08 m
Then by substitution
Ey = 2 ∫[K× λ×dx/ (h² + x²) ] × h / √h² + x²
reordering that equation:
Ey = 2×K×λ×h ∫ dx / [√ ( h² + x² ) ]³ (2)
To solve the integral we make use of a change of variables
x = h × tanα then x² = h² ×tan²α and dx = h× sec²α dα
plugging that values in equation (2)
Ey = 2×K×λ×h ∫ h× sec²α× dα / [√ ( h² + h²tan²α)]³
Ey = 2×K×λ×h² ∫ sec²α× dα / [ h × √ (1 + tan²α)]³ 1 + tan²α = sec²α
Ey = 2×K×λ×h²× ∫ (sec²α / h³× sec³α )×dα
Ey = 2×K×λ/h × ∫ ( 1 / secα dα
Ey = 2×K×λ/h × sinα now we αneed to come back to our original variables:
as x = h × tanα tanα = x/h then x is the opposite leg in a right triangle and h the adjacent one then the hypothenuse is √ (h² + x²) then sin α = x/ √ (h² + x²)
Ey = 2×K×λ/h × x/ √ (h² + x²) |₀⁰°⁰⁵
Ey = 2×8.9×10⁹× 150×10⁻⁹× 5×10⁻²/8× 10⁻²× √ 10⁻² ( 8 + 5 ) N/C
Ey = 4623 N/C
To answer the second question again we will make use of symmetries if you look at drawing ( Figure 2 ) you see that again the components in direction of x-axis cancel each other and the components in y-axis direction will add. Then
Ey = ∫ dE× cosθ
following the same procedure we will find:
Ey = ∫ [K×λ × dl/d²] × h/ d
The importan point here is that the radius of the circle is
2×π×r = 0.01 ( the length of the wire) ⇒ r = 0.16×10⁻² m
And we need to take into account that the integration is over the circle and the length of the circle is 0.01 m or ××2×π×r. All other factors are constant. Then by substitution
Ey = [K×λ ×h× / ( √ r² + h²)³ ] × 10⁻² N/C
Ey = 8.9 × 10⁹ × 150× 10⁻⁹ × 6× 10⁻² × 10⁻² / √ 10⁻² ( 0.16 + 6)
Ey = 0.8 × 10² / 6
Ey = 19.34 N/C
The ability of a metal to react with air and water is a
property.
O chemical, physical, respectively
O physical, chemical, respectively
o physical, physical, respectively
O chemical, chemical, respectively
[tex]\bold{{Answer}}[/tex]
O physical, chemical, respectively
please buddy correct me if im wrong
The ability of a metal to react with air and water is a chemical, chemical, respectively property
What is a chemical change?A chemical change occur when the substance's composition is changed or when bonds are broken and new ones are forms a chemical change occur .
When a metal burn in air or react with air it form metal oxide and when metal react with water it forms oxides or hydroxides and release hydrogen gas . In both the cases change in composition is taking place . hence , bot are chemical reactions
correct answer is O chemical, chemical, respectively
learn more about chemical change
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Which wave interaction results in a change in the direction of the wave as it passes through one medium to another medium?
Answer:
Refraction of waves involves a change in the direction of waves as they pass from one medium to another. Refraction, or the bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves.
In uplifting 750 newton load with the help of 2 meter long lever lever took 250 newton efforts . What will be the efficiency of a machine if the fulcrum is kept 50 cm from the load
Answer:
the efficiency of the machine is 100%
Explanation:
Given;
load, L = 750 N
length of the lever, L = 2 m
effort applied, E = 250 N
Position of the load from the fulcrum, = 50 cm
50cm
0↓--------------------------Δ-------------------------------↓---------200 cm
750 N x cm 250 N
Apply the principle of moment;
750(50) = 250(x)
x = (750 x 50) / (250)
x = 150 cm
the distance of the effort = 150 cm = 1.5 m
the distance of the load = 50 cm = 0.5 m
The velocity ratio of the machine = 1.5/0.5
= 3
The mechanical advantage of the machine is calculated as;
M.A = Load/effort
M.A = 750 / 250
M.A = 3
The efficiency of the machine is calculated as;
E = (M.A / V.R) x 100%
E = (3/3) x 100% = 100%
Therefore, the efficiency of the machine is 100%
Two uses of static electricity and dangers of static electricity
Answer: The uses of static electricity include:
--> photocopying machines
--> precipitators
The dangers of static electricity include:
--> sparks that can lead to explosion
--> Damage to electronic equipment
Explanation:
STATIC electricity is defined as the imbalance that exists between a positive and a negative charge either within or outside an object. This is because all physical objects are made up of atoms which contains protons, electrons and neutrons. The protons are positively charged, the electrons are negatively charged, and the neutrons are neutral. This shows that physical objects are made up of charges.
Opposite charges attract each other (negative to positive). Like charges repel each other (positive to positive or negative to negative). Most of the time positive and negative charges are BALANCED in an object, which makes that object neutral.
Applications( uses) of static electricity include:
--> photocopying machines: Inkjet photocopiers and printers use static electricity to guide a minute jet of ink to the page’s precise position.
--> precipitators: the static electricity is applied in an electrostatic precipitator whereby they remove smoke from waste gases before they pass out of the chimneys in power stations that burn fossil fuels.
The disadvantages (dangers) of static electricity include:
--> Sparks that can lead to explosion: sparks generated from static electricity can cause fires or explosions due to the ignition of flammable or explosive mixtures.
--> Damage to electronic equipment: this is due to components from electrostatic discharge.
Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?
The softball experiences
force as a result of Amy’s throw. As the ball moves, it experiences
from the air it passes through. It also experiences a downward pull because of
.
Answer:
1.the friction of air, gravity2.gravity
Answer:
The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences drag from the air it passes through. It also experiences a downward pull because of gravity.
Explanation:
Plato
A car's bumper is designed to withstand a 5.04-km/h (1.4-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.255 m while bringing a 890-kg car to rest from an initial speed of 1.4 m/s.
Answer:
3420.39 N
Explanation:
Applying,
Fd = 1/2(mv²-mu²)................. Equation 1
Where F = force on the bumber, d = distance, m = mass of the car, v = final velocity, u = initial velocity.
make F the subject of the equation
F = (mv²-mu²)/2d............... Equation 2
From the question,
Given: m = 890 kg, v = 0 m/s (to rest), u = 1.4 m/s, d = 0.255 m
Substitute these values into equation 2
F = [(890×0²)-(890×1.4²)]/(2×0.255)
F = -1744.4/0.51
F = -3420.39 N
The negative sign denotes that the force in opposite direction to the motion of the car.
Which items did Mendeleev write down on note cards while he was observing
different elements?
A. Buoyancy
B. Density
C. Color
D. Atomic mass
Mendeleev wrote down atomic mass
Answer:
A. Buoyancy.
hope it helps
stay safe healthy and happy..physics class 9 chapter 8 please tell please
Answer:
(a) The motion is uniform
(b) 11.11 m/s
Explanation:
(a)
From the table below, the motion of the bus is uniform.
(b)
Speed(s) = Δd/Δt
s = Δd/Δt............. Equation 1
From the table,
Given: Δd = 10 km = 10000 m, Δt = 15 minutes = (15×60) = 900 seconds
Substitute these values into equation 1
s = 10000/900
s = 11.11 m/s
Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10^- 8 N how far apart are centers of mass?
Answer:
Explanation:
The formula for this is
[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.
Filling in:
[tex]7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2}[/tex] and moving things around to solve for r:
[tex]r=\sqrt{\frac{6.67*10^{-11}(90.0)(30.0)}{7.5*10^{-8}} }[/tex] Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m
A sample from of water is heated with 5000 J of energy and its temperature goes up by 6 K. What is the mass of the sample?
Answer:
mass= 0.1993 kg
Explanation:
Using the formula c = Q / (mΔT)
How much heat is required to evaporate 0.15 kg of lead at 1750°C, the boiling point for lead? The heat of vaporization for lead is Lv = 871 × 103 J/kg.
Answer:
Heat required = mass× latent heat Q = 0.15 × 871 ×
The heat required to evaporate 0.15 kg of lead at 1750°C will be 130,650 J.
What is heat?The movement of energy from a hot to a cold item is characterized as heat. Heat energy flows from a hot material to a cold one.
This occurs because faster-vibrating molecules transmit their energy to slower-vibrating ones.
The given data in the problem is;
m is the mass of lead = 0.15 kg
T is the temperature = 1750°C,
The latent heat of vaporization for lead is, [tex]\rm L_V[/tex] = 871 × 10³ J/kg.
The heat is found as;
[tex]\rm Q= m \times L_V \\\\ \rm Q= 0.15 \times 871 \times 10^3 \\\\ Q=130,650 \ J[/tex]
Hence the heat required to evaporate 0.15 kg of lead at 1750°C will be 130,650 J.
To learn more about the heat refer to the link;
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(12 points) Analysis from the point where the block is released to the point where it reaches the maximum height i) Calculate the highest height reached by the block (or the largest distance travelled along the ramp.) ii) Calculate the work done by the gravitational force. iii) Calculate the work done by the normal force. iv) Calculate the work done by the friction force.
Answer:
i) a₁ = -g (sin θ + μ cos θ), x = v₀² / 2a₁
ii) W = mg L sin θ , iii) Wₙ = 0
iv) W = - μ m g L cos θ x
Explanation:
With a drawing this exercise would be clearer, I understand that you have a block on a ramp and it is subjected to some force that makes it rise, for example the tension created by a descending block.
The movement is that when the system is released, the tension forces are greater than the friction and the component of the weight and therefore the block rises up the ramp
At some point the tension must become zero, when the hanging block reaches the ground, as the block has a velocity it rises with a negative acceleration to a point and stops where the friction force and the weight component would be in equilibrium along the way. along the plane
i) Let's use Newton's second law
the reference system is with the x axis parallel to the ramp
Axis y
N - W cos θ = 0
X axis
T - W sin θ - fr = ma
the friction force is
fr = μ N
fr = μ mg cos θ
we substitute
T - m g sin sin θ - μ mg cos θ = m a
a = T / m - g (sin θ + μ cos θ)
With this acceleration we can find the height that the block reaches, this implies that at some point the tension becomes zero, possibly when a hanging block reaches the floor.
T = 0
a₁ = -g (sin θ + μ cos θ)
v² = v₀² - 2a1 x
v = 0 at the highest point
x = v₀² / 2a₁
ii) the work of the gravitational force is
W = F .d
W = mg sin θ L
iii) the work of the normal force
the force has 90º with respect to the displacement so cos 90 = 0
Wₙ = 0
iv) friction force work
friction force always opposes displacement
W = - fr d
W = - μ m g cos θ L
Which has lower ionization energy: an atom with a small radius or an atom with a large
radius? Why? (4 points)
Which of the following is defined as a force that pushes and pulls the current through the circuit? Group of answer choices D) resistance B) electricity A) current C) voltage
Answer:
C voltage
Explanation:
Voltage is the change in electric potential so basically current flows from high potential to low potential due to voltage.
An emergency relief plane is dropping a care package from a plane to a group of medical personnel working for a relief agency in an African village. The package is designed to land in a small lake, inflate an attached raft upon impact, and finally resurface with the raft side down. The plane will be moving horizontally with a ground speed of 59.1 m/s. The package will be dropped a horizontal distance of 521 m from the intended target location. At what altitude above the pond must the plane be flying in order to successfully accomplish this feat?
Answer:
The altitude of the plane is 379.5 m.
Explanation:
Initial horizontal velocity, u = 59.1 m/s
Horizontal distance, d = 521 m
let the time taken by the packet to cover the distance is t.
Horizontal distance = horizontal velocity x time
521 = 59.1 x t
t = 8.8 s
let the vertical height is h .
Use second equation of motion in vertical direction.
[tex]h = u t + 0.5 gt^2\\\\h = 0 + 4.9 \times 8.8\times8.8\\\\h= 379.5 m[/tex]
What is the current flowing through the circuit shown? (V= 110 V, R, = 200, R2 = 300, R3 = 10 0) (Ohm's law: V = IR)
A. 1.8 A
B. 20 A
C. 0.05 A
D. 0.55 A
Answer:
A
Explanation:
The first thing you have to do is go back and list the resistances correctly. R1 = 20 R2 = 30 and R3 = 10.
Leave the units off if you can't make an omega.
The resistance of a series circuit (that's what this is) is r1 + r2 + r3 = 10 + 20 + 30 = 60 ohms
Now use ohms law.
R = 60 ohms
V = 110 volts.
I = V / R
I = 110/60
I = 1.833 to the nearest 1/10 = 1.8
Complete the sentence-
Friction always acts
1 along the direction of the motion.
2 opposite to the motion.
3 both of these.
4 none of these.
Answer:
Friction always acts opposite to the motion.
In an experiment the mass of a calorimeter is 36.35 g . Express in micrometer ,millimetre and kg.
Answer:
1. 36.35 g = 36.35E15 micrometer.
II. 36.35 g = 363.5 millimetre.
III. 36.35 g = 0.03635 kilogram.
Explanation:
Given the following data;
Mass of calorimeter = 36.35 gramsTo convert the mass in grams (g) to;
I. Micrometer
Conversion:
1 g = 1 exp 15 um
36.35 g = X um
Cross-multiplying, we have;
X = 36.35 * 1 exp 15 = 36.35 exp 15 um
36.35 g = 36.35E15 micrometer
II. Millimetre
Conversion:
1 g = 1 milliliter
36.35 g = X milliliter
Cross-multiplying, we have;
X = 36.35 * 1 = 36.35 milliliter
Next, we would convert milliliter to millimetre;
1 milliliter = 10 millimetre
36.35 milliliter = X millimetre
Cross-multiplying, we have;
X = 36.35 * 10 = 363.5 millimetre
36.35 g = 363.5 millimetre
III. Kilogram
Conversion:
1000 grams = 1 kilogram
36.35 g = X kilogram
Cross-multiplying, we have;
X * 1000 = 36.35 * 1
Dividing both sides by 1000, we have;
X = 36.35/1000 = 0.03635 kilogram
36.35 g = 0.03635 kilogram
Note:
g is the symbol for grams.Exp (E) means exponential = 10um is the symbol for micrometer.If 6000 J of heat is added to 200 gm of water at 25° C. What will be its final
temperature?
Answer:
T₂ = 305.17 K
Explanation:
Given that,
Heat, Q = 6000 J
Mass, m = 200 gram
Initial temperature, T₁ = 25° C
We need to find its final temperature. Let it is T₂.
We know that,
[tex]Q=mc\Delta T[/tex]
Where
c is the specific heat of water, c = 4.18 J/g°C
So,
[tex]6000=200\times 4.18\times (T_2-298)\\\\\dfrac{6000}{200\times 4.18}=(T_2-298)\\\\7.17=(T_2-298)\\\\7.17+298=T_2\\\\T_2=305.17\ K[/tex]
So, the final temperature is equal to 305.17 K.
please help me with this question
Answer:
(4) A = 3 A, A₂ = 11 A
(5) 7 A
Explanation:
(4)
From the diagram,
A = 3+6+2
A = 11 A
V = A₂R
A₂ = V/R₂............ Equation 1
Given: V = 12 V, R₂ = 4 Ω
Substitute these values into equation 1
A₂ = 12/4
A₂ = 3 A
(5) Applying,
V = IR'
I = V/R'............ Equation 1
Where V = Voltage, I = cuurent, R' = total resistance.
But,
1/R' = (1/3)+(1/4)
1/R' = (3+4)/12
1/R' = 7/12
R' = 12/7 Ω
Given: V = 12 V
Substitute these values into equation 1
I = 12/(12/7)
I = 7 A
Therefore
A = 7 A
What is potential energy? What are some of its examples.
Answer:
the energy stored in an object because of its specific state or position is called its potential energyexamples:-a compressed springWater that is behind a dam.A car that is parked at the top of a hill. a moving car.Etc....
Explanation:
❣️jess bragoli❣️#keep learning!!
Explanation:
POTENTIAL energy is the energy that is stored in an object due to its position relative to some zero position.
a ball is thrown straight up into the air while the ball is traveleling upwards what are the magnitue and direction
Answer: hi your question is incomplete attached below is the complete question
answer :
magnitude of acceleration : | a | = g = 9.81 m/s^2
direction : a = - g j
Explanation:
Neglecting Air resistance
magnitude of acceleration :
| a | = g = 9.81 m/s^2
Direction of acceleration
a = - g j ( given that the direction of acceleration is against the acceleration due to gravity i.e. in the opposite direction )
A bus starts to move with the speed of 15m/s. Calculate the time taken by the bus to cover distance of 15000m.
Answer:
1000
Explanation:
the formula of time taken= d/s. t=15000÷15 t=1000. I hope it helps
formula of distance is S=Vt
so, 15000=15t
15000 divide by 15=t
1000=t or t=1000
A 25.0kg girl pushes a 50.0kg boy with a force of 100.0N. What is the acceleration of the girl?
Answer:
im pretty sure it should be 50.0
If a wire lies withina magnetic field what must be true for the magnetic field to produce an electric current in the wire
Answer:
The magnetic field through the wire must be changing
Explanation:
According to Faraday's law, the induced emf, ε in a metallic conductor is directly proportional to the rate of change of magnetic flux,Φ through it. This is stated mathematically as ε = dΦ/dt.
Now for the wire, the magnetic flux through it is given by Φ = ABcosθ where A = cross-sectional area of wire, B = magnetic field and θ = angle between A and B.
So, dΦ/dt = dABcosθ/dt
Since A and B are constant,
dΦ/dt = ABdcosθ/dt = -(dθ/dt)ABsinθ
Since dθ/dt implies a change in the angle between A and B, since A is constant, it implies that B must be rotating.
So, for an electric current (or voltage) to be produced in the wire, the magnetic field must be rotating or changing.
The boiling point of a substance is _72 degree Celsius. This temperature will be equivalent to Kelvin scale is-------.
Answer:
345 K
Explanation:
Temperature can be defined as a measure of the degree of coldness or hotness of a physical object.
Generally, it is measured with a thermometer and its units are Celsius (°C), Kelvin (K) and Fahrenheit (°F).
Given the following data;
Boiling point = 72°CTo convert the temperature in degree Celsius to Kelvin, we would use the following mathematical expression;
Kelvin = 273 + °C
Substituting into the formula, we have;
Kelvin = 273 + 72
Kelvin = 345 K
Therefore, the temperature of 72°C will be equivalent to 345 K on the Kelvin scale.