Answer:
The required diagram is shown in the figure. When an object is placed in front of the convex lens, i.e., between 2F
1
and F
1
, its image is formed beyond 2F
2
on the other side of the lens. The image is real, inverted and enlarged.
solution
answer this plzzzzzzzzzzzzz
Jerry runs 60 meters east and then 20 meters west in 10
seconds. His average velocity is
m/s.
Answer: 8 meters per second
Explanation: If you add 60 to 20 you get 80 meters and since he ran those 80 meters in 10 seconds you divide 80 by ten and get 8 and then you get 8m/s
Two tectonic plates moving toward one another are at a
ANSWER CHOICES
convergent boundary.
divergent boundary.
subduction boundary.
transform boundary.
Answer:
A. cause i just took the test
Explanation:
Answer:
its A
Explanation:
no explanation is needed, just trust me.
please help i’ll mark u branliest
Answer:
62
Explanation:
it doesn't need explanation
Using a launch speed of 40.0 m/s and any angle between 0 and 90 degrees, what would be the largest possible range for a projectile?
its 45
Answer:
The largest possible range of the projectile is 163.27 m.
Explanation:
Given;
launch speed, u = 40 m/s
angle of projection, θ; between 0⁰ and 90⁰
The range of a projection is given as;
[tex]R = \frac{u^2sin(2\theta )}{g}[/tex]
The largest possible range will occur at 45 degrees angle of projection;
[tex]R = \frac{u^2sin(2\theta )}{g} \\\\R = \frac{(40)^2sin(2\ \times \ 45^0 )}{9.8}\\\\R = \frac{(40)^2sin( 90^0 )}{9.8}\\\\R = \frac{(40)^2( 1 )}{9.8} \\\\R = 163.27 \ m\\\\[/tex]
Therefore, the largest possible range of the projectile is 163.27 m.
Starting at 1.3 m/s, a runner accelerates at a constant 0.22 m/s2 for 6.0 s. What is the runner’s displacement during this time interval?
Answer:
answer is 11.76 meter
Explanation:
use 2nd equation of motion
S=ut+1/2at^2
A group of students conduct an experiment with a block of wood sliding down an incline. They find that the final energy of the block is less than the initial energy of the block. Which statement best describes this situation?
A. The students made an error in deterring the final energy.
B. The students made an error in calculating the initial energy.
C. Some energy was transformed to other forms.
D. Some energy was destroyed or lost.
What net force is necessary to give a 2 kg mass that is initially at rest an acceleration of 5 m/s2?
Answer:
10 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 2 × 5
We have the final answer as
10 NHope this helps you
Calculate the rotational inertia of a meter stick, with mass 0.499 kg, about an axis perpendicular to the stick and located at the 35.2 cm mark. (Treat the stick as a thin rod.)
Answer:
The rotational inertia of the meter stick is 0.0618 kgm².
Explanation:
Given;
mass of the meter stick, m = 0.499 kg
perpendicular distance to the rotational axis, r = 35.2 cm = 0.352 m
The rotational inertia or moment of inertia for a point mass is given by;
I = mr²
where;
m is the point mass
r is the perpendicular distance
Substitute the givens and solve for moment of inertia.
I = (0.499)(0.352)²
I = 0.0618 kgm²
Therefore, the rotational inertia of the meter stick is 0.0618 kgm².
The asteroid 234 Ida has a mass of about 4 × 1016 kg and an average radius of about 16 km. What is the acceleration due to gravity on 234 Ida? Assume that the asteroid is spherical; use G = 6.67 × 10–11 Nm2/kg2.
A. 1 cm/s2
B. 2 cm/s2
C. 5 cm/s2
D. 6 cm/s2
Answer:
1 cm/s²
Explanation:
I just took the quiz
The asteroid 234 Ida has a mass of about 4×1016 kg and an average radius of about 16 km. The acceleration due to gravity will be 1.04 cm/s². Hence, option A is correct.
What is the acceleration due to gravity?The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s² is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on the earth's surface is 9.8 m/s².
The formula for the acceleration due to gravity is g=GM/r².
According to the question, the given values are :
Mass, M = 4 × 1016 kg or
M = 4 × 10¹⁶.
Radius, r = 16 km or,
r = 16000 meter.
G = 6.67 × 10⁻¹¹ Nm²/kg²
g = (6.67 × 10⁻¹¹ ) (4 × 10¹⁶) / 16000²
g = 0.0104 m/s² or,
g = 1.04 cm/s².
Hence, the acceleration due to gravity will be 1.04 m/s²
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please answer this question
If the hiker starts climbing at an elevation of 350 ft, what will their change in gravitational potential energy be, in joules, once they reach the top
Answer:
352,088.37888Joules
Explanation:
Complete question;
A hiker of mass 53 kg is going to climb a mountain with elevation 2,574 ft.
A) If the hiker starts climbing at an elevation of 350 ft., what will their change in gravitational potential energy be, in joules, once they reach the top? (Assume the zero of gravitational potential is at sea level)
Chane in potential energy is expressed as;
ΔGPH = mgΔH
m is the mass of the hiker
g is the acceleration due to gravity;
ΔH is the change in height
Given
m = 53kg
g = 9.8m/s²
ΔH = 2574-350 = 2224ft
since 1ft = 0.3048m
2224ft = (2224*0.3048)m = 677.8752m
Required
Gravitational potential energy
Substitute the values into the formula;
ΔGPH = mgΔH
ΔGPH = 53(9.8)(677.8752)
ΔGPH = 352,088.37888Joules
Hence the gravitational potential energy is 352,088.37888Joules
The change in gravitational potential energy be, once the hiker reach the top of the mountain is 352088 joules or 352.1 kJ.
What is gravitational potential energy?Gravitational potential energy is the energy which a body posses because of its position.
The gravitational potential energy of a body is given as,
[tex]U=mgh[/tex]
Here, (m) is the mass of the body, (g) is the gravitational force and (h) is the height of the body.
The mass of the hiker is 53 kg and the height of the climb is 2574 ft.
Now, the hiker starts climbing at an elevation of 350 ft. Thus, the net height of the hiker has to climb is,
[tex]h=2574-350\\h=2224\rm\; ft[/tex]
Convert this into the meter by multiplying 03048 as,
[tex]h=2224\times0.3048\\h=677.8752\rm\; m[/tex]
It is known that the value of g is 9.8 m/s². Plug in all the values as,
[tex]U=53\times9.8\times677.8752\\U=352088J\\U=352.1 \;\rm kJ[/tex]
Thus, the change in gravitational potential energy be, once the hiker reach the top of the mountain is 352088 joules or 352.1 kJ.
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A hunter aims directly at a target (on the same level) 120 m away. If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target
Answer:
The distance the bullet will miss the target is 1.13 m.
Explanation:
Given;
Initial velocity of the bullet = 250 m/s
Distance of the target = 120 m
Time of motion;
t = 120 / 250
t = 0.48 s
During this time the bullet is under the gravitational pull and the distance it will miss the target is given by;
Y = V₀y + ¹/₂gt²
where;
V₀y is the initial vertical velocity = 0
Y = 0+ ¹/₂gt²
Y = ¹/₂(9.8)(0.48)²
Y = 1.13 m
Therefore, the distance the bullet will miss the target is 1.13 m.
A jeweler's grinding wheel slows down at a constant rate from 185 rad/s to 105 rad/s while it rotates through 16.0 revolutions. How much time does this take?
Answer:
t = 0.6933 s
Explanation:
This is a rotational kinematics exercise, let's find the angular acceleration of the wheel
w² = w₀² + 2 α θ
α = (w² - w₀²) / 2 θ
Let's reduce the angles to the SI system
θ = 16 rev (2π rad / 1 rev) = 32π rad
let's calculate
α = (105² - 185²) / (2 32π)
α = -115.39 rad / s²
now let's use the relation
w = w₀ + α t
t = (w- w₀) /α
t = (105 - 185) / (- 115.39)
t = 0.6933 s
One object has a temperature twice as large as another object. If the objects have the same surface area, how much more power does the hotter object radiate than the cooler object
Answer:
The hotter object radiate more power than the cooler object 15 times i.e 15σeA[tex]T^{4}[/tex]
Explanation:
From Stefan's law, an object would radiate power with respect to its temperature.
i.e Radiative power, Q = σeA[tex]T^{4}[/tex]
where Q is the radiative power, σ is the constant, e is the emissivity of the object, A is the area of the object and T is the temperature.
Let the temperature of the cooler object be represented by T.
So that its radiative power = σeA[tex]T^{4}[/tex]
Given that the temperature of the hotter object is twice as large as that of the cooler object.
Temperature of hotter object = 2T
So that its radiative power = σeA[tex](2T)^{4}[/tex]
= 16σeA[tex]T^{4}[/tex]
Radiative power difference between the two objects = 16σeA[tex]T^{4}[/tex] - σeA[tex]T^{4}[/tex]
= 15σeA[tex]T^{4}[/tex]
The hotter object radiate more power than the cooler object 15 times.
The hotter object radiates 15 times more power than the power of cooler object.
Absolute Temperature of one object = [tex]T_1[/tex]
Absolute Temperature of second object =[tex]T_2[/tex] = [tex]2T_1[/tex]
The Power emitted by the an object is given by the equation (1)
[tex]P= A\epsilon \sigma\;T^4[/tex].......(1)
Equation (1) is called as Stephan Boltzmann Law
Where
P = Power emitted by the object in Joule
A = Surface area of the object
[tex]\epsilon[/tex] = Emissivity of the object
T = Absolute Temperature
Let us consider emissivities are equal
[tex]So \; P_1/P_2 = T_1^4/T_2^4\\[/tex] ( Areas of both objects are equal)
[tex]P_1/P_2= T_1^4/16T1^4= 1/16[/tex]
[tex]P_2= 16 P_1[/tex]
Difference in Power = [tex]16P_1- P_1[/tex] = [tex]15P_1[/tex]
So the hotter object radiate 15 times more power than the power of cooler object.
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Helicopters rotor blades, could spin at high speed of 510 rpm. Find the angular displacement in radian for 3 hour(s) operation
Answer:
The angular displacement of the blade is 576,871.2 radians
Explanation:
Given;
angular speed of the Helicopters rotor blades, ω = 510 rpm (revolution per minute)
time of motion, t = 3 hours
The angular speed of the Helicopters rotor blades in radian per second is given as;
[tex]\omega = \frac{510 \ rev}{mins} *\frac{2 \pi \ rad}{1 \ rev} *\frac{1 \ min}{60 \ s}\\\\\omega = 53.414 \ rad/s[/tex]
The angular displacement in radian is given as;
θ = ωt
where;
t is time in seconds
θ = (53.414)(3 x 60 x 60)\\
θ = 576,871.2 radians
Therefore, the angular displacement of the blade is 576,871.2 radians
Which vocabulary word would describe both a cell membrane and a screen door?
Answer:
flexible covering
Explanation:
A rolling ball has 8 joules of kinetic energy and is rolling 4m/s. Find it’s mass
Answer:
m = 1
Explanation:
K.E = 8J
v = 4m/s
m = ?
Now,
K.E = 1÷2mv^2
8 = 1÷2 × m × (4)^2
8 = 1÷2 × m × 16
8 = m × 8
m × 8 = 8
m = 8 ÷ 8 = 1
m = 1
VERIFICATION:K.E. = 1÷2mv^2
K.E = 1÷2 × 1 × 4^2
K.E. = 8J
-TheUnknownScientist
I’m testing if this app works.What is gravity?
Answer:
Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another. On Earth, gravity gives weight to physical objects, and the Moon's gravity causes the ocean tides.
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.330 mm wide. The diffraction pattern is observed on a screen 2.55 m away. Define the width of a bright fringe as the distance between the minima on either side.
a. What is the width of the central bright fringe?
b. What is the width of the first bright fringe on either side of the central one?
Answer:
a) y_total = 19.916 10⁻⁵ m , b) Δθ = 1.91 10-⁻³ rad
Explanation
This is a diffraction exercise that is described by the expression
a sin θ = m λ
the first minimum occurs for m = 1
a sin θ = λ
sin θ = λ / a
θ = sin⁻¹ (633 10⁻⁹ / 0.330 10⁻³)
θ = 1.918 10⁻³ rad
let's use trigonometry
tan θ = y / x
y = x tan θ
y = 2.55 tan (3.936 10-3)
y = 5.75 10⁻- m
this value is from the central maximum to one extreme of the value,
y_ total = 2 y
y_total = 2 (5.75e1)
y_total = 19.916 10⁻⁵ m
b) For the second point and constructive inference we have m = 2
sin θl = m λ
θ = sin⁻¹ (lat / a)
θ = sin⁻¹ (2 633 10-9 / 0.33010-3) = son-1 (3.836 10-3)
θ = 3.84 10-3 give
The width of this maximum is
Δθ = 1.3 10-3
Δθ = 3.84 10⁻³- 1.918 10⁻³
Δθ = 1.91 10-⁻³ rad
The legs includes which anatomical features? Select all that apply. CD occipital region patellar region plantar region crural region cranial region lumbar region DONE.f
Answer:
Patellar region
Plantar region
Crural region
Answer:
B, C, D
Patellar, Plantar, Crural regions
Explanation:
Legs are the lower limbs of the body, consisting of all these parts. Just did assignment too.
5)
Using only the information available in the periodic table, consider the elements calcium and chlorine. From their
location on the periodic table, identify the oxidation state and number of valence electrons for calcium and chlorine.
Which statement most accurately describes the compound formed by calcium and chlorine?
C
A)
B)
Calcium, a nonmetal with an oxidation number of +2 will form a covalent
bond with chlorine, a halogen (nonmetal) with an oxidation number of -1
called calcium chloride (CaCl2).
Calcium, an alkaline earth metal with an oxidation number of +2 will form
covalent bond with chlorine, also a metal with an oxidation number of -1
called calcium dichloride (CaCla)
Calcium, an alkaline earth metal with an oxidation number of +2 will form
an ionic bond with chlorine, a halogen in group VILA with an oxidation
number of -1 called calcium chloride (CaCl2).
Calcium, an alkaline earth metal with an oxidation number of 2 will share
electrons to form an lonic bond with chlorine, a nonmetal with an
xidation number of -1 called calcium dichloride (CaCl).
D)
Answer:C,(Calcium,an alkaline earth metal with an oxidation number of +2 will form an ionic bond with chlorine,a halogen in group VllA with an oxidation number of -1 called calcium chloride (CaCl2)
Explanation:
on USAtestprep !!
Calcium, an alkaline earth metal with an oxidation number of +2 will form
an ionic bond with chlorine, a halogen in group VIIA with an oxidation
number of -1 called calcium chloride (CaCl₂). This is correct statement.
What is oxidation number?Simply said, the number assigned to each element in a chemical combination is the definition of an oxidation number. The total number of electrons that an atom in a molecule can share, lose, or gain while forming a chemical bond with an atom of a different element is known as the oxidation number.
Also known as oxidation state, oxidation number is a numerical value. But depending on whether we take into account the atoms' electronegativity or not, these phrases might occasionally have a different meaning. In coordination chemistry, the term "oxidation number" is often used.
According to Periodic table: calcium is a alkaline earth metal with an oxidation number of +2 whereas chlorine is a halogen in group VIIA with an oxidation number of -1. When they reacts chemically, they form an ionic compound named calcium chloride having chemical formula CaCl₂.
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Determine the focal length of a plano-concave lens (refractive index n =1.5) with 24 cm radius of curvature on its curve surface
1)-96 cm
2)-24 cm
3)-48 cm
4)-72 cm
Answer:
Option 3: -48 cm
Explanation:
We are given:
refractive index; n = 1.5
radius of curvature; r2 = 24 cm
Formula for the focal length is given as;
1/f = (n - 1) × [(1/r1) - (1/r2)]
As r1 tends to infinity, 1/r1 = 0
Thus,we now have;
1/f = (n - 1) × (-1/r2)
Plugging in the relevant values;
1/f = (1.5 - 1) × (-1/24)
1/f = -0.02083333333
f = -1/0.02083333333
f = -48 cm
You compress a spring by x, and then release it. Next you compress the spring by 2x. How much more work did you do the second time than the first
Answer:
Work done is [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex].
Explanation:
The work done by the spring is the same as the potential energy stored in the spring.
So that,
work done = potential energy = [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
where k is the spring constant of the material of the spring, and x is the compression.
When the spring is compressed by x;
work done = [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
When the spring is compressed by 2x;
work done = [tex]\frac{1}{2}[/tex] k[tex](2x)^{2}[/tex]
= [tex]\frac{1}{2}[/tex] k(4[tex]x^{2}[/tex])
= 2k[tex]x^{2}[/tex]
Therefore,
The work done the second time more than the first = 2k[tex]x^{2}[/tex] - [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
= [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex]
The work done the second time more than the first is [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex].
Spring work is equivalent to the effort done to extend the spring, Work did you do the second time than the first time will be [tex]\rm \frac{3}{2} Kx^2[/tex].
What is spring work?Spring work is equivalent to the effort done to extend the spring, which is dependent on both the spring constant k and the distance stretched.
The potential energy stored as a result of the deformation of an elastic item, such as spring stretching, is referred to as elastic potential energy.
Work done by spring = potential energy
[tex](PE)_{spring }= \frac{1}{2} Kx^2[/tex]
Case 1
spring is compressed by x
[tex](PE)_1{spring }= \frac{1}{2} Kx^2[/tex]
Case 2
spring is compressed by 2x
[tex]\rm (PE)_2{spring }= \frac{1}{2} K(2x)^2\\\\\rm (PE)_2{spring }= \frac{1}{2}\times 4K(x)^2\\\\\rm (PE)_2{spring }= 2K(x)^2[/tex]
The difference in the potential energy is found by;
[tex](PE)_2-(PE)_1=2Kx^2-\frac{1}{2} Kx^2=\frac{3}{2} Kx^2[/tex]
Hence spring work did you do the second time than the first time will be [tex]\rm \frac{3}{2} Kx^2[/tex].
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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4. Determine the magnitude of force at point and determine if the ladder will slip.
This question is incomplete, the complete question;
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.
Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb
Answer:
- the magnitude of force at point A is 79.1033 lb
- since FA < FA_max; Ladder WILL NOT slip
Explanation:
Given that;
∑'MA = 0
⇒ NB [Lsin∅] - W[L/2.cos∅] = 0
NB = W / 2tan∅ -------let this be equation 1
∑Fx = 0
⇒ FA - NB = 0
FA = NB
therefore from equation 1
FA = NB = W / 2tan∅
we substitute in our values
FA = NB = 76 / 2tan(60°) = 21.9393 lb
Now ∑Fy = 0
NA - W = 0
NA = W = 76 lb
Net force at A will be
FA' = √( NA² + FA²)
= √( (W)² + (W / 2tan∅)²)
we substitute in our values
FA' = √( (76)² + (21.9393)²)
= √( 5776 + 481.3328)
= √ 6257.3328
FA' = 79.1033 lb
Therefore the magnitude of force at point A is 79.1033 lb
Now maximum possible frictional force at A
FA_max = μ × NA
so, FA_max = 0.4 × 76
FA_max = 30.4 lb
So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.
Therefore since FA < FA_max; Ladder WILL NOT slip
Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artificial gravity, human growth is stunted and biological functions break down.
An effective way to create artificial gravity is through the use of a rotating enclosed cylinder, as shown in the figure. Humans walk on the inside of the outer edge of the cylinder, which has a diameter of =3335 m that is large enough such that its curvature is not readily noticeable to the inhabitants. (The space station in the figure is not drawn to scale.)
Once the space station is rotating at the necessary angular speed to create an artificial gravity of 1 , how many minutes would it take the space station to make one revolution?
I did [tex]2\pi \sqrt{\frac{1667.6}{9.8} } =81.96[/tex] then I converted 81.96 to minutes which was 1.4 but it still got marked wrong. 81.96 was wrong as well. Am I using the wrong equation for it? I'm not sure what to do.
The period of the enclosed cylinder is approximately 115.866 seconds.
The rotating enclosed cylinder is rotating at constant angular speed ([tex]\omega[/tex]), in radians per second, which means that experiments a constant radial angular acceleration ([tex]\alpha[/tex]), in radians per square second. Then, we derive an expression for the period of the cylinder, this is, the time needed by the cylinder to make one revolution:
[tex]g = \omega^{2}\cdot R[/tex] (1)
Where:
[tex]g[/tex] - Gravitational acceleration, in meters per square second. [tex]R[/tex] - Radius of the enclose cylinder, in meters.[tex]g = \frac{4\pi^{2}\cdot R}{T^{2}}[/tex]
[tex]T = 2\pi\cdot \sqrt{\frac{R}{g} }[/tex] (2)
Where [tex]T[/tex] is the period, in seconds.
If we know that [tex]R = 3335\,m[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the period of the enclosed cylinder is:
[tex]T = 2\pi\cdot \sqrt{\frac{3335\,m}{9.807\,\frac{m}{s^{2}} } }[/tex]
[tex]T \approx 115.866\,s\,(1.931\,min)[/tex]
The period of the enclosed cylinder is approximately 115.866 seconds.
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What is your worldview? Explain?
Answer:
My worldview is that the world is a beautiful place. IT has many wonders, and many kind people. There are also very nice places, governments, tech and more.
1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms
2. When an electric soldering iron is used in a 110 V circuit, the current flowing through the iron is
2 A. What is the resistance of the iron?
3. A current of 0.2 A flows through an electric bell having a resistance of 65 ohms. What must be
the applied voltage in the circuit?
Answer:
(1) 0.04 ohms (2) 55 ohms (3) 13 volt
Explanation:
(1) The resistance of an electric device is 40,000 microhms.
We need to convert it into ohms.
[tex]1\ \mu \Omega =10^{-6}\ \Omega[/tex]
To covert 40,000 microhms to ohms, multiply 40,000 and 10⁻⁶ as follows :
[tex]40000 \ \mu \Omega =40000 \times 10^{-6}\ \Omega\\\\=0.04\ \Omega[/tex]
(2) Voltage used, V = 110 V
Current, I = 2 A
We need to find the resistance of the iron. Using Ohms law to find it as follows :
V = IR, where R is resistance
[tex]R=\dfrac{V}{I}\\\\R=\dfrac{110}{2}\\\\R=55\ \Omega[/tex]
(3) Current, I = 0.2 A
Resistance, R = 65 ohms
We need to find the applied voltage in the circuit. Using Ohms law to find it as follows :
V=IR
V = 0.2 × 65
V = 13 volt
Answer:
1. 0.04 Ohms
2. 55 Ohms
3. 13 Volts
Explanation:
Penn Foster
To remove a stain using a solvent the stain has to become dissolved in the solvent
True
False
Answer:
True
Explanation:
have a good day:)
Answer: This statement is True
A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what isthe current in the primary?and WHy?
Answer:
The correct solution will be "12.0 A".
Explanation:
The given values are:
[tex]N_p= 500 \ turn[/tex]
[tex]N_s= 200 \ turn[/tex]
[tex]I_s= 3.0 \ A[/tex]
By using the transformer formula, we get
⇒ [tex]\frac{N_p}{N_s} =\frac{I_s}{I_p}[/tex]
⇒ [tex]I_p = I_s\times \frac{N_s}{N_p}[/tex]
On substituting the given values, we get
⇒ [tex]=3.0 \ A\times \frac{2000}{500}[/tex]
⇒ [tex]=12.0 \ A[/tex]