Let a be a set containing 10 distinct elements, then the total -Turito

General

Easy

Maths-

Let $A$ be a set containing 10 distinct elements, then the total number of distinct functions from $A$ to $A$ is-

Maths-General

$10^{10}$
$10!$
$2^{10} - 1$
$2^{10}$

Answer:The correct answer is: $10 to the power of 10 end exponent$

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Related Questions to study

General

maths-

The number of bijective functions from set A to itself when a contains 106 elements-

maths-General

General

physics-

A particle of mass $m$ moving with a velocity $u$ makes an elastic one dimensional collision with a stationary particle of mass $m$ establishing a contact with it for extremely small time $T$ . Their force of contact increases from zero to $F subscript 0 end subscript$ linearly in time $T divided by 4$ , remains constant for a further time $T divided by 2$ and decreases linearly from $F subscript 0 end subscript$ to zero in further time $T divided by 4$ as shown. The magnitude possessed by $F subscript 0 end subscript$ is

Change in momentum = Impulse
= Area under force-time graph

therefore m v equals

Area of trapezium

rightwards double arrow m v equals fraction numerator 1 over denominator 2 end fraction open parentheses T plus fraction numerator T over denominator 2 end fraction close parentheses F subscript 0 end subscript rightwards double arrow m v equals fraction numerator 3 T over denominator 4 end fraction F subscript 0 end subscript rightwards double arrow F subscript 0 end subscript equals fraction numerator 4 m u over denominator 3 T end fraction

A particle of mass $m$ moving with a velocity $u$ makes an elastic one dimensional collision with a stationary particle of mass $m$ establishing a contact with it for extremely small time $T$ . Their force of contact increases from zero to $F subscript 0 end subscript$ linearly in time $T divided by 4$ , remains constant for a further time $T divided by 2$ and decreases linearly from $F subscript 0 end subscript$ to zero in further time $T divided by 4$ as shown. The magnitude possessed by $F subscript 0 end subscript$ is

physics-General

Change in momentum = Impulse
= Area under force-time graph

therefore m v equals

Area of trapezium

rightwards double arrow m v equals fraction numerator 1 over denominator 2 end fraction open parentheses T plus fraction numerator T over denominator 2 end fraction close parentheses F subscript 0 end subscript rightwards double arrow m v equals fraction numerator 3 T over denominator 4 end fraction F subscript 0 end subscript rightwards double arrow F subscript 0 end subscript equals fraction numerator 4 m u over denominator 3 T end fraction

General

physics-

Given below is a graph between a variable force $left parenthesis F right parenthesis$ (along $y$ -axis) and the displacement $left parenthesis X right parenthesis$ (along $x$ -axis) of a particle in one dimension. The work done by the force in the displacement interval between $0 blank m$ and $30 blank m$ is

physics-General

General

physics-

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 blank m$ is

Work done

W equals

area under

F minus S

graph
= area of trapezium

A B C D plus

area of trapezium

C E F D

equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses 10 plus 15 close parentheses cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times left parenthesis 10 plus 20 right parenthesis cross times 5

equals 125 plus 75 equals 200 blank J

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 blank m$ is

physics-General

Work done

W equals

area under

F minus S

graph
= area of trapezium

A B C D plus

area of trapezium

C E F D

equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses 10 plus 15 close parentheses cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times left parenthesis 10 plus 20 right parenthesis cross times 5

equals 125 plus 75 equals 200 blank J

General

physics-

If $w subscript 1 end subscript comma w subscript 2 blank a n d blank W subscript 3 end subscript end subscript$ represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively(as shown)in the gravitational field of a point mass m. Find the correct relation between $w subscript 1 end subscript comma w subscript 2 end subscript a n d blank w subscript 3 end subscript$

Gravitational field is a conservative force field. In a conservative force field work done is path independent.

therefore blank W subscript 1 end subscript equals W subscript 2 end subscript equals W subscript 3 end subscript

If $w subscript 1 end subscript comma w subscript 2 blank a n d blank W subscript 3 end subscript end subscript$ represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively(as shown)in the gravitational field of a point mass m. Find the correct relation between $w subscript 1 end subscript comma w subscript 2 end subscript a n d blank w subscript 3 end subscript$

physics-General

Gravitational field is a conservative force field. In a conservative force field work done is path independent.

therefore blank W subscript 1 end subscript equals W subscript 2 end subscript equals W subscript 3 end subscript

General

physics-

A particle is acted upon by a force $F$ which varies with position $x$ as shown in figure. If the particle at $x equals 0$ has kinetic energy of 25 J, then the kinetic energy of the particle at $x equals 16 blank m$ is

Work done=area between the graph force displacement curve and displacement

W equals fraction numerator 1 over denominator 2 end fraction cross times 6 cross times 10 minus 5 cross times 4 plus 5 cross times 4 minus 5 cross times 2

W equals 20 blank J

According to work energy theorem

increment equals K subscript E end subscript equals W

K subscript E end subscript subscript f end subscript equals W plus increment K

=20+25
=45J

A particle is acted upon by a force $F$ which varies with position $x$ as shown in figure. If the particle at $x equals 0$ has kinetic energy of 25 J, then the kinetic energy of the particle at $x equals 16 blank m$ is

physics-General

Work done=area between the graph force displacement curve and displacement

W equals fraction numerator 1 over denominator 2 end fraction cross times 6 cross times 10 minus 5 cross times 4 plus 5 cross times 4 minus 5 cross times 2

W equals 20 blank J

According to work energy theorem

increment equals K subscript E end subscript equals W

K subscript E end subscript subscript f end subscript equals W plus increment K

=20+25
=45J

General

physics-

A vertical spring with force constant $K$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$ . The net work done in the process is

Gravitational potential energy of ball gets converted into elastic potential energy of the spring

m g open parentheses h plus d close parentheses equals fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent

Net work done

equals m g open parentheses h plus d close parentheses minus fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent equals 0

A vertical spring with force constant $K$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$ . The net work done in the process is

physics-General

Gravitational potential energy of ball gets converted into elastic potential energy of the spring

m g open parentheses h plus d close parentheses equals fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent

Net work done

equals m g open parentheses h plus d close parentheses minus fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent equals 0

General

physics-

Figure shows the $F$ - $x$ graph. Where $F$ is the force applied and $x$ is the distance covered

By the body along a straight line path. Given that $F$ is in $n e w t o n$ and $x blank$ in $m e t r e$ , what is the work done?

Work done =area under curve and displacement axis

equals 1 cross times 10 minus 1 cross times 10 plus 1 cross times 10 equals 10 blank J

Figure shows the $F$ - $x$ graph. Where $F$ is the force applied and $x$ is the distance covered

By the body along a straight line path. Given that $F$ is in $n e w t o n$ and $x blank$ in $m e t r e$ , what is the work done?

physics-General

Work done =area under curve and displacement axis

equals 1 cross times 10 minus 1 cross times 10 plus 1 cross times 10 equals 10 blank J

General

maths-

Domain of definition of the function $f left parenthesis x right parenthesis equals fraction numerator 3 over denominator 4 minus x to the power of 2 end exponent end fraction plus l o g subscript 10 end subscript invisible function application open parentheses x to the power of 3 end exponent minus x close parentheses$ , is

maths-General

General

physics-

Velocity-time graph of a particle of mass 2 kg moving in a straight line is as shown in figure. Work done by all forces on the particle is

Initial velocity of particle,

v subscript i end subscript equals 20 blank m s to the power of negative 1 end exponent

Final velocity of the particle,

v subscript f end subscript equals 0

According to work-energy theorem,

W subscript n e t end subscript equals increment K E equals K subscript f end subscript minus K subscript i end subscript

equals fraction numerator 1 over denominator 2 end fraction m left parenthesis v subscript f end subscript superscript 2 end superscript minus v subscript i end subscript superscript 2 end superscript right parenthesis

equals fraction numerator 1 over denominator 2 end fraction cross times 2 left parenthesis 0 to the power of 2 end exponent minus 20 to the power of 2 end exponent right parenthesis

equals negative 400 blank J

Velocity-time graph of a particle of mass 2 kg moving in a straight line is as shown in figure. Work done by all forces on the particle is

physics-General

Initial velocity of particle,

v subscript i end subscript equals 20 blank m s to the power of negative 1 end exponent

Final velocity of the particle,

v subscript f end subscript equals 0

According to work-energy theorem,

W subscript n e t end subscript equals increment K E equals K subscript f end subscript minus K subscript i end subscript

equals fraction numerator 1 over denominator 2 end fraction m left parenthesis v subscript f end subscript superscript 2 end superscript minus v subscript i end subscript superscript 2 end superscript right parenthesis

equals fraction numerator 1 over denominator 2 end fraction cross times 2 left parenthesis 0 to the power of 2 end exponent minus 20 to the power of 2 end exponent right parenthesis

equals negative 400 blank J

General

physics-

A frictionless track $A B C D E$ ends in a circular loop of radius $R$ . A body slides down the track from point $A$ which is it $a$ height $h equals 5 blank c m$ . Maximum value of $R$ for the body to successfully complete the loop is

Condition for vertical looping

h equals fraction numerator 5 over denominator 2 end fraction r equals 5 c m blank therefore r equals 2 blank c m

A frictionless track $A B C D E$ ends in a circular loop of radius $R$ . A body slides down the track from point $A$ which is it $a$ height $h equals 5 blank c m$ . Maximum value of $R$ for the body to successfully complete the loop is

physics-General

Condition for vertical looping

h equals fraction numerator 5 over denominator 2 end fraction r equals 5 c m blank therefore r equals 2 blank c m

General

physics-

The force acting on a body moving along $x$ -axis varies with the position of the particle as shown in the fig

The body is in stable equilibrium at

When particle moves away from the origin then at position

x equals x subscript 1 end subscript

force is zero and at

x greater than x subscript 1 end subscript

, force is positive (repulsive in nature) so particle moves further and does not return back to original position

i. e.

the equilibrium is not stable Similarly at position

x equals x subscript 2 end subscript

force is zero and at

x greater than x subscript 2 end subscript

, force is negative (attractive in nature) So particle return back to original position

i. e.

the equilibrium is stable

The force acting on a body moving along $x$ -axis varies with the position of the particle as shown in the fig

The body is in stable equilibrium at

physics-General

When particle moves away from the origin then at position

x equals x subscript 1 end subscript

force is zero and at

x greater than x subscript 1 end subscript

, force is positive (repulsive in nature) so particle moves further and does not return back to original position

i. e.

the equilibrium is not stable Similarly at position

x equals x subscript 2 end subscript

force is zero and at

x greater than x subscript 2 end subscript

, force is negative (attractive in nature) So particle return back to original position

i. e.

the equilibrium is stable

General

physics-

The pointer reading $v divided by s$ load graph for a spring balance is as given in the figure. The spring constant is

Spring constant

k equals fraction numerator F over denominator x end fraction equals

Slope of curve

therefore k equals fraction numerator 4 minus 1 over denominator 30 end fraction equals fraction numerator 3 over denominator 30 end fraction equals 0.1 blank k g divided by c m

The pointer reading $v divided by s$ load graph for a spring balance is as given in the figure. The spring constant is

physics-General

Spring constant

k equals fraction numerator F over denominator x end fraction equals

Slope of curve

therefore k equals fraction numerator 4 minus 1 over denominator 30 end fraction equals fraction numerator 3 over denominator 30 end fraction equals 0.1 blank k g divided by c m

General

physics-

The force $F$ acting on a particle moving in a straight line is shown in figure. What is the work done by the force on the particle in the 1^st meter of the trajectory

Work done

open parentheses W close parentheses equals

Area under curve of

F

x

graph
= Area of triangle

O A B equals fraction numerator 1 over denominator 2 end fraction cross times 5 cross times 1 equals 2.5 blank J

The force $F$ acting on a particle moving in a straight line is shown in figure. What is the work done by the force on the particle in the 1^st meter of the trajectory

physics-General

Work done

open parentheses W close parentheses equals

Area under curve of

F

x

graph
= Area of triangle

O A B equals fraction numerator 1 over denominator 2 end fraction cross times 5 cross times 1 equals 2.5 blank J

General

maths-

Range of function f(x) = $fraction numerator x to the power of 2 end exponent plus 2 x plus 3 over denominator x end fraction$ , $x element of times R$ is given by -

]
(C)

left parenthesis negative infinity comma 2 minus 2 square root of 3 right square bracket union left square bracket 2 plus 2 square root of 3 comma infinity right parenthesis

Range of function f(x) = $fraction numerator x to the power of 2 end exponent plus 2 x plus 3 over denominator x end fraction$ , $x element of times R$ is given by -

maths-General

]
(C)

left parenthesis negative infinity comma 2 minus 2 square root of 3 right square bracket union left square bracket 2 plus 2 square root of 3 comma infinity right parenthesis

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Can’t find what you’re looking for?

Let $A$ be a set containing 10 distinct elements, then the total number of distinct functions from $A$ to $A$ is-

$10^{10}$
$10!$
$2^{10} - 1$
$2^{10}$

Answer:The correct answer is: $10 to the power of 10 end exponent$

Book A Free Demo

Related Questions to study

The number of bijective functions from set A to itself when a contains 106 elements-

The number of bijective functions from set A to itself when a contains 106 elements-

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 blank m$ is

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 blank m$ is

A particle is acted upon by a force $F$ which varies with position $x$ as shown in figure. If the particle at $x equals 0$ has kinetic energy of 25 J, then the kinetic energy of the particle at $x equals 16 blank m$ is

A particle is acted upon by a force $F$ which varies with position $x$ as shown in figure. If the particle at $x equals 0$ has kinetic energy of 25 J, then the kinetic energy of the particle at $x equals 16 blank m$ is

A vertical spring with force constant $K$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$ . The net work done in the process is

A vertical spring with force constant $K$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$ . The net work done in the process is

Figure shows the $F$ - $x$ graph. Where $F$ is the force applied and $x$ is the distance covered

By the body along a straight line path. Given that $F$ is in $n e w t o n$ and $x blank$ in $m e t r e$ , what is the work done?

Figure shows the $F$ - $x$ graph. Where $F$ is the force applied and $x$ is the distance covered

By the body along a straight line path. Given that $F$ is in $n e w t o n$ and $x blank$ in $m e t r e$ , what is the work done?

Velocity-time graph of a particle of mass 2 kg moving in a straight line is as shown in figure. Work done by all forces on the particle is

Velocity-time graph of a particle of mass 2 kg moving in a straight line is as shown in figure. Work done by all forces on the particle is

A frictionless track $A B C D E$ ends in a circular loop of radius $R$ . A body slides down the track from point $A$ which is it $a$ height $h equals 5 blank c m$ . Maximum value of $R$ for the body to successfully complete the loop is

A frictionless track $A B C D E$ ends in a circular loop of radius $R$ . A body slides down the track from point $A$ which is it $a$ height $h equals 5 blank c m$ . Maximum value of $R$ for the body to successfully complete the loop is

The force acting on a body moving along $x$ -axis varies with the position of the particle as shown in the fig

The body is in stable equilibrium at

The force acting on a body moving along $x$ -axis varies with the position of the particle as shown in the fig

The body is in stable equilibrium at

The pointer reading $v divided by s$ load graph for a spring balance is as given in the figure. The spring constant is

The pointer reading $v divided by s$ load graph for a spring balance is as given in the figure. The spring constant is

The force $F$ acting on a particle moving in a straight line is shown in figure. What is the work done by the force on the particle in the 1^st meter of the trajectory

The force $F$ acting on a particle moving in a straight line is shown in figure. What is the work done by the force on the particle in the 1^st meter of the trajectory

Range of function f(x) = $fraction numerator x to the power of 2 end exponent plus 2 x plus 3 over denominator x end fraction$ , $x element of times R$ is given by -

Range of function f(x) = $fraction numerator x to the power of 2 end exponent plus 2 x plus 3 over denominator x end fraction$ , $x element of times R$ is given by -

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Can’t find what you’re looking for?

Let be a set containing 10 distinct elements, then the total number of distinct functions from to is-

Answer:The correct answer is:

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Related Questions to study

The number of bijective functions from set A to itself when a contains 106 elements-

The number of bijective functions from set A to itself when a contains 106 elements-

Given below is a graph between a variable force (along -axis) and the displacement (along -axis) of a particle in one dimension. The work done by the force in the displacement interval between and is

Given below is a graph between a variable force (along -axis) and the displacement (along -axis) of a particle in one dimension. The work done by the force in the displacement interval between and is

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of is

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of is

If represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively(as shown)in the gravitational field of a point mass m. Find the correct relation between

If represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively(as shown)in the gravitational field of a point mass m. Find the correct relation between

A particle is acted upon by a force which varies with position as shown in figure. If the particle at has kinetic energy of 25 J, then the kinetic energy of the particle at is

A particle is acted upon by a force which varies with position as shown in figure. If the particle at has kinetic energy of 25 J, then the kinetic energy of the particle at is

A vertical spring with force constant is fixed on a table. A ball of mass at a height above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance . The net work done in the process is

A vertical spring with force constant is fixed on a table. A ball of mass at a height above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance . The net work done in the process is

Figure shows the -graph. Where is the force applied and is the distance coveredBy the body along a straight line path. Given that is in and in , what is the work done?

Figure shows the -graph. Where is the force applied and is the distance coveredBy the body along a straight line path. Given that is in and in , what is the work done?

Domain of definition of the function, is

Domain of definition of the function, is

Velocity-time graph of a particle of mass 2 kg moving in a straight line is as shown in figure. Work done by all forces on the particle is

Velocity-time graph of a particle of mass 2 kg moving in a straight line is as shown in figure. Work done by all forces on the particle is

A frictionless track ends in a circular loop of radius . A body slides down the track from point which is it height . Maximum value of for the body to successfully complete the loop is

A frictionless track ends in a circular loop of radius . A body slides down the track from point which is it height . Maximum value of for the body to successfully complete the loop is

The force acting on a body moving along -axis varies with the position of the particle as shown in the figThe body is in stable equilibrium at

The force acting on a body moving along -axis varies with the position of the particle as shown in the figThe body is in stable equilibrium at

The pointer reading load graph for a spring balance is as given in the figure. The spring constant is

The pointer reading load graph for a spring balance is as given in the figure. The spring constant is

The force acting on a particle moving in a straight line is shown in figure. What is the work done by the force on the particle in the 1st meter of the trajectory

The force acting on a particle moving in a straight line is shown in figure. What is the work done by the force on the particle in the 1st meter of the trajectory

Range of function f(x) = , is given by -

Range of function f(x) = , is given by -

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Let $A$ be a set containing 10 distinct elements, then the total number of distinct functions from $A$ to $A$ is-

Answer:The correct answer is: $10 to the power of 10 end exponent$

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 blank m$ is

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 blank m$ is

A particle is acted upon by a force $F$ which varies with position $x$ as shown in figure. If the particle at $x equals 0$ has kinetic energy of 25 J, then the kinetic energy of the particle at $x equals 16 blank m$ is

A particle is acted upon by a force $F$ which varies with position $x$ as shown in figure. If the particle at $x equals 0$ has kinetic energy of 25 J, then the kinetic energy of the particle at $x equals 16 blank m$ is

A vertical spring with force constant $K$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$ . The net work done in the process is

A vertical spring with force constant $K$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$ . The net work done in the process is

Figure shows the $F$ - $x$ graph. Where $F$ is the force applied and $x$ is the distance covered

By the body along a straight line path. Given that $F$ is in $n e w t o n$ and $x blank$ in $m e t r e$ , what is the work done?

Figure shows the $F$ - $x$ graph. Where $F$ is the force applied and $x$ is the distance covered

By the body along a straight line path. Given that $F$ is in $n e w t o n$ and $x blank$ in $m e t r e$ , what is the work done?

A frictionless track $A B C D E$ ends in a circular loop of radius $R$ . A body slides down the track from point $A$ which is it $a$ height $h equals 5 blank c m$ . Maximum value of $R$ for the body to successfully complete the loop is

A frictionless track $A B C D E$ ends in a circular loop of radius $R$ . A body slides down the track from point $A$ which is it $a$ height $h equals 5 blank c m$ . Maximum value of $R$ for the body to successfully complete the loop is

The force acting on a body moving along $x$ -axis varies with the position of the particle as shown in the fig

The body is in stable equilibrium at

The force acting on a body moving along $x$ -axis varies with the position of the particle as shown in the fig

The body is in stable equilibrium at

The pointer reading $v divided by s$ load graph for a spring balance is as given in the figure. The spring constant is

The pointer reading $v divided by s$ load graph for a spring balance is as given in the figure. The spring constant is

The force $F$ acting on a particle moving in a straight line is shown in figure. What is the work done by the force on the particle in the 1^st meter of the trajectory

The force $F$ acting on a particle moving in a straight line is shown in figure. What is the work done by the force on the particle in the 1^st meter of the trajectory

Range of function f(x) = $fraction numerator x to the power of 2 end exponent plus 2 x plus 3 over denominator x end fraction$ , $x element of times R$ is given by -

Range of function f(x) = $fraction numerator x to the power of 2 end exponent plus 2 x plus 3 over denominator x end fraction$ , $x element of times R$ is given by -