Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

If $2 to the power of 2006 end exponent minus 2006$ divided by 7, the remainder is

0
1
2
4

The correct answer is: 0

Related Questions to study

The remainder when $2 3 to the power of 23 end exponent$ is divided by 53 is

The remainder when $2 3 to the power of 23 end exponent$ is divided by 53 is

If $C subscript r end subscript equals 32 C subscript r end subscript$ , then $not stretchy sum subscript r equals 0 end subscript superscript 5 end superscript C subscript 6 r end subscript equals$

If $C subscript r end subscript equals 32 C subscript r end subscript$ , then $not stretchy sum subscript r equals 0 end subscript superscript 5 end superscript C subscript 6 r end subscript equals$

$bullet$ If $C subscript r end subscript equals to the power of 30 end exponent C subscript r end subscript$ , then $C subscript 0 end subscript plus C subscript 4 end subscript plus C subscript 8 end subscript plus$ $plus C subscript 28 end subscript equals$

$bullet$ If $C subscript r end subscript equals to the power of 30 end exponent C subscript r end subscript$ , then $C subscript 0 end subscript plus C subscript 4 end subscript plus C subscript 8 end subscript plus$ $plus C subscript 28 end subscript equals$

parallel

A particle goes from point A to B. Its displacement is X and pathlength is y. So $x over y$

A particle goes from point A to B. Its displacement is X and pathlength is y. So $x over y$

A car goes from one end to the other end of a semicircular path of diameter ' d '. Find the ratio between path length and displacement.

A car goes from one end to the other end of a semicircular path of diameter ' d '. Find the ratio between path length and displacement.

A particle moves from A to P and then it moves from P to B as shown in the figure. Find path length and displacement.

A particle moves from A to P and then it moves from P to B as shown in the figure. Find path length and displacement.

parallel

If the thickness of the wire is doubled, then the breaking force in the above question will be

proportional to pi r to the power of 2 end exponent

If thickness (radius) of wire is doubled then breaking force will become four times

If the thickness of the wire is doubled, then the breaking force in the above question will be

physics-General

proportional to pi r to the power of 2 end exponent

If thickness (radius) of wire is doubled then breaking force will become four times

The Young’s modulus of the material of a wire is $2 cross times 10 to the power of 10 end exponent N m to the power of negative 2 end exponent.$ If the elongation strain is $1 percent sign$ , then the energy stored in the wire per unit volume in $J m blank to the power of negative 3 end exponent$ is

Per unit volume energy stored

equals fraction numerator 1 over denominator 2 end fraction cross times Y cross times left parenthesis s t r a i n right parenthesis to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction cross times Y cross times open parentheses fraction numerator l over denominator L end fraction close parentheses to the power of 2 end exponent

l equals L cross times 1 percent sign

l equals fraction numerator L over denominator 100 end fraction cross times s t o r e d blank e n e r g y

Y equals fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 to the power of 10 end exponent cross times open parentheses fraction numerator L over denominator 100 L end fraction close parentheses to the power of 2 end exponent equals 10 to the power of 6 end exponent J m to the power of negative 3 end exponent

The Young’s modulus of the material of a wire is $2 cross times 10 to the power of 10 end exponent N m to the power of negative 2 end exponent.$ If the elongation strain is $1 percent sign$ , then the energy stored in the wire per unit volume in $J m blank to the power of negative 3 end exponent$ is

physics-General

Per unit volume energy stored

equals fraction numerator 1 over denominator 2 end fraction cross times Y cross times left parenthesis s t r a i n right parenthesis to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction cross times Y cross times open parentheses fraction numerator l over denominator L end fraction close parentheses to the power of 2 end exponent

l equals L cross times 1 percent sign

l equals fraction numerator L over denominator 100 end fraction cross times s t o r e d blank e n e r g y

Y equals fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 to the power of 10 end exponent cross times open parentheses fraction numerator L over denominator 100 L end fraction close parentheses to the power of 2 end exponent equals 10 to the power of 6 end exponent J m to the power of negative 3 end exponent

The following data were obtained when a wire was stretched within the elastic region Force applied to wire 100 N
Area of cross-section of wire $10 to the power of negative 6 end exponent m to the power of 2 end exponent$
Extensional of wire $2 cross times 10 to the power of negative 9 end exponent m$
Which of the following deductions can be correctly made from this data?
I. The value of Young’s Modulus is $10 to the power of 11 end exponent N blank m to the power of negative 2 end exponent$
II. The strain is $10 to the power of negative 3 end exponent$
III. The energy stored in the wire when the load is applied is 10 J

fraction numerator 100 N over denominator 10 to the power of negative 6 end exponent m to the power of 2 end exponent end fraction equals 10 to the power of 8 end exponent blank N m to the power of negative 2 end exponent

equals blank fraction numerator 2 cross times 10 to the power of negative 3 end exponent over denominator 2 end fraction equals 10 to the power of negative 3 end exponent

Young modulus
=

fraction numerator 10 to the power of 8 end exponent over denominator 10 to the power of negative 3 end exponent end fraction blank N m to the power of negative 2 end exponent equals 10 to the power of 11 end exponent blank N m to the power of negative 2 end exponent

Energy stored =

fraction numerator 1 over denominator 2 end fraction cross times 100 blank cross times 2 cross times 10 to the power of negative 3 end exponent blank J

minus

The following data were obtained when a wire was stretched within the elastic region Force applied to wire 100 N
Area of cross-section of wire $10 to the power of negative 6 end exponent m to the power of 2 end exponent$
Extensional of wire $2 cross times 10 to the power of negative 9 end exponent m$
Which of the following deductions can be correctly made from this data?
I. The value of Young’s Modulus is $10 to the power of 11 end exponent N blank m to the power of negative 2 end exponent$
II. The strain is $10 to the power of negative 3 end exponent$
III. The energy stored in the wire when the load is applied is 10 J

physics-General

fraction numerator 100 N over denominator 10 to the power of negative 6 end exponent m to the power of 2 end exponent end fraction equals 10 to the power of 8 end exponent blank N m to the power of negative 2 end exponent

equals blank fraction numerator 2 cross times 10 to the power of negative 3 end exponent over denominator 2 end fraction equals 10 to the power of negative 3 end exponent

Young modulus
=

fraction numerator 10 to the power of 8 end exponent over denominator 10 to the power of negative 3 end exponent end fraction blank N m to the power of negative 2 end exponent equals 10 to the power of 11 end exponent blank N m to the power of negative 2 end exponent

Energy stored =

fraction numerator 1 over denominator 2 end fraction cross times 100 blank cross times 2 cross times 10 to the power of negative 3 end exponent blank J

minus

parallel

Which of the following relations is true

Y equals 2 eta open parentheses 1 plus sigma close parentheses rightwards double arrow sigma equals fraction numerator 0.5 Y minus eta over denominator eta end fraction

Which of the following relations is true

physics-General

Y equals 2 eta open parentheses 1 plus sigma close parentheses rightwards double arrow sigma equals fraction numerator 0.5 Y minus eta over denominator eta end fraction

A particle moves from A to B and then it moves from to as shown in figure. Calculate the ratio between path length and displacement.

A particle moves from A to B and then it moves from to as shown in figure. Calculate the ratio between path length and displacement.

If $A equals open square brackets table attributes columnalign left left left columnspacing 1em end attributes row cell 1 0 1 end cell row cell 2 1 0 end cell row cell 3 2 1 end cell end table close square brackets$ then det A=

A equals open square brackets table attributes columnalign left left left columnspacing 1em end attributes row cell 1      0      1 end cell row cell 2      1      0 end cell row cell 3      2      1 end cell end table close square brackets

open vertical bar table row 1 0 1 row 2 1 0 row 3 2 1 end table close vertical bar equals open square brackets 1 left parenthesis 1 cross times 1 minus 2 cross times 0 right parenthesis close square brackets minus open square brackets 0 left parenthesis 2 cross times 1 minus 3 cross times 0 right parenthesis close square brackets plus open square brackets 1 left parenthesis 2 cross times 2 minus 3 cross times 1 right parenthesis close square brackets space space space space space space space space space space space space space space space equals open square brackets 1 cross times 1 close square brackets minus 0 plus open square brackets 1 cross times 1 close square brackets space space space space space space space space space space space space space space space equals 1 minus 0 plus 1 equals 2

If $A equals open square brackets table attributes columnalign left left left columnspacing 1em end attributes row cell 1 0 1 end cell row cell 2 1 0 end cell row cell 3 2 1 end cell end table close square brackets$ then det A=

A equals open square brackets table attributes columnalign left left left columnspacing 1em end attributes row cell 1      0      1 end cell row cell 2      1      0 end cell row cell 3      2      1 end cell end table close square brackets

open vertical bar table row 1 0 1 row 2 1 0 row 3 2 1 end table close vertical bar equals open square brackets 1 left parenthesis 1 cross times 1 minus 2 cross times 0 right parenthesis close square brackets minus open square brackets 0 left parenthesis 2 cross times 1 minus 3 cross times 0 right parenthesis close square brackets plus open square brackets 1 left parenthesis 2 cross times 2 minus 3 cross times 1 right parenthesis close square brackets space space space space space space space space space space space space space space space equals open square brackets 1 cross times 1 close square brackets minus 0 plus open square brackets 1 cross times 1 close square brackets space space space space space space space space space space space space space space space equals 1 minus 0 plus 1 equals 2

parallel

$open vertical bar table attributes columnalign center center center columnspacing 1em end attributes row 1 w cell w squared end cell row w cell w squared end cell 1 row cell w squared end cell 1 w end table close vertical bar equals$

space space space space open vertical bar table attributes columnalign center center center columnspacing 1em end attributes row 1 w cell w squared end cell row w cell w squared end cell 1 row cell w squared end cell 1 w end table close vertical bar equals open vertical bar table row cell 1 plus w plus w squared end cell w cell w squared end cell row cell 1 plus w plus w squared end cell cell w squared end cell 1 row cell 1 plus w plus w squared end cell 1 w end table close vertical bar space space space space space space space open square brackets C subscript 1 rightwards arrow C subscript 1 plus C subscript 2 plus C subscript 3 close square brackets equals open parentheses 1 plus w plus w squared close parentheses open vertical bar table row 1 w cell w squared end cell row 1 cell w squared end cell 1 row 1 1 w end table close vertical bar space space space space space space space space equals open parentheses 1 plus w plus w squared close parentheses cross times 0 space space space space space space space space space space space space space open square brackets C o l l i n e a r i t y close square brackets equals 0

$open vertical bar table attributes columnalign center center center columnspacing 1em end attributes row 1 w cell w squared end cell row w cell w squared end cell 1 row cell w squared end cell 1 w end table close vertical bar equals$

space space space space open vertical bar table attributes columnalign center center center columnspacing 1em end attributes row 1 w cell w squared end cell row w cell w squared end cell 1 row cell w squared end cell 1 w end table close vertical bar equals open vertical bar table row cell 1 plus w plus w squared end cell w cell w squared end cell row cell 1 plus w plus w squared end cell cell w squared end cell 1 row cell 1 plus w plus w squared end cell 1 w end table close vertical bar space space space space space space space open square brackets C subscript 1 rightwards arrow C subscript 1 plus C subscript 2 plus C subscript 3 close square brackets equals open parentheses 1 plus w plus w squared close parentheses open vertical bar table row 1 w cell w squared end cell row 1 cell w squared end cell 1 row 1 1 w end table close vertical bar space space space space space space space space equals open parentheses 1 plus w plus w squared close parentheses cross times 0 space space space space space space space space space space space space space open square brackets C o l l i n e a r i t y close square brackets equals 0

Frame of reference is a ... and a ···.. from where an observer takes his observation,

Frame of reference is a ... and a ···.. from where an observer takes his observation,

To locate the position of the particle we need

To locate the position of the particle we need

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.