Two blocks of masses 2kg and 1kg are in contact with each other on a friction less table. When a horizontal force of 3N is applied on the block of mass 2kg, the value of the force of contact between the two blocks is.

Two masses m₁ and m₂ are suspended from a rigid support through a light string. The tension in the string between support and mass m₁ is.

If the angle of friction is 45 $degree$ , then the coefficient of static friction is.

A block of mass 2kg lies on a horizontal surface and it is subjected to a vertical downward force 30N as shown in figure. The contact force between the block and surface is.

The free body diagram of block 'm' is

If $vertical line z vertical line equals 1$ and $z not equal to 1$ , then all the values of $fraction numerator z over denominator 1 minus z to the power of 2 end exponent end fraction$ lie on

Therefore the correct option is choice 4

If $vertical line z vertical line equals 1$ and $z not equal to 1$ , then all the values of $fraction numerator z over denominator 1 minus z to the power of 2 end exponent end fraction$ lie on

Therefore the correct option is choice 4

If $z$ is a complex number such that $vertical line equals vertical line greater or equal than 2$ then the minimum value of $∣ z plus fraction numerator 1 over denominator 2 end fraction$ is is strictly

Therefore the correct option is choice 4

If $z$ is a complex number such that $vertical line equals vertical line greater or equal than 2$ then the minimum value of $∣ z plus fraction numerator 1 over denominator 2 end fraction$ is is strictly

Therefore the correct option is choice 4

If $open vertical bar fraction numerator z subscript 1 end subscript over denominator z subscript 2 end subscript end fraction close vertical bar equals 1$ and $a r g invisible function application open parentheses z subscript 1 end subscript z subscript 2 end subscript close parentheses equals 0$ then

Therefore the correct option is choice 3

If $open vertical bar fraction numerator z subscript 1 end subscript over denominator z subscript 2 end subscript end fraction close vertical bar equals 1$ and $a r g invisible function application open parentheses z subscript 1 end subscript z subscript 2 end subscript close parentheses equals 0$ then

Therefore the correct option is choice 3

One of the values of $open parentheses cis invisible function application fraction numerator pi over denominator 6 end fraction close parentheses to the power of fraction numerator 1 over denominator 2 end fraction end exponent plus open parentheses cis invisible function application fraction numerator negative pi over denominator 6 end fraction close parentheses to the power of fraction numerator 11 over denominator 2 end fraction end exponent$

If the distance between the points $left parenthesis a c o s invisible function application theta comma a s i n invisible function application theta right parenthesis$ , $left parenthesis a c o s invisible function application ϕ comma a s i n invisible function application ϕ right parenthesis$ is $2 a$
then $theta equals$ 6
Assertion (A): If $2 s i n invisible function application fraction numerator theta over denominator 2 end fraction equals square root of 1 plus s i n invisible function application theta end root plus square root of 1 minus s i n invisible function application theta end root$ , then $fraction numerator theta over denominator 2 end fraction$ lies between
$2 n pi plus fraction numerator pi over denominator 4 end fraction text and end text 2 n pi plus fraction numerator 3 pi over denominator 4 end fraction left parenthesis n element of z right parenthesis$
Reason $left parenthesis R right parenthesis colon$ If $fraction numerator theta over denominator 2 end fraction element of open parentheses fraction numerator pi over denominator 4 end fraction comma fraction numerator 3 pi over denominator 4 end fraction close parentheses$ , then $s i n invisible function application fraction numerator theta over denominator 2 end fraction greater than 0$

If $c o s e c to the power of 6 end exponent invisible function application q minus c o t to the power of 6 end exponent invisible function application q equals a c o t to the power of 4 end exponent invisible function application q plus b c o t to the power of 2 end exponent invisible function application q plus c$ then a+b+c=

So here we have used the trigonometric functions and trigonometric formulas to solve this, the algebraic expressions were used to formulate it. Here the answer of a+b+c is 7.

If $c o s e c to the power of 6 end exponent invisible function application q minus c o t to the power of 6 end exponent invisible function application q equals a c o t to the power of 4 end exponent invisible function application q plus b c o t to the power of 2 end exponent invisible function application q plus c$ then a+b+c=

So here we have used the trigonometric functions and trigonometric formulas to solve this, the algebraic expressions were used to formulate it. Here the answer of a+b+c is 7.

If a,b,c are the sides of the triangle ABC such that $open parentheses 1 plus fraction numerator b minus c over denominator a end fraction close parentheses to the power of a end exponent open parentheses 1 plus fraction numerator c minus a over denominator b end fraction close parentheses to the power of b end exponent open parentheses 1 plus fraction numerator a minus b over denominator c end fraction close parentheses to the power of c end exponent greater or equal than 1$
Then the triangle $A B C$ must be

If $s i n invisible function application y equals x s i n invisible function application left parenthesis a plus y right parenthesis$ and $fraction numerator d y over denominator d x end fraction equals fraction numerator A over denominator 1 plus x to the power of 2 end exponent minus 2 x c o s invisible function application a end fraction$ then the value of $blank to the power of ´ end exponent A$ ' is

Therefore the correct option is choice 3

If $s i n invisible function application y equals x s i n invisible function application left parenthesis a plus y right parenthesis$ and $fraction numerator d y over denominator d x end fraction equals fraction numerator A over denominator 1 plus x to the power of 2 end exponent minus 2 x c o s invisible function application a end fraction$ then the value of $blank to the power of ´ end exponent A$ ' is

Therefore the correct option is choice 3

The maximum value of $open parentheses c o s invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o s invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis horizontal ellipsis. open parentheses c o s invisible function application alpha subscript n end subscript close parentheses$ under the restriction
$0 less or equal than alpha subscript 1 end subscript comma alpha subscript 2 end subscript horizontal ellipsis horizontal ellipsis horizontal ellipsis subscript n end subscript less or equal than fraction numerator pi over denominator 2 end fraction & c o t invisible function application alpha subscript 1 end subscript times c o t invisible function application alpha subscript 2 end subscript horizontal ellipsis horizontal ellipsis c o t invisible function application alpha subscript n end subscript equals 1$ is

The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2, and 6. Then the other two are