parallelogram law is applicable only for 2/3/4/any no. of forcestriangle law is applicable only for 2/3/4/any no. of forces.polygon law is applicable only for 2/3/4/any no.of forcesthe resultant force of two coplanar collinear forces will be maximum when they act a. same direction, b. different direction, c.at a right angle, d. at an angle​

Answer:

√64 = 8

√256 = 16

Step-by-step explanation:

Answer:

a=25 d=a2-a1

Step-by-step explanation:

do as in attachment

we can do it in another method that is an=a+(n-1)d

substitute 15,25 in 'n'

Given: The A.P. series is 6, 9, 12, 15, ……​

            First-term(a)= 6  and The common difference(d)=(9-6)=3

To find The value of a₂₅-a₁₅

Solution: The given series of numbers are considered to be in Arithmetic Progressions or in A.P. because the difference between the numbers is the same that is (9-6)=3, (12-9)=3, (15-12)=3, so we can see the common difference is 3 for this series. The nth term of an A.P. is given as

aₙ=a+(n-1)×d.

From above a=6, d=3

∴ a₂₅=a+(25-1)×d  [∵here n=25]

⇒ a₂₅ = 6+(24×3)

⇒ a₂₅ = 6+72

⇒ a₂₅ = 78

Now, a₁₅=a+(15-1)×d  [∵here n=15]

⇒ a₁₅=6+(14×3)

⇒ a₁₅=6+42

⇒ a₁₅=48

∴ a₂₅-a₁₅

=78-48

=30

Hence the value of a₂₅-a₁₅ is 30.

Answer:

The Sower' is a poem about a farmer who is sowing seeds. He has seen many bad harvests in his life. However he is so optimistic of having a good harvest every time he tosses grain. ... The poem has a very optimistic tone and it shows the worthy results of dedication and hard- work.

Answer:

The Dimension of Work Done

= Unit of work done is Joule. Therefore, the dimension formula of work is represented as [M1 L2 T-2].

Answer:

Dimension of work is M¹L⁻²T⁻²

Explanation:

In context to the question;

we have to find the dimensional formula of work,

DERIVATION OF DIMENSIONAL FORMULA OF WORK;

We know that;

work (W) = force x displacement ..(1)

since; force = mass x acceleration = [M¹ x (L¹T⁻²)]

so the dimensional formula of force is  [M¹L¹T⁻²]

similarly;

the dimensional formula of force is only in meters so [M⁰L¹T⁰]

now by putting the both the formula in equation (1)

we get,

Work =  [M¹L¹T⁻²] x [M⁰L¹T⁰]

Work = M¹L⁻²T⁻²

Hence, Dimension of work is M¹L⁻²T⁻²