difference between संयोग चिन्ह and अपसारण चिन्ह ​

Answer:

The thickness of base plate has been determined by equating moment due to upward pressure at critical section of base plate to the flexural (bending) strength of base plate.

Answer:

Horizontal range R = u^2sin2θ / g

= 400× sin60° = 20root3

time of flight = 2usinθ / g

2×20×1/2 / 10 = 10s

Answer:

The number 0.000125 in scientific notation is written as 1.25 × 10-4.

Given,

The number

0.000125

To find,

Standard form of the given number

Solution,

We know that,

To convert any decimal number into it's standard form we must shift the decimal to the right or left until we get at least one non zero number at the left hand side of the decimal.

So in above number we should shift decimal by 4 in order to get 1 at RHS

⇒[tex]00001.25[/tex]

Now to balance the number we should put the number of times we shifted the decimal to the power of 10

⇒[tex]1.25[/tex]×[tex]10^{-4}[/tex]

Here power is negative because we shifted to the right.

Therefore [tex]0.000125=1.25[/tex]×[tex]10^{-4}[/tex]

Answer:

The average thrust of the sand on the body of mass 2kg will be 1980N.

Explanation:

The values given in the question are,

Mass(m)=2Kg ,

Height(h)=60m,

Depth into the sand(s)=0.6m

Just before hitting the ground

[tex]v^{2}[/tex]=[tex]u^{2}[/tex]+2gh=0+2 × 9.8 × 60    (∵[tex]u^{2}[/tex]=0 as it is dropped from rest)

[tex]v^{2}[/tex]=1176m/s

When the stone penetrates the sound

[tex]v^{2}[/tex]=[tex]u^{2}[/tex]+2as

0=[tex]u^{2}[/tex]+2 × a × 0.6         (∵[tex]v^{2}[/tex]=0 as it is brought to rest )

-1176=1.2a

a=-1176[tex]/[/tex]1.2=-980m[tex]s^{-2}[/tex]

Here (-) indicates the direction of acceleration.

Average thrust on the sand(F)=m(g+a)      (∵gravitational force and frictional force are acting in opposite directions)

F=2(9.8+980)=2 × 989.8=1979.6N≈1980N

Hence, the average thrust of the sand on 2kg mass will be 1980N.