All Answers 358514 Answers

mgh = energy of hammer delivered to pile = 150*9.8*5 = 7350 j

Amount delivered to the pile  400kg x 9.8 x .0075=400*9.8*.0075 = 29.4 j

The difference is delivered to the ground (vibrations, ground movement, heat of friction etc) and is the resistance (?)

7350-29.4 = 7320.6 j

(a little unsure)

That is true...I used a slide rule back then !   Really....

Thanks! never used one of these before. 30 years ago we relyied on tables.

First....look in the Lower Left corner to make sure the calculator is set to ' deg'

then sin = opp/hyp = 4.5/11.75 = .382978

The press the ' 2nd ' key on the calculator then the ' asin ' key ....which is arc sin  which is sin inverse...

= 22.52 degrees

This answer has  10,243  decimal digits  !!!  [ http://www.wolframalpha.com/input/?i=1024%5E1024*(-49)%5E7035*(-625)%5E2561%2F(-343%5E4690)

In other words.......I wouldn't worry about it.....!!!!

-11  - 2/3   =   -11 2/3 = -35/3 = -11.6666667

I think we will not know until the Universe ends......   and then there will be nobody around to know...    

  sorry......well, kind of sorry !

Following BEDMAS, 6*7 =42    6/2=3      Therefor 42-3=39

:)

No...the E means the exponent of 10   so it would be 3.76375 x 10^22      NOT 10^25

P  = 300 (2)^(D/11)

D          P

0          300

11        600

22       1200

33       2400

44       4800

etc.

http://www.wolframalpha.com/input/?i=300+(2)%5E(D%2F11)++for+D+%3D+0,+11,+22,+33,44

Tan x = .0563

ARCtan(x) = x = Arctan(.0563) =3.222 degrees =.0562 R

We can't factor this  in a "normal"  way  because it has no real roots

The facorizaion  is   [ x -  (- 1 -  i √  7  ) / 2 ]  [  x - (- 1 +  i √  7  ) / 2 ]

Expanding the above we get   →   x^2 + x + 2  ....exactly what we started wih  ....!!!!

Thank you for the answer CPhill, but I would like to know how to factor the equation using the FOIL method.

[4/3](3.14)(70^3)  ≈  1,436,026.67  units^3

sin 60 cos 45 + cos 30 sin 30

[√3/2] [√2 / 2]  + [√3/2] [1/2]  =

( √3  [ √2 +  1 ] )  / 4  ≈ 1.04539

Pls help i don't understand this..

A crate with a mass of 500kg is lifted from rest with a uniform acceleration by a crane such that after 5 seconds it has a velocity of 8 m/s. Calculate the tension (force) in the cable.

\(a=\frac{v}{t}=\frac{8m/s}{5s}=1.6m/s^2\)

\(F=m\times( a+g)=500kg\times (1.6+9.81)m/s^2\times\frac{Ns^2}{kgm}\)

\(F=5705N\)

\(The \ pulling \ force \ in \ the \ cable \ is \ 5705 N.\)

  !

x^2 + x + 2.= 0    use the quad formula    a = 1, b=  1   c = 2

So we have

x  =  (- 1 ± √ [ (1)^2  - 4(1) (2)  ]  ) / 2  = 

(- 1 ± √ [ (1)  - 8  ]  ) / 2  =

(- 1 ±  i √  7  ) / 2

Easy that would be (16/21)

Hi, gotta ask a question to get an answer. 

x+15+18=90 

x+33=90 

90-33=x 

Good question, think of it like this, velocity is a change in position over time, measured in meters per second or m/s. 

Acceleration, is a change in velocity over time, which would be meters per seond per second, or m/s² 

Well, 3x-3x=0x 

0x+2y=2y 

Well, 1/5 of 100 equals 20, so think of it like 1/5 of 1 which would equal 0.2. Now you need to find 4/5, so add 0.2 together 4 times and you will find your answer. 

Hi :)

You do realize that this is an international forum?

You are certainly welcome to ask your question but a little more info (like your country/state/province ) would be good for those of us who are spectating.

I, for one, have no idea what you are talking about ://