Answer:
18.03 N
Explanation:
From the fiqure below,
Using parallelogram law of vector
R² = 15²+5²-2×5×15cos(180-60)
R² = 225+25-150cos120°
R² = 250-150(-0.5)
R² = 250+75
R² = 325
R = √325
R = 18.03 N
Hence the resultant force of the object is 18.03 N
I believe the answer is Microwaves (just to clarify, not the oven). They're widely used for communications.<span />
Шада и я тебя понимаю с чего начать с того ни другого человека дал в долг у меня в здоровье
Answer:
Explanation:
Normal length of spring = 28.3 cm
stretched length of spring = 38.2 cm
length of extension = 38.2 - 28.3 = 9.9 cm
= 9.9 x 10⁻² m
force applied to stretch = .55 x 9.8 ( mg )
= 5.39 N
Force constant = force applied / extension
= 5.39 / 9.9 x 10⁻²
= .5444 x 10² N /m
= 54.44 N/m
Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
