HELP ASAP!!

What is the component of d1 along the direction of d2?

aniked [119]

I don't know lol hahahahhahahahahahahahahahahahahaha

8 0

3 years ago

When mantle rocks near the radioactive core are heated, they become less dense than the cooler, upper mantle rocks. These warmer

avanturin [10]

C.convection cells is the right answer

4 0

3 years ago

Read 2 more answers

Why steel ships becomes megnetized as it is constructed

Ksivusya [100]

Bcjcghfggcvhyzzfu6ytehnfnffnnxbzbxbdnsnddbbf

3 0

4 years ago

Water is circulating through a closed system of pipes in a two floor apartment. On the first floor, the water has a gauge pressu

anastassius [24]

Answer:

The value of gauge pressure at outlet = -38557.224 pascal

Explanation:

Apply Bernoulli' s Equation

$\frac{P_{1}}{9810}$ + $\frac{V_{1} ^{2}}{19.62}$ + $h_{1}$ = $\frac{P_{2}}{9810}$ + $\frac{V_{2} ^{2}}{19.62}$ + $h_{2}$ --------------(1)

Where

$P_{1}$ = Gauge pressure at inlet = 3.70105 pascal

$V_{1}$ = velocity at inlet = 2.4 $\frac{m}{sec}$

$P_{2}$ = Gauge pressure at outlet = we have to calculate

$V_{2}$ = velocity at outlet = 3.5 $\frac{m}{sec}$

$h_{2} - h_{1}$ = 3.6 m

Put all the values in equation (1) we get,

⇒ $\frac{3.70105}{9810}$ + $\frac{2.4 ^{2}}{19.62}$ = $\frac{P_{2}}{9810}$ + $\frac{3.5 ^{2}}{19.62}$ + 3.6

⇒ 0.294 = $\frac{P_{2}}{9810}$ + 0.6244 + 3.6

⇒ $\frac{P_{2}}{9810}$ = 0.294 - 0.6244 - 3.6

⇒ $\frac{P_{2}}{9810}$ = - 3.9304

⇒ $P_{2}$ = - 38557.224 pascal

This is the value of gauge pressure at outlet.

3 0

3 years ago

A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the

Mademuasel [1]

Answer:

a)

$u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)$

b)

m = 48lb

c)

b = 144.76lb

Explanation:

The general equation of a damping oscillate motion is given by:

$u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)$ (1)

uo: initial position

m: mass of the block

b: damping coefficient

w: angular frequency

α: initial phase

a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:

$|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s$

Then, you obtain by replacing in (1):

6in = 0.499 ft

$u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)$

b.

mass, m = 48lb

c.

b = 144.76 lb/s

8 0

3 years ago