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Aloiza [94]
3 years ago
13

Based on your book reading, what is the difference between ecosystem serveries and natural resources? Explain your response.

Physics
1 answer:
chubhunter [2.5K]3 years ago
8 0

Answer:

Answer:This organism may be identified by its color, the spines on its back, the antennae, and therefore the long, thin body. There are many other characteristics that might even be wont to identify this organism.

Explanation:

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Neutrons are also known as____
PilotLPTM [1.2K]
C.alpha particles....
4 0
3 years ago
Read 2 more answers
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
A rocket is launched at an angle of 39◦ above the horizontal with an initial speed of 90 m/s. It moves for 7 s along its initial
vagabundo [1.1K]

Answer:

Y=1370.23m

Explanation:

The motion have two moments the first one the time the initial velocity is accelerating then when the engines proceeds to move as a projectile

a=19 \frac{m}{s^{2} } \\voy=vo*sin(\alpha )\\voy=90*sin(39 )\\y_{o}=0m\\y_{f}=y_{o}+v_{oy}*t+\frac{1}{2}*a*t^{2}\\y_{f}=90*sin(39)*7s+\frac{1}{2}*19\frac{m}{s^{2} }*(7)^{2}\\y_{f}=861.97m

Now the motion the rocket moves as a projectile so:

v_{fy}=v_{iy}+a*t\\v_{fy}=90+9.8*7\\v_{fy}=158.6 sin(39)

Now the final velocity is the initial in the second one

v_{fy}^{2}=v_{fi}^{2}+2*a*yf \\\\a=g\\

The maximum altitude Vf=0

0=v_{fi}^{2}+2*a*yf \\\\yf=\frac{(158.6 sin(39))^{2} }{2*9.8\frac{m}{s^{2} } } \\yf=508.26m

So total altitude is both altitude of the motion so:

Y=508.2m+861.97m\\Y=1370.23m

6 0
3 years ago
The spaceship Intergalactic landed on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist
max2010maxim [7]

Answer:

<em>18808.7 m/s^2</em>

Explanation:

Given

Length of the pendulum L = 1.44 m

Number of complete cycles of oscillation n = 1.10 x 10^2

total time of oscillation t = 2.00 x 10^2 s

The period of the T = n/t

T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s

The period of a pendulum is gotten as

T = 2\pi \sqrt{\frac{L}{g} }

where g is the acceleration due to gravity

substituting values, we have

0.55 = 2\pi \sqrt{\frac{1.44}{g} }

0.0875 = \sqrt{\frac{1.44}{g} }

squaring both sides of the equation, we have

7.656 x 10^-3 = 144/g

g = 144/(7.656 x 10^-3) = <em>18808.7 m/s^2</em>

3 0
3 years ago
A 755 N diver drops from a board 10.0 m above the water's surface. Find
MrRa [10]

Answer:

F = 755 N and h = 10.0 m and a = 9.8 m/s 2 and m=? and v = ?

5 0
3 years ago
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