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Taya2010 [7]
3 years ago
12

What does a biomass pyramid show?

Physics
2 answers:
Ilia_Sergeevich [38]3 years ago
7 0

Answer:

The distribution of matter in an ecosystem

Explanation:

I took a quiz with this question, and this was the correct option

lora16 [44]3 years ago
5 0

Answer:

I SAVE D1K BY GIVING IT CPR

Explanation:

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Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock
pashok25 [27]

Answer:

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

c) H=294300\ m

d) t_T=544.95\ s

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, a=2g=2\times 9.81=19.62\ m.s^{-2}

time for which the rocket accelerates, t_a=100\ s

<u>For the course of upward acceleration:</u>

using eq. of motion,

s_a=ut+\frac{1}{2}at_a^2

where:

u= initial velocity of the rocket at the launch =0

s_a= height the rocket travels just before its fuel finishes off

so,

s_a=0+\frac{1}{2}\times 19.62\times 100^2

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

v_a=u+at_a

v_a=0+19.62\times 100

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

v^2=v_a^2-2gh

where:

g= acceleration due to gravity

v= final velocity of the rocket at the top height

0^2=1962^2-2\times 9.81\times h

h=196200\ m

c) So the total height at which the rocket gets:

H=h+s

H=196200+98100

H=294300\ m

d)

Time taken by the rocket to reach the top height after the fuel is over:

v=v_a+g.t

0=1962-9.81t

t=200\ s

Now the time taken to fall from the total height:

H=v.t'+\frac{1}{2}\times gt'^2

294300=0+0.5\times 9.81\times t'^2

t'=244.95\ s

Hence the total time taken by the rocket to strike back on the earth:

t_T=t_a+t+t'

t_T=100+200+244.95

t_T=544.95\ s

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0
3 years ago
True or false? Thanks,
vichka [17]

Answer: The answer is true

7 0
1 year ago
Read 2 more answers
A 90kg mountain climber hangs from a nylon rope and streches it ny 25.0cm. if the rope was originally 30.0m long and it's diamet
bixtya [17]

Answer:

Explanation:

E = σ/ε = (F/A) / (ΔL/L)

E = (mg/(πd²/4) / (ΔL/L)

E = (4mg/(πd²) / (ΔL/L)

E = 4Lmg/(πd²ΔL)

E = 4(30.0)(90)(9.8)/(π(0.01²)0.25)

E = 1.35 x 10⁹ Pa  or 1.35 GPa

7 0
3 years ago
True of false efficiency compared the output work to the output force
Lerok [7]
The statement about "<span>efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.</span>
8 0
4 years ago
Russell bradley carried 207 kg of bricks 3.65 m up a ladder. If the amount of work required to perform that task is used to comp
MrRa [10]

Answer:

Explanation:

Work done in carrying bricks

mgh

= 207 x 9.8 x 3.65

-= 7404.4 J

Work done in compressing gas

PΔV

Pressure x change in volume

1.8 x 10⁶ ΔV = 7404.4

ΔV  = 7404.4  / 1.8 x 10⁶m³

= 4113.33 x 10⁻⁶ m³

= 4113.33 cc

8 0
3 years ago
Read 2 more answers
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