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Amanda [17]
3 years ago
14

Listed in the Item Bank are some important labels for sections of the image below. To find out more information about labels, so

me have more details available when you click on them. Drag and drop each label to the corresponding area it identifies in the image.
ITEM BANK: Move to Bottom
BoilingGas/VaporLiquidMeltingSolidTemperatureTime
0
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
5 0

Answer:reactant,active site,enzyme below,substrate,products

Explanation:

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In this graph, calculate the speed of<br> segment A in m/s?
andrew11 [14]

Answer:

The answer is Speed=2m/s

Explanation:

S=D/T

S=10m/5s

S=2m/s

5 0
3 years ago
In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin
alexandr402 [8]

Answer:

a) v=0.999124c

b) E=7.566*10^{22}

c) E_a=760 times\ larger

Explanation:

From the question we are told that

Distance to Betelgeuse d_b=430ly

Mass of Rocket M_r=20000

Total Time in years traveled T_d=36years

Total energy used by the United States in the year 2000 E_{2000}=1.0*10^20

Generally the equation of speed of rocket v mathematically given by

v=\frac{2d}{\triangle t}

v=860ly/ \triangle t

where

\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\sqrt{(860)^2+(36)^2}

\triangle t=860.7532

Therefore

v=\frac{860ly}{ 860.7532}

v=0.999124c

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2

E=7.566*10^{22}

c)Generally the equation of the energy E_a required to accelerate the rocket mathematically given by

E_a=\frac{E}{E_{2000}}

E_a=\frac{7.566*10^{22}}{1.0*10^{20}}

E_a=760 times\ larger

8 0
3 years ago
A nautical mile is _____.
Leviafan [203]
Grater than a statute mile 
5 0
3 years ago
How can an experiment support or fail to support a hypothesis?
Alisiya [41]

a hypothesis is given to explain a phenomena which has not been explained earlier. it can proved  by an experiment if that experiment gets the other results regarding that particular phenomena in agreement with that being predicted by the hypothesis and if  result doesnt match the prediction then the hypothesis is again  totally discarded at times.
6 0
3 years ago
Read 2 more answers
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
iogann1982 [59]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

6 0
3 years ago
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