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Minchanka [31]
2 years ago
15

Which biome has the most extreme conditions

Physics
1 answer:
Lelu [443]2 years ago
3 0
Well this all depends on the region you would like to know about. One biome would be The Tundra. This biome is a very bitter cold. Some times the temperature can drop to -45f! So your answer more than likely would be Tundra.

Have a wonderful day user!
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A minivan is tested for acceleration and braking. In the street-start acceleration test, the elapsed time is 8.6 s for a velocit
Varvara68 [4.7K]

The question is incomplete. Here is the complete question:

A minivan is tested for acceleration and braking. In the street-start acceleration test, the elapsed time is 8.6 s for a velocity increase from 10 km/h to 100 km/h. In the braking test, the distance traveled is 44 m during braking to a stop from 100 km/h. Assume constant values of acceleration and deceleration. Determine  

(a) the acceleration during the street-start test,  

(b) the deceleration during the braking test.

Answer:

(a) 37500 km/h²

(b) 113636.36 km/h²

Explanation:

part (a)

Because it is given that we can assume constant acceleration therefore we can use the following equation of motion:

<em>v = u + (a)(t) </em>

where <em>v </em>is final velocity, <em>u </em>is initial velocity, <em>a </em>is acceleration and <em>t </em>is time change

Given in the question:

v = 100km/h

u = 10 km/h

t = 8.6 sec (changing to hours)

t = 0.0024 hours (round off to 4 decimal places)

100 = 10 + (a x 0.0024)

Rearranging the equation to find value of a

a = (100 – 10) / 0.0024

a = 37500 km/h² (Answer)

part (b)

Now we can use the following equation to find deceleration

<em>2(a)(s) = v² – u²</em>

Where a is acceleration, s is distance travelled, v is final velocity and u is initial velocity

Given in the question

s = 44 m

changing to km

s = 0.044 km

v = 0 km/h (because it stops)

u = 100 km/h

2(a)(0.044) = (0)² – (100)²

0.088(a) = 0 - 10000  

a = - 10000/0.088

a = - 113636.36 km/h2  

The negative sign in the answer shows that it is deceleration

Therefore deceleration = 113636.36 km/h² (Answer)

6 0
3 years ago
When NASA's Skylab reentered the Earth's atmosphere on July 11, 1979, it broke into a myriad of pieces. One of the largest fragm
Burka [1]

Answer:

Kinetic Enerrgy = 12744000 J or 12.744 MJ

Explanation:

Given Data:

Mass of fragment = m =1770 kg    ;

Estimated speed = v = 120 m/sec ;

Solution:

As we know that

Kinetic energy = K.E = (1/2)mv²

By putting the values, we get

              K.E = 0.5*1770*120²

              K.E = 12744000 J

                            or

              K.E = 12.744 MJ

So, the kinetic energy of lead-lined vault when it landed was 12.744MJ approximately.

6 0
3 years ago
What does it mean when there is a curved line going upwards on a graph?<br><br>science 8th grade :)
yulyashka [42]

Answer: it is the asymptote

Explanation: a line that continually approaches a given curve but does not meet it at any finite distance.

8 0
3 years ago
To calculate the ideal mechanical advantage of a lever divide the input arm by the
Ivanshal [37]
To calculate the ideal mechanical advantage of a lever divide the input arm by the output arm. 
Mechanical advantage is the amount by which a machine can multiply an input force, calculated by dividing output Force in newtons by input force in newtons, while the ideal mechanical advantage is the mechanical advantage of a machine that has no friction, calculated by dividing the input distance by the output distance. 
6 0
3 years ago
The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f
tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

4 0
2 years ago
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