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2 years ago

____ would describe an airplane as "Earthen metal infused with Air and propelled by elemental Fire"

Physics

1 answer:

svlad2 [7]2 years ago

8 0

Answer:

Aristotle

Explanation:

Aristotle would describe an airplane as "Earthen metal infused with Air and propelled by elemental Fire".

Their presence spread in to the Renaissance from the Antiquity and Early medieval Ages, and was not consistently displaced until the Inquisition and philosophies such as classical physics.Aristotle's views on biblical theory were influenced by physical science. Many of the zoological findings of Aristotle found in his physiology, such as on the octopus ' hectocotyl (reproductive) head, remained incredible until the 19th century.

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The weight of a 72.0 kg astronaut on the Moon, where g = 1.63 m/s2 is (5 points) Select one: a. 112 N b. 117 N c. 135 N d. 156 N

kipiarov [429]

Answer: The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

Explanation:

Mass of the astronaut on the moon , m= 72 kg

Acceleration due to gravity on moon,g = 1.63 $m/s^2$

According to Newton second law of motion: F = ma

This will changes to = Weight = mass × g

$Weight=72 kg\times 1.63m/s^2=117.36 N$

The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

7 0

2 years ago

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A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu

melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

$|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)$

By taking into account the angles between the vectors r and v in each case we obtain:

v=(2,0)

r=(0,1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

r=(0,-1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

r=(1,1)

angle = 45°

$L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}$

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

r=(-1,-1)

angle = 135°

$L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}$

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0

2 years ago

A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi

Temka [501]

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

Initial speed (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

$h = ut + \frac{1}{2}\times a \times t^2$

$h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2$

h = 313.6 m

Hence, the depth of the well is 313.6 m

4 0

2 years ago

For each property listed, identify the type of element it describes. Very good electrical conductivity: Amphoteric, able to form

kow [346]

The elements that is very good in electrical conductivity are gold and copper: elements that is amphoteric are copper, zinc, tin, lead, aluminum and beryllium: elements that is gaseous at room temperature are hydrogen, nitrogen, oxygen, fluorine and chlorine: elements that is solid at room temperature are all metal except mercury and perhaps some unseen radioactive elements. Lastly, elements that is brittle are hydrogen, carbon, nitrogen, oxygen, phosphorus, sulfur and selenium

4 0

2 years ago

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In a "worst-case" design scenario, a 2000 kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushion

Whitepunk [10]

Answer:

A. V =3.65m/s

B. a = 4m/s^2

Explanation:

Determine force of gravity (f) on the elevator.

f = mg

(m = 2000kg, g = 9.8m/s

2000kg × 9.8m/s^2= 19600N

Given,

Force of opposing friction clampforce of gravity = 17000N

the Net force on the elevator

= force of gravity - Force of opposing friction clamp

=19600 - 17000

= 2600 N

Lets determine the kinetic energy of the elevator at the point of contact with the spring

K.E = 1/2 m v^2

(m = 2000kg, v = 4.00m/s)

= (1/2) × 2000kg × (4m/s)^2

= 16000J

kinetic energy and energy gain will be absorbed by the spring across the next 2m

Therefore,

E = K.E + P.E

K.E = 16000J,

P.E of spring = net force absorbed × distance at compression

net force absorbed = 2600N and distance at compression = 2.0m)

P.E = 5200J

E = 16000J + 5200J

E = 21200 J

Note, spring constant wasn't given

Lets determine it's value

Using,

E = (1/2) × k × (x)^2

Where:

E = energy = 21200J, K = ?, X = 2m

21200J=(1/2) × k × (2m)^2

21200J × 2 =(4m)k

K = 42400J/4m

K = 10600 N/m

Therefore,

acceleration at 1m compression = ?

Using F = K × X

(F is force provided by the spring = 10600N/m, K = 10600 N/m and X = 1m)

= 10600N/m × 1m = 10600 N ( upward)

A. The speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?

Using.

original Kinetic energy + net force on the elevator = final kinetic energy + spring energy

16000N + 2600N = (1/2)mv^2 + (1/2)k x^2

18600 = (1/2)(2000)(v^2) + (1/2)(10600N)(1^2)

18600 = 1000(v^2) + 5300

18600 - 5300 = 1000(v^2)

13300 = 1000(v^2)

V^2 = 13.300

V =3.65m/s

The acceleration of the elevator is 1.00 {\rm m} below point where it first contacts a spring

Spring constant = net force on the elevator + resultant force

(Spring constant = 10600N, net force on the elevator = 2600N, resultant force = ?)

10600N = 2600N + resultant force

resultant force = 10600N - 2600N

=8000N

Therefore

F = ma

a = f/m

(a = ?, f =8000N and m =2000kg)

= 8000 / 2000

a = 4m/s^2

(It's accelerating upward, since acceleration is positive

5 0

2 years ago

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