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3 years ago

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

Physics

1 answer:

svlad2 [7]3 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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What is the advantage of nuclear plants?

DerKrebs [107]

They emit low leavls of carbon dioxide (CO2) and the emissions of green house gases and therefore the contribution of nuclear power plants to global warming is therefore pretty little

8 0

3 years ago

Read 2 more answers

In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel

insens350 [35]

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t² (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t² (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t² (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

$p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125$

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

5 0

3 years ago

On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1

Ad libitum [116K]

Answer:

D) The worker will accelerate at 2.17 m/s² and the box will accelerate at 1.08 m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m = W/g

Where:

W : Weight (N)

g : acceleration due to gravity (m/s²)

Data

W₁ =1.8 kN : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁ m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

$a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}$

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

$a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}$

a₂= -2.17 m/s² : acceleration of the worker

5 0

3 years ago

Answer ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

ikadub [295]

$\sf \huge \purple{ Question : - }$

Initial velocity of a car is 36 km/h . Find the distance after min, if it goes with acceleration 2 m/s².

$\sf \huge \color {gold}{ AnSwer :- }$

Initial velocity, u = 36 km/h

$\sf u = 36 \times \dfrac{5}{18} {ms}^{ - 1} = 10 {ms}^{ - 1}$

Time, t = 1min

t = 60s

Acceleration, a = 2m/s²

Apply 2nd equation of motion

$\longrightarrow \sf s = ut + \dfrac{1}{2} at ^{2} \\ \\ \longrightarrow \sf s =(10)(60) + \frac{1}{2} (2)(60) {}^{2} \\ \\ \longrightarrow \sf s =600 + 3600 \\ \\ \longrightarrow \sf \blue{ s =4200m} \: \: \large \blue \star$

6 0

3 years ago

In each of the given pairs, choose which element will have the bigger atom. Give reasons for your choice. (a) Mg (atomic number

Dafna1 [17]

Answer:

Mg (atomic number 12)

K (atomic number 19)

Explanation:

The size of an atom is estimated in terms of its atomic radius.

The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance (d) between two nuclei in the solid - state of metals.

Across a period, atomic radii decrease progressively from left to right.
This is due to the progressive increase in the nuclear charge without increase in the number of electronic shells.
Down a group, atomic radii increase progressively due to the successive shells of electrons being added which have been compensated for by the increase in nuclear charge.

Cl is further right of Mg in the third period

K is below Na in the first group

3 0

3 years ago