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Triss [41]
3 years ago
6

9. Kokio dydžo Archimedo jėga veikia 0,5 m² tūrio medinį rastą vandenyje?​

Physics
1 answer:
ICE Princess25 [194]3 years ago
6 0

Answer:

i dont knowwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

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What sound signal tells a river lock attendant that you wish to go through the lock?
Hoochie [10]
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6 0
3 years ago
A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle
Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of <em>θ</em> = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height <em>y</em> at time <em>t</em> would be

<em>y</em> = (40 m/s) <em>t</em> - 1/2 <em>g t</em> ²

Set <em>y</em> = 160 m, and you'll find that there is no (real) solution for<em> t</em>, so the ball never attains the given maximum height.

From another perspective: recall that

<em>v </em>² - <em>v</em>₀² = 2<em>a </em>∆<em>y</em>

where

• <em>v</em>₀ = initial velocity

• <em>v</em> = final velocity

• <em>a</em> = acceleration

• ∆<em>y</em> = displacement

At its maximum height, the ball has zero vertical velocity, and ∆<em>y</em> = maximum height = 160 m. The ball is in free fall once it's launched, so <em>a</em> = -<em>g</em>.

So we have

0² - (40 m/s)² = -2<em>g </em>(160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0
3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

5 0
4 years ago
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