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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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To find the volume of a box of Tissue (also known as a rectangular prism) just multiply the area of the base (length times width) by the height. The formula is V = l• w• h.

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This is the simplest type of fold where the rock bends like this:
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During a clinic visit a4 year old girl suddenly yells don't sit on erin the parents wispers that erin is an imaginary friend wha
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3 years ago
roblem 10: In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following p
miskamm [114]

Answer:

W= -2.5 (p₁*0.0012) joules

Explanation:

Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then

In adiabatic compression, work done by mixture during compression is

W= \int\limits^f_i {p} \, dV  where f= final volume and i =initial volume, p=pressure

p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ

W= \int\limits^f_i {K/V^} \, dV

W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)

W=1/1-γ (p₁Vf-p₀Vi)

W= 1/1-1.40 (p₁*6/5 -p₀*0)  

W= -2.5 (p₁*6/5*0.001)   changing liters to m³

W= -2.5 (p₁*0.0012) joules

3 0
3 years ago
A brick falls to the ground. if the time for the collision of the brick and the ground is increased by a factor of 4, the force
melomori [17]

Answer:

By a factor of 1/4.

Explanation:

The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,

\\\begin{aligned}\\\small F &=\small \frac{\Delta (mV)}{\Delta T}\end{aligned}

in which \small \Delta (mV) , \small \Delta t represent the change in momentum and the time taken for that change.

If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by \small F_1, the following manipulation confirms the answer to this question.

\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}

Here \small F is the force that was applied to the object previously.

#SPJ4

4 0
2 years ago
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