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Leviafan [203]
2 years ago
6

Is 1 milliliter/second a small or large flow? Why?

Physics
2 answers:
ozzi2 years ago
7 0

Answer:

1 ml/second is a small flow

Explanation:

NeTakaya2 years ago
7 0

It dependependsdependsdepends


Are you a blue whale or a brine shrimp ?

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nikklg [1K]

Answer:

<u>0.04 °C⁻¹</u>

Explanation:

First, we need to calculate linear expansivity, then after finding that value, we can move on to finding the area expansivity.

<u />

=============================================================

Finding Linear Expansivity :

⇒ α = Final length - Original length / (Original length × ΔT)

⇒ α = 9 - 4 / (4 × 70 - 20)

⇒ α = 5 / 5 × 50

⇒ α = <u>0.02</u>

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Finding Area Expansivity :

⇒ Area Expansivity = 2 × Linear Expansivity

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⇒ β = <u>0.04 °C⁻¹</u>

3 0
1 year ago
Read 2 more answers
A ball rolls down a ramp. A small amount of friction heats up the ball. What energy transformations are taking place?
Natasha2012 [34]

Answer:kinetic energy converted to heat energy

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As the ball rolls down kinetic energy is converted to heat energy

6 0
3 years ago
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A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0
3 years ago
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