Help! xoxo thank you

monitta

Answer:

SI unit is an international system of measurements that are used universally in technical and scientific research to avoid the confusion with the units. Having a standard unit system is important because it helps the entire world to understand the measurements in one set of unit system.

4 0

3 years ago

What occurs to the arrangement of water molecules as water melts

Nikolay [14]

The arrangements of the water molecules they begin to spread apart. As the water(ice) melts the molecules develop fewer compact and start juddering harder until the ice has changed state into liquid.

7 0

3 years ago

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WILL GIVE BRAINLIEST PLZ HELP

MatroZZZ [7]

Answer:

Question 1: protons and electrons

Question 2: atomic number

Explanation: trust me

4 0

3 years ago

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Assuming that the cost of 16-gauge steel sheet metal is $25.00 per square meter, what is the ideal radius and height of a cylind

Anton [14]

Answer: Radius = 32.7 cm and height = 65.5 cm

Explanation:

An oil drum must have a volume of 218 liters or 218*1000cm^3 = 218,000 cm^3

The price is per square meter, so if we reduce the surface of the oil drum, we will pay less:

So we want to play with the measures of the oil drum in such a way that the surface is minimized.

Now, first, the volume of the oil drum (a cylinder) is:

V = pi*r^2*h

where pi = 3.14, r is the radius and h is the height.

and the surface is:

S = 2*pi*r^2 + 2*pi*r*h

we know that:

pi*r^2*h = 218,000 cm^3

r^2*h = 218,000cm^3/3.14 = 70,096.5 cm^3

now we can write

h = 70,096.5 cm^3/r^2

now we can replace it in the surface equation:

S = 2*pi*r^2 + 2*pi*r*h

= 6.28*(r^2 + 70,096.5 cm^3/r)

So we want to minimize this, we can derivate it and find the zero:

S' = 6.28(2*r - 70,096.5 cm^3/r^2) = 0

2r = 70,096.5 cm^3/r^2

r^3 = (70,096.5 cm^3)/2

r = ∛( (70,096.5 cm^3)/2 ) = 32.7cm

And then the height is:

h = 70,096.5 cm^3/r^2 = 70,096.5 cm^3/(32.7cm)^2 = 65.5cm

4 0

3 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

3 years ago