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Katena32 [7]
4 years ago
5

Which organism did we virtually dissect today? Question 1 options: frog worm starfish pig

Physics
1 answer:
Harman [31]4 years ago
6 0
We dissect frogs
Worms are to small for a scalpel and nothing to look at rlly
Starfish are illegal to dissect
Pigs are to big for a classroom and they can’t be cryogenics frozen to be dissected
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A 25 kg circular disk has a diameter of 2.5 feet and a thickness of 2.5 cm. Find the density of the disk in kg/m3. Next, find th
Gre4nikov [31]

Answer:

Assume that \rm g= 9.81\; N\cdot kg^{-1}; \rho(\text{Water}) = \rm 1000\;kg\cdot m^{-3}.

Density of the disk: approximately \rm 2.19\times 10^{3}\; kg\cdot m^{-3}.

Weight of the disk: approximately \rm 245\;N.

Buoyant force on the disk if it is submerged under water: approximately \rm 112\; N.

The disk will sink when placed in water.

Explanation:

Convert the dimensions of this disk to SI units:

  • Diameter: d = \rm 25\; inches = (25\times 0.3048)\; m = 0.762\;m.
  • Thickness h = \rm 2.5\; cm = (2.5\times 0.01)\; m = 0.025\;m.

The radius of a circle is 1/2 its diameter:

\displaystyle r = \rm \frac{1}{2}\times 0.762\;m = 0.381\; m.

Volume of this disk:

V(\text{disk}) = \pi\cdot r^{2}\cdot h = \pi\times 0.381^{2}\times 0.025 \approx 0.0114009\; m^{3}.

Density of this disk:

\displaystyle \rho(\text{disk}) = \frac{m}{V} = \rm \frac{25\; kg}{0.0114009\; m^{3}} = 2.19\times 10^{3}\;kg\cdot m^{-3}.

\rho(\text{disk}) >\rho(\text{water}) indicates that the disk will sink when placed in water.

Weight of the object:

W(\text{disk}) = m\cdot g = \rm 25\times 9.81 = 245.25\; N.

The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately \rm 0.0114009\; m^{3} of water. The buoyant force on the disk will be:

\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1\times 10^{3}\; kg\cdot m^{-3}\times 0.0114009\; m^{3}\times 9.81\; N\cdot kg^{-1}\\ &\approx \rm 112\; N\end{aligned}.

The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.

3 0
3 years ago
Two parallel plates are a distance apart with a potential difference between them. A point charge moves from the negatively char
IgorC [24]

Answer:

K' = 1200 J

Explanation:

To find the kinetic energy you first take into account the formula for the kinetic energy of the charge:

K=\frac{1}{2}mv^2 = 800J   (1)

m: mass of the charge

v: final speed of the charge when it reaches the positively charged plate.

Furthermore, you have that the acceleration of the charge is obtained by using the second Newton law:

F=ma=qE\\\\a=\frac{qE}{m} (2)

a: acceleration

E: electric field

q: charge

The electric field between two parallel plates is V/d, being V the potential difference and d the separation between plates. You replace E in (2) and obtain:

a=\frac{qV}{md}

Next, you take into account the following formula for the calculation of the final speed of the charge:

v^2=v_o^2+2ad\\\\v_o=0m/s\\\\v=\sqrt{\frac{2qVd}{md}}=\sqrt{\frac{2qV}{m}}

Next, you replace this value of v in (1):

K=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{2qV}{m})=qV = 880J   (3)

If the distance between plates is tripled, and the potential difference is halved, you have for the new final speed:

v'^2=v'_o^2+2a(3d)\\\\v_o=0m/s\\\\v'=\sqrt{6ad}=\sqrt{6(\frac{q}{md})\frac{V}{2}d}=\sqrt{\frac{3qV}{m}}

And the kinetic energy becomes:

K'=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{3qV}{m})=\frac{3}{2}qV    (4)

You calculate the ratio between both kinetic energies K and K', that is, you divide equations (3) and (4), in order to find the new kinetic energy:

K=qV=800J\\\\K'=\frac{3}{2}qV\\\\\frac{K}{K'}=\frac{qV}{3/2\ qV}=\frac{2}{3}\\\\K'=\frac{3}{2}K=\frac{3}{2}(800J)=1200J

hence, the kinetic energy of the charge incresases to 1200J

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Answer:

Most likely, it will be harder to get strong magnets to change phase because they have more density.

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satellites and weather fronts

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