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3 years ago

Which statement accurately describes malleability? HELP ASAP!!!

Physics

2 answers:

vovangra [49]3 years ago

8 0

Answer:

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Explanation:

Malleability is when a certain substance can be easily pressed, bent, or pulled out of shape permanently. An example of a malleable substance is gold, as it can be permanently pressed into a sheet and unable to return to its original state.

inysia [295]3 years ago

6 0

Answer:

Malleability is a substance's ability to deform under pressure (compressive stress). If malleable, a material may be flattened into thin sheets by hammering or rolling. Malleable materials can be flattened into metal leaf. ... Many metals with high malleability also have high ductility. that's the answer ok

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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

A 1,200 kg subcompact car accelerates at a rate of 3.0m/s

ludmilkaskok [199]

Missing question (found on internet):
"what is the force of the car?"

Solution:
According to Newton's second law, the force of the car is equal to the product between its mass and its acceleration:
$F=ma$
For the car in the problem,
$m=1200 kg$
$a=3.0 m/s^2$
So the force that accelerates the car is
$F=(1200 kg)(3.0 m/s^2)=3600 N$

6 0

3 years ago

Pilots often take advantage of the ____,which are highly-speed winds between 7km and 16 km above earths surface.

lukranit [14]

I believe it is called or referred to as the "Jet Stream". During World War II, allied pilots encountered high speed winds in the upper air. They named those winds after the fastest planes they came up against: fighters equipped with jet engines! Jet stream winds in winter time can reach up to 300 MPH as well!

6 0

3 years ago

Solve for n: 2 + 4tan(1 - n) = 1

Anettt [7]

Answer:

SOLVE FOR N:

n=πn1+arcsin(1717)+1

n1∈Z

----------------------------------------

Trigonometry

2+4 tan (1-9)=1

I hope this helps ( ﾟдﾟ)つ Bye

8 0

3 years ago

Standing waves can be established within a cylindrical volume. However, it is not known if the volume is closed at both ends, op

Evgen [1.6K]

The frequencies are missing in the question. The three successive resonance frequencies within the volume are $f_1, f_2 \text{ and}\ f_3$ .

Solution :

Let the volume be : v

The frequency for one end open and one end closed is given by :

So, $f_n = \frac{(2n-1)v}{4L}$

Therefore,

$f_1 = \frac{v}{4L}$ , $f_2 = \frac{3v}{4L}$ , $f_3 = \frac{5v}{4L}$

So, $f_2 - 3f_1$ and $f_3=\frac{5}{3}f_2$

Therefore, the ratio of $\frac{f_3}{f_2}=\frac{5}{3}$ which is not a whole number.

Now the frequency for open volume and closed at both end

$f_n=\frac{nv}{4L}$

So, $f_1=\frac{v}{4L}$ , $f_2 = 2f_1$ , $f_2=3f_1$

From above formulae we can see that ratio of is not a whole number that is a identification for the frequency of the volume at one end open and one end closed.

Also ratio of the consecutive frequency of the volume at open from both side and closed from both side is always a whole number.

4 0

3 years ago