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2 years ago

A 15-g block of aluminum has a volume of 5.5 cm. what is the density

Physics

1 answer:

Mnenie [13.5K]2 years ago

4 0

Answer: 2.72727272......g/cm3^

Explanation:

D = g/cm^3

soo d = 15/5.5

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A 10-kg piece of aluminum sits at the bottom of a lake, right next to a 10-kg piece of lead, which is much denser than aluminum.

zalisa [80]

Answer:

Aluminium

Explanation:

When a body is immersed in a liquid partly or wholly it experiences an upward force which is called buoyant force.

The amount of buoyant force depends on the volume of body immersed, density of liquid and the value of acceleration due to gravity.

Here, the density of liquid is same in both the cases and g be the same. So, here the amount of buoyant force depends on the volume of body immersed.

As the density of lead is more than the density of aluminium, so the volume of aluminium is more than lead, as volume is equal to mass divided by density. So, the buoyant force acting on the aluminium is more than lead.

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3 years ago

The photeselestric effect is observed when light of a sufficiently high frequency is focused onto a polished metal surface, emit

Helga [31]

Answer:

$3.4\cdot 10^{-19} J$

Explanation:

In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

$1 eV = 1.6 \cdot 10^{-19} J$

In our problem, the work function of cesium is

$E=2.1 eV$

so, we can convert it into Joules by using the following proportion:

$1 eV : 1.6\cdot 10^{-19} J = 2.1 eV : x\\x=\frac{(1.6\cdot 10^{-19} J)(2.1 eV)}{1 eV}=3.4\cdot 10^{-19} J$

8 0

3 years ago

In Linnean taxonomy, the most unique level is _____. species genus kingdom breed

Nookie1986 [14]

The answer is species, I hope this helps!

6 0

3 years ago

Read 2 more answers

What system will break the chewed food down to a form your cell can use

horsena [70]

If I'm not mistaken it should be the digestive system due to the fact that our mouths and stomachs break down food and our intestines absorb any water and nutrients

4 0

3 years ago

Suppose we take a 1 m long uniform bar and support it at the 33 cm mark. Hanging a 0.15 kg mass on the short end of the beam res

MakcuM [25]

Answer:

The mass of the beam is 0.074 kg

Explanation:

Given;

length of the uniform bar, = 1m = 100 cm

Set up this system with the given mass and support;

0-----------------33cm-----------------------------------100cm

↓ Δ ↓

0.15kg m

Where;

m is mass of the uniform bar

Apply the principle of moment to determine the value of "m"

sum of anticlockwise moment = sum of clockwise moment

0.15kg(33 - 0) = m(100 - 33)

0.15(33) = m(67)

$m = \frac{0.15kg(33 \ cm)}{67 \ cm}\\\\m = 0.074 \ kg$

Therefore, the mass of the beam is 0.074 kg

8 0

3 years ago