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3 years ago

Compute the density in g/cm^3 of a piece of metal that had mass of 0.485 kg and a volume of 52cm^3

Physics

1 answer:

steposvetlana [31]3 years ago

8 0

Answer:

9.3 g/cm³

Explanation:

First, convert kg to g:

0.485 kg × (1000 g / kg) = 485 g

Density is mass divided by volume:

D = (485 g) / (52 cm³)

D = 9.33 g/cm³

Rounding to two significant figures, the density is 9.3 g/cm³.

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Lostsunrise [7]

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2 years ago

Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.

mr_godi [17]

Answer:

The magintude of the acceleration for both objects is $3.11m/s^2$

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

$T_A-m_A*g=m_A*a$

and for mB (descending):

$T_B-m_B*g=-m_B*a$

$T_A=T_B$

because the two boxes has the same acceleration because they are attached together:

$m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2$

So the magintude of the acceleration for both objects is $3.11m/s^2$

5 0

3 years ago

An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is

Alik [6]

We use a fundamental kinematic equation as follows:

V = Vo + g*t.
<span>Tr = (V-Vo)/g = (0-10)/-10 = 1 s. = </span><span>time to reach max. height </span>

<span>Tf = Tr = 1 s. = Fall time or time to fall back to edge of bldg. </span>

<span>3-Tr-Tf = 3-1-1 = 1 s. Below edge of bldg. </span>

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<span>d = 10*1 + 5*1^2 = 15 m. <---- OPTION C</span>

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3 years ago

A power company charges its customers for electricity based upon which one of the following?a. powerb. energyc. power per unit o

Vlad [161]

B. Energy

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3 years ago

A 42 Watt light bulb is connected to a 12 volt source of potential

abruzzese [7]

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Explanation:

just the way it is

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3 years ago