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natta225 [31]
2 years ago
15

A lead ball has a mass of 55.0 grams and a density of 11.4 g/cm3. what is the volume of the ball?

Physics
2 answers:
lesantik [10]2 years ago
6 0
Density=mass/volume therefore volume=mass/density; 55g/11.4g/cm^3= 4.82cm^3
densk [106]2 years ago
6 0

Answer:

Volume,V=4.82\ cm^3

Explanation:

Given that,

Mass of the lead ball, m = 55 grams

Density of lead ball, d=11.4\ g/cm^3

We need to find the volume of the ball. The formula of density is given by :

d=\dfrac{m}{V}

V=\dfrac{m}{d}

V=\dfrac{55\ g}{11.4\ g/cm^3}

V=4.82\ cm^3

So, the volume of the the lead ball is

V=4.82\ cm^3. Hence, this is the required solution.

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Answer:

1st – Place the film canister on the <u>scale</u>.

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4th – Slide the <u>small </u>weight on the front beam until the <u>lines</u> match up.

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The triple beam balance is an instrument that is used in measuring the mass of substances to a very high degree of precision. The reading error is given by ±0.05 grams. The triple beam balance as the name implies has three beams that measure substances of different mass levels.

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5 0
3 years ago
A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa
DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0
3 years ago
Mass is a measure of weight. True False
8_murik_8 [283]

Answer: False

Explanation: Mass is a measure of the amount of matter in an object.

5 0
3 years ago
Read 2 more answers
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
Lubov Fominskaja [6]

Answer:

The moment of inertia is I=\frac{M}{12} a^{2}

Explanation:

The moment of inertia is equal:

I=\int\limits^a_b {r^{2} } \, dm

If r is -\frac{a}{2}

and dm=\frac{M}{a} dr

I=\int\limits^a_b {r^{2}\frac{M}{a}  } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}

I=\frac{M}{a} \int\limits^a_b {r^{2}  } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\  I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}

7 0
3 years ago
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