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Bas_tet [7]
3 years ago
11

You make the following measurements of an object 42kg and 22m3 what would the objects density be

Physics
2 answers:
miv72 [106K]3 years ago
4 0
The density of the object is approximately 1.91 kg per m³.

42 kg is a measure of mass, and 22 m³ is a measure of volume. Knowing this, you can use the relationship $$density = mass / volume$$ to solve for the object's density.

42 kg \div 22 m³ \approx 1.91 kg per m³.
sammy [17]3 years ago
4 0
<h2>Answer: </h2>

Density = 1.909 kg/m3

<h3>Explanation: </h3>

Given data:

Mass of the object = 42 kg

Volume of the object = 22 m3

Density of the object = ?

solution

Density = mass/volume

Density =42/22 = 1.909

Density = 1.909 kg/m3

Hence the density of the object is 1.909 and its unit is kg/m3 which is system international unit of density.


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The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres
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Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure(P_1)=210+101.325=311.325

T_1=25^{\circ}\approx 298 K

V=0.025 m^3

T_2=50^{\circ}\approx 323 K

as the volume remains constant therefore

\frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{311.325}{298}=\frac{P_2}{323}

P_2=337.44 KPa

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}

m_1=0.91 kg

Final mass m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}

m_2=0.839

Therefore m_1-m_2=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

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3 years ago
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The magnitude is doubled. The direction doesn't change.
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Solve each problem. Be sure to show your work and give a final answer with units rounded to the given number of significant figu
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Answer:

Coefficient of friction = 0.836

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tangential acceleration = 5.8 m/s²

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