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jarptica [38.1K]
3 years ago
15

A student submerges an irregularly object in a graduated cylinder half filled with water. The level of the water in the cylinder

rises from 17.8 ml to 49.7 ml. If the density of water is 0.9984 kg/liter, calculate the mass of the object in grams. The use of correct number of significant figures is required in answering all questions.
Physics
1 answer:
lakkis [162]3 years ago
7 0

solution:

volume of cylinder = change in level of water

volume of cylineder=(49.7-17.8)mL

=31.9mL

=0.0319liter

Mass= density x volume

=09984 x 0.0319

=0.0318kg

=31.8gm


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Horizontal force is applied to the object causing the spring to stretch. the acceleration of gravity is 9.8 m/s
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What makes silver and gold different by using the terms atom and element
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Calculate the capacitance of a system that stores 9.4 x 10-10 C of charge at
pantera1 [17]

The capacitance of the system is 1.9\cdot 10^{-11}F

Explanation:

The capacitance of a capacitor is given by the following equation:

C=\frac{Q}{V}

where

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V is the potential difference across the capacitor

In this problem, we have:

Q=9.4\cdot 10^{-10}C is the charge stored

V=50.0 V is the potential difference across it

Substituting into the equation, we find the capacitance:

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Read 2 more answers
A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a
tia_tia [17]

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer (\dot Q), measured in watts, in the hollow cylinder is:

\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})

Where:

k - Thermal conductivity, measured in watts per meter-Celsius.

L - Length of the cylinder, measured in meters.

D_{i} - Inner diameter, measured in meters.

D_{o} - Outer diameter, measured in meters.

T_{i} - Temperature at inner surface, measured in Celsius.

T_{o} - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)

If we know that \dot Q = 40.8\,W, L = 0.6\,m, T_{i} = 50\,^{\circ}C, T_{o} = 20\,^{\circ}C, D_{i} = 0.01\,m and D_{o} = 0.04\,m, the thermal conductivity of the biomaterial is:

k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)

k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

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