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Damm [24]
2 years ago
15

You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfo

rtunately, you have the density of mercury in units of kilogram-meter^3 and the density of silicon in other units: 2.33 gram-centimeter^3. You decide to convert the density of silicon into units of kilogram-meter^3 to perform the comparison.
By which combination of conversion factors will you multiply 2.33 gram-centimeter^3 to perform the unit conversion?
Physics
1 answer:
dolphi86 [110]2 years ago
3 0

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³  / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

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seraphim [82]
Either theory or evidence
8 0
2 years ago
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If a 20.0 g object at a temperature of 35.0∘C has a specific heat of 2.89Jg∘C, and it releases 450. J into the atmosphere, what
nataly862011 [7]

Answer:

The final temperature of the object will be 42.785 °C

Explanation:

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change), so that the temperature varies.

The equation for calculating the heat exchanges in this case is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

  • Q= 450 J
  • c= 2.89 \frac{J}{g*C}
  • m= 20 g
  • ΔT= Tfinal - Tinitial= Tfinal - 35 °C

Replacing:

450 J= 2.89 \frac{J}{g*C} *20 g* (Tfinal - 35°C)

Solving for Tfinal:

\frac{450 J}{2.89\frac{J}{g*C}*20g} =Tfinal -35C

7.785 °C=Tfinal - 35°C

7.785 °C + 35°C= Tfinal

42.785 °C=Tfinal

<u><em>The final temperature of the object will be 42.785 °C</em></u>

8 0
2 years ago
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Nerve impulses are carried along axons, the elongated fibers that transmit neural signals. We can model an axon as a tube with a
jeka94

Answer:

The resistance of the axon is 1.27\times 10^7\ \Omega.

Explanation:

Given that,

Inner diameter of the model of an axon, d=10\ \mu m

Radius of the model, r=5\ \mu m=5\times 10^{-6}\ m

Resistivity of the fluid inside the tube wall, \rho=0.5\ \Omega -m

Length of the axon, l = 2 mm = 0.002 m

We know that the resistance in terms of resistivity of an object is given by :

R=\rho\dfrac{l}{A}\\\\R=0.5\times \dfrac{0.002}{\pi (5\times 10^{-6})^2}\\\\R=1.27\times 10^7\ \Omega

So, the resistance of the axon is 1.27\times 10^7\ \Omega. Hence, this is the required solution.

8 0
2 years ago
A ledge on a building is 18 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up ove
True [87]

Answer:t=5.07 s

Explanation:

Given

height of Building h=18 m

mass of Paint can m_1=4 kg

mass of second can m_2=3 kg

let T be the Tension in the rope

For  4 kg can

4g-T=4a

T=4(g-a)----1

For 3 kg can

T-3g=3a

T=3(g+a)----2

From 1 and 2

4(g-a)=3(g+a)

g=7a

a=\frac{g}{7}

So time taken to cover 18 m is

h=ut+\frac{at^2}{2}

18=0+\frac{g\cdot t^2}{7\times 2}

t^2=\frac{18\times 2\times 7}{g}

t=5.07 s

5 0
2 years ago
suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?
guajiro [1.7K]
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6 0
2 years ago
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