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3 years ago

GPS consists of three segments what are they?

Physics

1 answer:

WARRIOR [948]3 years ago

4 0

Answer:

GPS consist of the space segment, the control segment, and the user segments

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Two bodies made of the same material have same external dimension and appearance, but one is compact solid and other is hollow.

Bumek [7]

Answer:

Different

Explanation:

The hollow one will expand even more making it have a larger volume then the solid one so they are different

5 0

3 years ago

A damped harmonic oscillator consists of a mass on a spring, with a small damping force that is proportional to the speed of the

exis [7]

Answer:

2.19 N/m

Explanation:

A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:

T = 2π√(m/K)

Where T is the period, m is the mass (in kg), and K is the damping constant. So:

2.4 = 2π√(0.320/K)

√(0.320/K) = 2.4/2π

√(0.320/K) = 0.38197

(√(0.320/K))² = (0.38197)²

0.320/K = 0.1459

K = 2.19 N/m

4 0

3 years ago

Which of the following is most closely related to an activated complex?

Anon25 [30]

Among the choices above, the one that is most closely related to an activated complex is the transition state. The answer is letter D. This formation forms quickly and does not stay in a way compound is. It usually forms during the enzyme – substrate reaction.

5 0

3 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$