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2 years ago

Which of the following units would need to be converted before being used for a calculation?

Physics

1 answer:

Evgen [1.6K]2 years ago

7 0

Explanation:

To do this, one uses a conversion factor. In mathematics, specifically algebra, a conversion factor is used to convert a measured quantity to a different unit of measure without changing the relative amount. To accomplish this, a ratio ( fraction ) is established that equals one (1).

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A. Write two or three sentences to describe the conductivity of an insulator. Explain its conductivity in terms of the electrons

aliya0001 [1]

PART A)

Conductivity of insulator is very small as there is no free electrons to conduct the current trough that medium

So here number of conduction electrons are very less in insulators

PART B)

Resistance is the property of a conducting medium which will oppose the flow of current trough it

Resistance of wire directly depends on its length so resistance of long wire will be more than the resistance of short wire

Resistance inversely depends on the area so if a wire has more crossectional area then its resistance must be small

PART C)

power of light bulb is defined as rate of electrical energy

it is given by formula

P = i V

here we know that

i = 1.46 A

V = 120 volts

so power is given as

$P = 1.46 \times 120$

$P = 175.2 Watt$

3 0

2 years ago

Read 2 more answers

Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation d

Schach [20]

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

$h_{fg}$ = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

$T_{L}$ = -18.77°C = 254.23 K

At saturation pressure 800 kPa, the saturation temperature is

$T_{H}$ = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, $q_{reject}$ = $h_{fg}$ = 171.82 kJ/kg

We know COP of heat pump

COP = $\frac{T_{H}}{T_{H}-T_{L}}$

= $\frac{304.31}{304.31-254.23}$

= 6.076

Therefore, Work out put, W = $\frac{q_{reject}}{COP}$

= 171.82 / 6.076

= 28.27 kJ/kg

8 0

3 years ago

Trees in the mid-west change colors and lose their leaves in the cooler parts of the year.

wariber [46]

Deciduous
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3 0

3 years ago

Read 2 more answers

A cylinder of radius R and height H is floating upright in

emmainna [20.7K]

Answer:

Pressure difference between Top and Bottom of the cylinder is given as

$\Delta P = \frac{gH}{2}(\rho_A + \rho_B)$

Explanation:

As we know that the force due to pressure is balanced by the weight of the cylinder

So we will have

$F = mg$

so we have

$\Delta P \pi R^2 = mg$

so we have

$\Delta P \pi R^2 = \pi R^2(\rho_A(\frac{H}{2}) + \rho_B(\frac{H}{2}))g$

so we have

$\Delta P = \frac{gH}{2}(\rho_A + \rho_B)$

6 0

3 years ago

The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t

Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

$PE=\frac{GMm}{r}$

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

PART A) Potential energy when the ocean is at its furthest point to the moon,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) Potential energy when the ocean is at its closest point to the moon

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

$\Delta KE = \Delta PE$

$\frac{1}{2}mv^2 = PE_2-PE_1$

$v=\sqrt{2(PE_2-PE_1)/m}$

$v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}$

$v = 29.4m/s$

8 0

3 years ago