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3 years ago

How does the vacuum between the inner and outer walls of a thermos bottle limit energy loss through conduction and convection?

1 answer:

3 0

By having no solid matter between the walls (like styrofoam) conduction is eliminated since conduction requires matter to be in contact with other matter. The vacuum state eliminates convection since that is the heating of gas that then carries the energy to another body by traveling there and exchanging heat with it.

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What unit is used to measure force?

jarptica [38.1K]

Answer:

Explanation:

6 0

3 years ago

Read 2 more answers

Two parallel wires are separated by 5.60 cm, each carrying 2.65 A of current in the same direction. (a) What is the magnitude of

ololo11 [35]

Explanation:

It is given that,

The separation between two parallel wires, r = 5.6 cm = 0.056 m

Current in both the wires is 2.65 A

(a) We need to find the magnitude of the force per unit length between the wires. It can be given by :

$\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.65\times 2.65}{2\pi \times 0.056}\\\\\dfrac{F}{l}=2.5\times 10^{-5}\ N/m$

(b) As the current is in same direction, the wires will attract each other.

5 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

Faraday discovered that a current could be induced in a solenoid (a coil of wire) when _____.

igomit [66]

Faraday discovered that a current could be induced in a solenoid (a coil of wire) when "<span>a magnetized rod is being moved through the coil"

Hope this helps!</span>

6 0

3 years ago

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Read the intro to lab 10. We are going to assume that the experiment was done in a styrofoam cup, so we are not going to include

artcher [175]

Answer:

e% = 3.4%

Explanation:

This is a calorimetry problem where the heat released equals the heat absorbed

m $c_{e1}$ (T₀ - T_f) = M c_{e2} (T₁ - T_f)

Index 1 refers to water and index 2 to metal, in this case it asks for the specific heat of the metal (c_{e2})

c_{e2} = m / M c_{e1} (T_f -T₀) / (T₁ - T_f)

Let's calculate

c_{e} = 60/100 4.19 (24-20) / (100-24)

c_{e2} = 0.1323 j / gC

This metal is possibly lead, which is its specific heat is 0.128 J / gC

The percentage error is

e% = (c_{e2} - 0.128) /0.128 100

e% = 3.4%

8 0

2 years ago