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3 years ago

Based on its location on the periodic table, which element is a metal?

Physics

2 answers:

soldier1979 [14.2K]3 years ago

8 0

A. Sodium, it is the only type of metal listed I believe!

goldfiish [28.3K]3 years ago

6 0

Sodium (Na) is a Alkali metal so that's the answer

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The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c

Natalka [10]

Answer:

$EMF = 1684.67 Volts$

Explanation:

As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time

So we can say

$EMF = \frac{d\phi}{dt}$

now we have

$\phi = BA$

here since magnetic field is constant so we have

$EMF = A\frac{dB}{dt}$

now we have

$A = (190 \times 10^3)(190 \times 10^3)$

$A = 3.61 \times 10^{10} m^2$

now we have

$EMF = 3.61\times 10^{10} (\frac{2.8 \times 10^{-6}T}{60 s})$

$EMF = 1684.67 Volts$

6 0

3 years ago

The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of t

oee [108]

The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.

By definition the gravitational potential energy is given by,

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

Where,

m = Mass of Moon

G = Gravitational Universal Constant

M = Mass of Ocean

r = Radius

First we calculate the mass through the ratio given by density.

$m = \rho V$

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

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

PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon

Now we define the radius at the most distant point

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

Then the potential energy at this point would be,

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

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

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

PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.

At the nearest point we perform the same as the previous process, we calculate the radius

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

The we calculate the Potential gravitational energy,

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

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

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

7 0

4 years ago

A child has an ear canal that is 1.3 cm long. At what sound frequencies in the audible range will the child have increased heari

ElenaW [278]

Answer:

The answer to this is

6600 Hz to 19,800 Hz

Explanation:

The shape of the human ear is analogous to a closed ended pipe hence

we have λ = 4L or wavelength = 4 * length of the child ear

The frequency c/λ where c = speed of sound = 343 m/s

hence the child's audible range is multiples of 343/(4*0.13) =6600Hz

or 13200 Hz or 19,800 Hz

The generally quoted range of human hearing is 20 Hz to 20 kHz

7 0

3 years ago

Which is not part of the first law of thermodynamics?

Nataly [62]

C. Energy can change forms

5 0

4 years ago

A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp

VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir $T_L=281K$

We know that efficiency of Carnot engine is given by

$\eta =1-\frac{T_L}{T_H}$

$0.26 =1-\frac{281}{T_H}$

$T_H=379.72K$

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so $T_H=379.72K$

So $0.35=1-\frac{T_L}{379.72}$

$T_L=246.818K$

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0

4 years ago