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1 year ago

Which of the following statements accurately describes the atmospheric patterns that influence local weather?

Physics

1 answer:

timurjin [86]1 year ago

4 0

Answer: A

Explanation:

Well the high and lows effect the humidity the more humidity the more hot it is so the high brings higher temperatures.

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Gravitational potential energy is often released by burning substances. true or false

jolli1 [7]

Gravitational potential energy is caused when an object is resting above the ground. It is released when the object is falling, not by burning substances.

4 0

2 years ago

Read 2 more answers

Magnetic fields within a sunspot can be as strong as 0.4T. (By comparison, the earth's magnetic field is about 1/10,000 as stron

WARRIOR [948]

Answer:

The speed of ejection is $2.06\times 10^{4}\ m/s$

Solution:

As per the question:

Magnetic field density, B = 0.4 T

Density of the material in the sunspot, $\rho = 3\times 10^{4}\ kg/m^{3}$

Now,

To calculate the speed of ejection of the material, v:

The magnetic field energy density is given by:

$U_{B} = \frac{B^{2}}{2\mu_{o}}$

This energy density equals the kinetic energy supplied by the field.

Thus

$KE = U_{B}$

$\frac{1}{2}mv^{2} = \frac{B^{2}}{2\mu_{o}}$

where

m = mass of the sunspot in $1\ m^{3}$ = $3\times 10^{- 4}\ kg/m^{3}$

$v = \frac{B}{\sqrt{\mu_{o}m}}$

$v = \frac{0.4}{\sqrt{4\pi \times 10^{- 7}\times 3\times 10^{- 4}}} = 2.06\times 10^{4}\ m/s$

3 0

2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

2 years ago

An air-filled capacitor is formed from two long conducting cylindrical shells that are coaxial and have radii of 30 mm and 80 mm

Licemer1 [7]

Answer:

Explanation:

4 0

2 years ago

Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged part

erica [24]

Answer:

Two possible points

x= 0.67 cm to the right of q1

x= 2 cm to the left of q1

Explanation:

Electrostatic Forces

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

$\displaystyle f=k\frac{q_1\ q_2}{d^2}$

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

$\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}$

$\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}$

$\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}$

$\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}$

$\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}$

Equating

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}$

Operating and simplifying

$\displaystyle (0.02-x)^2=4x^2$

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

$\displaystyle 0.02-x=\pm 2x$