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2 years ago

54. Which of the following behaviors is most closely

Physics

1 answer:

Vitek1552 [10]2 years ago

8 0

Answer:

Explanation:

Foot-in-the-door technique is a compliance tactic that aims at getting a person to agree to a large request by having them agree to a modest request first

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Given that an electric field of 3×106V/m3×106V/m is required to produce an electrical spark within a volume of air, estimate the

Andre45 [30]

Answer:

Length, l = 33.4 m

Explanation:

Given that,

Electrical field, $E=3\times 10^6\ V/m$

Let the electrical potential is, $V=10^8\ V$

We need to find the length of a thundercloud lightning bolt. The relation between electric field and the electric potential is given by :

$V=E\times d\\\\d=\dfrac{V}{E}\\\\d=\dfrac{10^8}{3\times 10^6}\\\\d=33.4\ m$

So, the length of a thundercloud lightning bolt is 33.4 meters. Hence, this is the required solution.

5 0

3 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

3 years ago

Land plants are adapted to survive on land through stomata to regulate water regulation, specialized cells, structures that coll

Law Incorporation [45]

Answer:

stomata to regulate water

specialized cells

Explanation:

Land plants are able to successfully adapt to staying on land through the use of stomata to regulate water loss and through the development of specialized cells that give support to grow erect.

Water regulation is important for the survival of plants on land, otherwise, they may face desiccation as a result of the heat from the sun. The stomata is a structure found on the leaves of land plants that open and close to regulate the amount of water lost to the environment through evapotranspiration. When there is adequate water and the plant can afford to lose some water, the stomata open, otherwise, the stomata close.

Support is also important for land plant's survival and the development of specialized cells in the form of sclerenchyma cells take care of this. The fibers and sclerids are two types of sclerenchyma cells that primarily function to support the land plants.

8 0

2 years ago

Pls help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

KatRina [158]

Answer:

Its answer d honey

Explanation:

Plzzzzzzz give 5 stars

4 0

2 years ago

Acceleration is the rate at which happens

EastWind [94]

Velocity and time inceases

4 0

3 years ago