Wave base is the most affected by a wave in deep water. :]
The positively charged atmosphere attracts negatively charged spider silk, might electrostatic force play in spider dispersal, according to a recent study.
Answer: Option C
<u>Explanation:</u>
The positive charge present in upper of the atmosphere and the negative charge on planet’s surface. During cloudless skies days, the air possesses a voltage of nearly around 100 volts for each and every meter from above the ground.
Ballooning spiders process within this planetary electric field. When their silk relieve their bodies then it picks up a negative charge. This oppose the similar negative charges on the surfaces on which the spiders settles and create sufficient force to lift them into the air. And spiders can hike those forces by climbing onto blades of grass,twigs, or leaves.
A firm current ratio is 1. 0 and its quick ratio is 1. 0. If current liabilities are 12300 then its inventories will be 12300
Inventory is the accounting of items, component parts and raw materials that a company either uses in production or sells
The quick and current ratios are liquidity ratios that help investors and analysts gauge a company's ability to meet its short-term obligations. The current ratio divides current assets by current liabilities. The quick ratio only considers highly-liquid assets or cash equivalents as part of current assets.
current ratio = current assets / current liabilities
current assets = current ratio * current liabilities
= 1 * 12300 = 12300
since , inventory is a current asset for accounting purpose , hence inventories will be 12300
To learn more about current ratios
brainly.com/question/19579866?referrer=searchResults
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The part of the ear where the sound wave converted into electrical impulse would be the cochlea. This part is the auditory portion of the inner ear which produces nerve impulses in response to sound vibrations. Hope this answers the question. Have a nice day.
Answer:
the electric field strength on the second one is 2.67 N/C.
Explanation:
the electric fiel on the first one is:
E1 = k×q/(r^2)
r^2 = k×q/(E1)
= (9×10^9)×(q)/(24.0)
= 375000000q
then the electric field on the second one is:
E2 = k×q/(R^2)
we know that R = 3r
R^2 = 9×r^2
E2 = k×q/(9×r^2)
= k×q/(9×375000000q)
= k/(9×375000000)
= (9×10^9)/(9×375000000)
= 2.67 N/C
Therefore, the electric field strength on the second one is 2.67 N/C.
