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IgorLugansk [536]
3 years ago
14

The net Forward force on the propeller of a 3.2 KG model airplane is 7.0 N. What is the acceleration of the airplanes

Physics
1 answer:
Alina [70]3 years ago
3 0

Answer:

2.1875 ms^{-2}

Explanation:

Newton's second law: F=ma Let's substitute and solve for whatever is left behind:7.0N= 3.2kg *a[tex]a =\frac {7.0}{3.2} \frac{N}{kg} = 2.1875 \frac{kg\frac{m}{s^2}}{kg}=2.1875 \frac{m}{s^2}

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A firm current ratio is 1. 0 and its quick ratio is 1. 0. If current liabilities are 12300, what are its inventories?
Anna007 [38]

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

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4 0
2 years ago
For spring break you are traveling to San Diego. The trip is 1,746.4 miles and the trip takes 20.1 hours. How fast will you be t
postnew [5]

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86 mph

Explanation:

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Trong thí nghiệm Y-âng về giao thoa ánh sáng : khoảng cách giữa hai khe là 0,5 mm ; khoảng cách từ mặt phẳng chứa hai khe đến mà
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Answer: 3.1158

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2 years ago
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
muminat

Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

And since the charge of one electron is

e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

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2 years ago
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