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3 years ago

What does the prefix for kilo- mean

1 answer:

3 0

The word/prefix " $kilo$ " means $one$ $thousad$

Good luck on your assignment and enjoy your day!

~ $MeIsKaitlyn :)$

You might be interested in

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

brainly.com/question/19579866?referrer=searchResults

#SPJ4

4 0

2 years ago

What is the biggest barrier to the use of renewable energy in the United States?

storchak [24]

The biggest barrier to the use of renewable energy in the United States is the Citizen opposition to negative environmental impact.
So, the answer is B.

7 0

2 years ago

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

atroni [7]

(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

4 0

3 years ago

The density of iron is 7.87 g/cm3. If an iron cannonball has a volume of 105 cm3

Reptile [31]

the answer is 0.284 lb/in3

6 0

2 years ago

A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity

Viktor [21]

Answer:

-26 m/s.

Explanation:

Hello,

In this case, since the vertical initial velocity is 26 m/s and the vertical final velocity is 0 m/s at P, we compute the time to reach P:

$t=\frac{0m/s-26m/s}{-9.8m/s^2} =2.65s$

With which we compute the maximum height:

$y=26m/s*2.65s-\frac{1}{2}*9.8m/s^2*(2.65s)^2 \\\\y=34.5m$

Therefore, the final velocity until the floor, assuming P as the starting point (Voy=0m/s), turns out:

$v_f=\sqrt{0m/s-(-9.8m/s^2)*2*34.5m}\\ \\v_f=-26m/s$

Which is clearly negative since it the projectile is moving downwards the starting point.

Regards.

3 0

3 years ago