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monitta
3 years ago
7

What does the prefix for kilo- mean

Physics
1 answer:
artcher [175]3 years ago
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
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