What percent of air by volume is nitrogen

What is meant by the statement '' density of water is 1000 kilograms per cubic metre ''?

LenKa [72]

It means that if you had a cubic meter of water it would weigh 1000 kilograms

4 0

3 years ago

A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the

shutvik [7]

Acceleration=(change in speed)/(time for the change). 43/0.28 = 153.6 m/s^2.

6 0

2 years ago

Read 2 more answers

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

dem82 [27]

Answer:

Explanation:

a )

Reaction force of the ground

R = mg

= 160 N

Maximum friction force possible

= μ x R

= μ x 160

= .4 x 160

= 64 N .

b )

160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,

Taking moment about top point of ladder

160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3

240 + 444 + 4f = 2700

f = 504 N

c )

Let x be the required distance.

Taking moment about top point of ladder

160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4 = 900 x 3 ( .4 x 900 is the maximum friction possible )

240 + 444 x + 1440 = 2700

x = 2.3 m

so man can go upto 2.3 at which maximum friction acts .

8 0

3 years ago

Through which medium would sound travel the fastest, water, a steel bar, or nitrogen gas explain

777dan777 [17]

It would travel faster through a steel bar, because sound waves travel by vibrating molecules in a medium. In solids, these molecules are spaced closer together, causing sound to travel faster through it. so that means the second fastest medium is liquid and the slowest is gas

3 0

2 years ago

The gravitational force of attraction be-

Sliva [168]

Answer:

2.15 m

Explanation:

Newton's Law of Universal Gravitation:

$\displaystyle F_g = G \frac{m_1 m_2}{r^2}$

$\displaystyle F_g$ is the gravitational force of attraction

$\displaystyle G$ is the universal gravitational constant

$m_1$ and $m_2$ are the two masses of the two objects

$\displaystyle r$ is the distance between the centers of the two objects.

List the known values:

$\displaystyle F_g = 3.47 \cdot 10^-^8 \ \text{N}$
$\displaystyle G = 6.673 \cdot 10^-^1^1 \ \frac{Nm^2}{kg^2}$
$\displaystyle m_1 = 44.8 \ \text{kg} \\ m_2 = 53.9 \ \text{kg}$
$\displaystyle r =\ ?$

Plug these values into the equation:

$\displaystyle 3.47 \cdot 10^-^8 \ \text{N} = 6.673 \cdot 10^-^1^1 \ \frac{Nm^2}{kg^2} \frac{(44.8 \ \text{kg})(53.9 \ \text{kg})}{r^2}$

Notice that the units $\displaystyle \text{N}$ , $\displaystyle \text{kg}^2$ , and $\displaystyle \text{m}$ cancel out. We are left with the unit $\displaystyle \text{m}$ for radius r.