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Radda [10]
3 years ago
14

What part of this equation represents the products? 2NH3 + 4O2?2NO + 3H2O

Physics
1 answer:
oee [108]3 years ago
6 0
C.
remember
reactants ----> products
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A bullet at 800 m/s horizontally hits a target that is 180m away. How far does the bullet fall before it hits the target?
ale4655 [162]
The bullet falls 0.25m

The answer is -0.25m because it is falling, therefore it is negative.
Explanation:

t = dy/vy = 180m/800m/s = 0.23 secs

dy = 1/2 gt^2 = -0.25m
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A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?
allsm [11]

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side.  See more details further below.

Explanation:

Start with the formula for pressure

(pressure P) = (Force F) / (Area A)

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

F_g = m \cdot g

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3

The density of water is 997 kg/m^3, so the mass of the barrel is:

m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg

and so the weight is

F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N

and so the pressure is

P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.  

7 0
3 years ago
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