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3 years ago

What is the subscript for nitrogen in hydrazine, N2H4? 2 N H 4

Physics

2 answers:

Ksenya-84 [330]3 years ago

5 0

2................. hope this helps

Ad libitum [116K]3 years ago

3 0

The subscript for nitrogen in hydrazine, N2H4 is 2.
In chemistry, when writing the chemical formula of chemical compounds, subscripts refers to the numerical numbers that are sometimes put at the base of some chemical symbols. The subscripts are used to indicate the number of atoms that are present in each element. For example, in hydrazine, the subscript 2 indicates that two atoms of nitrogen are present, while the subscript 4 indicates that four atoms of hydrogen are presents.

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Velocity (m/s)

alekssr [168]

Explanation:

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(a) The segment A shows acceleration as velocity increases with the increase in time.

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(b) The segment C shows the object is slowing down as the time increases in segment C, the velocity decreases and afterwards it comes to rest.

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(c) The velocity is segment B is 40m/s. And in the diagram there is no change in velocity.

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(d) The acceleration of segment B is zero. As there in no change in curve and it is moving with uniform velocity.

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<h2>Thank you!</h2>

3 0

1 year ago

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a

Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

$d = v*t \rightarrow v = \frac{d}{t}$

Where:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

$t = \sqrt{\frac{2d}{g}}$

Where:

g: is gravity = 9.8 m/s²

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s$

Now, the horizontal velocity of the rock is:

$v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s$

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.

2. To calculate the distance at which the projectile will land, first, we need to find the time:

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s$

So, the distance is:

$d = v*t = 6.67 m/s*1.43 s = 9.54 m$

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!

7 0

3 years ago

What is the kinetic energy of an object with a mass of 50kg and is and it is traveling at a rate of 60m/s.

liq [111]

Answer:

90,000 J

Explanation:

Kinetic energy can be found using the following formula.

$KE=\frac{1}{2}mv^2$

where m is the mass in kilograms and v is the velocity in m/s.

We know the object has a mass of 50 kilograms. We also know it is a traveling at a rate of 60 m/s. Velocity is the speed of something, so the velocity of the object is 60 m/s.

m=50

v=60

Substitute these values into the formula.

$KE=\frac{1}{2}*50*60^2$