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2 years ago

13

Given that the rms of a helium atom at a certain temperature is 1350 m s-1, determine the rms speed of an oxygen molecule at thi

s temperature. The molar mass of O2 is 32 g mol-1 and for He is 4 g mol-1.

1 answer:

monitta2 years ago

5 0

Answer:

Isko moreno vs Alfredy I will gonna go to sleep soon though you were able and willing and ready to play some games ☺️☺️

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While trees are not the largest level in the pyramid of numbers, they are still the base for both the pyramids of biomass and py

Oxana [17]

It is true because <span>A pyramid of biomass is a representation of the amount of energy contained in biomass, at different trophic levels for a given point in time . The amount of energy available to one trophic level is limited by the amount stored by the level below. Because energy is lost in the transfer from one level to the next, there is successively less total energy as you move up trophic levels. Tree is a base as it provides food and energy.</span>

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2 years ago

A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes

Alchen [17]

<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=54.5m/s$ is the projectile's initial speed

$\theta=35\°$ is the angle

$t=2.80s$ is the time since the projectile is launched until it strikes the target

$x$ is the final horizontal position of the projectile (the value we want to find)

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the projectile (we are told it was launched at ground level)

$y$ is the final height of the projectile (the value we want to find)

$g=9.8m/s^{2}$ is the acceleration due gravity

Having this clear, let's begin with x (1):

$x=(54.5m/s)cos(35\°)(2.8s)$ (3)

$x=125m$ (4) This is the horizontal final position of the projectile

For y (2):

$y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}$ (5)

$y=48.308m$ (6) This is the vertical final position of the projectile

4 0

3 years ago

How do prokaryotic cells replicate?

Alika [10]

Answer:

The usual method of prokaryote cell division is termed binary fission. The prokaryotic chromosome is a single DNA molecule that first replicates, then attaches each copy to a different part of the cell membrane. When the cell begins to pull apart, the replicate and original chromosomes are separated.

6 0

3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

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3 years ago

Why are you more likely to get a concussion when hit by a field hockey ball than a

Aleksandr-060686 [28]

Answer:

because it can be hard

Explanation:

I said that because they be on bed rest

6 0

3 years ago