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2 years ago

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If an oxygen molecule traveling at the rms speed bounces back and forth between opposite sides of a cubical vessel of 0.10 m on

a side, what is the average force the molecule exerts on one of the walls of the container? Assume the molecule’s velocity is perpendicular to the walls it hits.

1 answer:

Alisiya [41]2 years ago

5 0

Answer:

1.25x10^-19N

Explanation:see attached file pls

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7. A force of 100 N acting on a body gives it a speed of 200 m/s in 2

alekssr [168]

Answer:

Choice a. 1 kg, assuming that all other forces on the object (if any) are balanced.

Explanation:

By Newton's Second Law,

$\displaystyle a = \frac{\Sigma F}{m}$ ,

where

$a$ is the acceleration of the object in $\text{m}\cdot\text{s}^{-2}$ ,
$\Sigma F$ is the net force on the object in Newtons, and
$m$ is the mass of the object in kilograms.

As a result,

$\displaystyle m = \frac{\Sigma F}{a}$ .

Assume that all other forces on this object are balanced. The net force on the object will be $100\;\text{N}$ . The net force is constant. Acceleration should also be constant and the same as the average acceleration in the two seconds.

<h3>What is the average acceleration of this object?</h3>

$\displaystyle \begin{aligned}\text{Acceleration} &= \text{Average Acceleration}=\frac{\text{Change in Velocity}}{\text{Time Taken}}\end{aligned}$ .

$\displaystyle {a} = \frac{200\;\text{m}\cdot\text{s}^{-1}}{2\;\text{s}}=100\;\text{m}\cdot\text{s}^{-2}$ .

<h3>Apply Newton's Second Law to find the mass of the object.</h3>

$\displaystyle m = \frac{\Sigma F}{a} = \frac{100\;\text{N}}{100\;\text{m}\cdot\text{s}^{-2}} = 1\;\text{kg}$ .

6 0

3 years ago

Read 2 more answers

a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik

grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

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

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

7 0

3 years ago

Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and fi

steposvetlana [31]

Answer:

$W=K_f-K_i$

Explanation:

The work done on a particle by external forces is defined as:

$W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,$

According to Newton's second law $F=ma$ . Thus:

$W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\$

Acceleration is defined as the derivative of the speed with respect to time:

$W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,$

Speed is defined as the derivative of the position with respect to time:

$W=m\int\limits^{v_f}_{v_i} v \cdot dv \,$

Kinetic energy is defined as $K=\frac{mv^2}{2}$ :

$W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i$

3 0

3 years ago

Speed is a component of skill related fitness what does speed enable you to do.

Mrrafil [7]

I'd say move faster, unless it's asking something else.

7 0

2 years ago

A toy car moves 8 min 4 s at the constant velocity. What is the car's velocity?

Schach [20]

Explanation:

<u>Formula:</u>

$velocity = (d \div t)$

<u>d = distance given</u>

<u>t</u><u> </u><u>=</u><u> </u><u>the amount of time </u><u>given</u>

<u>Substitute the given values into the formula for velocity</u><u>:</u>

$v = 8 \div 4$

velocity is shortened for v.

8 (distance) divided by 4 (time) equals the velocity.

<u>Solve:</u>

$2 = 8 \div 4$

The velocity of the toy car equals: B. 2 m/s.

6 0

2 years ago