Ask question

Ask question

All categories

2 years ago

5

Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very

few collisions with other molecules. Express your answer using two significant figures.

1 answer:

timofeeve [1]2 years ago

4 0

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ $v=\frac{502.1}{\sqrt{3} }$

$=289.9 \ m/s$

The time will be:

⇒ $t=\frac{d}{v}$

$=\frac{2\times 6}{289.9}$

$=\frac{12}{289.9}$

$=0.041 \ sec$

hence,

⇒ $N=\frac{1}{t}$

$=\frac{1}{0.041}$

$=24.39 \ per \ sec$

You might be interested in

Proved that<br>V = u+at<br>

kondaur [170]

Answer:

$\sf Proof \ below$

Explanation:

We know that acceleration is change in velocity over time.

$\sf a=\frac{\triangle v}{t}$

$\sf a=\frac{v-u}{t}$

v is the final velocity and u is the initial velocity.

Solve for v.

Multiply both sides by t.

$\sf at=v-u$

Add u to both sides.

$\sf at + u=v$

3 0

2 years ago

Read 2 more answers

Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,

klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble= $2\hat{i}-3\hat{k}$ N

Position vector= $r=0.5\hat{j}-2\hat{k}$ m

a.We have to find the resulting torque on the pebble about origin.

Torque= $r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque= $-k-1.5i-4j$ N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

$\tau=-6j-k-1.5i+2j=-1.5i-4j-k$ N-m

6 0

2 years ago

The temperature of a solution will be estimated by taking n independent readings and averaging them. Each reading is unbiased, w

Viktor [21]

Answer:

68 readings.

Explanation:

We need to take this problem as a statistic problem where the normal distribution table help us.

We can start considerating that X is the temperature of the solution, then

$0.9 = P(|\bar{x}-\mu|$

$0.9 = P(\frac{|\bar{x}-\mu|}{\frac{\sigma}{\sqrt{n}}}$

$0.9 = P(|Z|$

For a confidence level of 90% our $Z_{critic}$ is 1.645

Therefore,

$\frac{0.1}{\frac{\sigma}{\sqrt{n}}} = 1.645$

Substituting for $\sigma = 5$ and re-arrange for n, we have that n is equal to

$n=(\frac{1.645\sigma}{0.1})^2$

$n=\frac{(1.645)^2(0.5)^2}{0.1^2}$

$n=67.65$

$n=68$

We need to make 68 readings for have a probability of 90% and our average is within $0.1\°\frac$

3 0

1 year ago

What does "Position (m)” represent in the graph?

SCORPION-xisa [38]

<span>Position (m)” represent <u>t</u></span><u>he dependent variable</u> in the graph.

5 0

2 years ago

Read 2 more answers

A 10.1 g bullet leaves the muzzle of a rifle with a speed of 425 m/s. What constant force is exerted on the bullet while it is t

PIT_PIT [208]

Answer

Force will be 1520.2604 N

Explanation:

We have given mass of the bullet m = 10.1 gram = 0.0101 kg

Distance S= 0.6 m

Final velocity v = 425 m/sec

Initial velocity u will be 0 m/sec

From third equation of motion $v^2=u^2+2as$

$425^2=0^2+2\times a\times 0.6$

$a=150520.833m/sec^2$

We know that force is given by F = ma

So force = 0.0101×150520.833 = 1520.2604 N

5 0

2 years ago