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3 years ago

Determine how many Altuves you could travel in a day if you were moving at 45.000 mph all day. Altuve = 1.65 m

Physics

1 answer:

dimaraw [331]3 years ago

8 0

Answer:

The answer is below

Explanation:

Speed is the ratio of total distance traveled to the total time taken. It is given by the formula:

Speed = Distance / time taken

Given that the speed = The polynomial 45.000 mph.

In a day there are 24 hours, hence, the distance traveled is:

Speed = Distance / time taken

45 mph = Distance / 24 hours

Distance = 45 mph × 24 h = 1080 miles

But 1 mile = 1609.34 m

1080 miles = $1080\ miles*\frac{1609.34 \ m}{1\ miles}=1738092\ m$

Given that Altuve = 1.65 m

Hence, the number of Altuve traveled = 1738092 m / 1.65 m = 1053389

1053389 Altuves was traveled

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

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3 years ago

We know today that atoms cannot be divided into smaller parts true or false

djverab [1.8K]

Hi , the answer is false ,atoms can be divided into smaller parts , electrons , protons and neutrons.

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3 years ago

We are going to make an imaginary engine using water. We are going to heat 100 grams of water to 120 C from its initial temperat

Svetach [21]

Answer:

The work done by this engine is 800 cal

Explanation:

Given:

100 g of water

120°C final temperature

22°C initial temperature

30°C is the temperature of condensed steam

Cw = specific heat of water = 1 cal/g °C

Cg = specific heat of steam = 0.48 cal/g °C

Lw = latent heat of vaporization = 540 cal/g

Question: How much work can be done using this engine, W = ?

First, you need to calculate the heat that it is necessary to change water to steam:

$Q_{1} =m_{w} C_{w} (100-22)+m_{w}L_{w}+m_{w}C_{g}(120-100)$

Here, mw is the mass of water

$Q_{1} =(100*1*78)+(100*540)+(100*0.48*20)=62760cal$

Now, you need to calculate the heat released by the steam:

$Q_{2} =m_{w}C_{g}(120-100)+m_{w}L_{w}+m_{w}C_{w}(100-30)=(100*0.48*20)+(100*540)+(100*1*70)=61960cal$

The work done by this engine is the difference between both heats:

$W=Q_{1}-Q_{2}=62760-61960=800cal$

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3 years ago

Does gravity cause physical weathering or chemical weathering

Kazeer [188]

I think it causes physical weathering

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3 years ago

A pendulum completes 30 cycles every minute. What is the frequency of the pendulum

Strike441 [17]

Answer:

1 cycle every 2 seconds if that is what it is asking. So you have to do 2 x 30 and you will get 60 so ur awnser is 60

Explanation:

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3 years ago