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3 years ago

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Water boils at 100°C and turns into steam. Which similarities or differences are there between water in these two states? (1 poi

nt)
The particles will have more space between them and will be moving at higher speeds as steam.

The particles will have more space between them as steam, but they will be moving at the same speed in both states.

The particles will have more space between them as a liquid, but they will be moving at the same speed in both states.

The particles will have more space between them as a liquid, but they will be moving faster as steam.

1 answer:

photoshop1234 [79]3 years ago

3 0

Answer:

My answer was A not sure if its correct as i am still doing the rest of the questions.

Explanation:

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What happens when you drink water that was in the car heated then you drink it. Make sure to explain your answer to make you as

katovenus [111]

When you drink water that was in the car heated, you can get headaches and it would make you dizzy. I know a little bit because it happen to my dad.

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3 years ago

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A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90

svetoff [14.1K]

Answer:

$\tau=22.13Nm$

Explanation:

information we have:

mass: $m=0.9kg$

lenght: $L=0.95m$

frequency: $f=2.6rev/s$

time: $t=0.2s$

and from the information we have we can calculate the angular velocity $\omega$ . which is defined as

$\omega=2\pi f$

$\omega=2\pi (2.6rev/s)\\\omega=16.336 rev/s$

----------------------------

Now, to calculate the torque

We use the formula

$\tau=I \alpha$

where $I$ is the moment of inertia and $\alpha$ is the angular acceleration

moment of inertia of a uniform rod about the end of it:

$I=\frac{1}{3}mL^2$

substituting known values:

$I=\frac{1}{3} (0.9kg)(0.95m)^2\\I=0.271kg/m^2$

for the torque we also need the acceleration $\alpha$ which is defined as:

$\alpha=\frac{\omega}{t}$

susbtituting known values:

$\alpha=\frac{16.336rev/s}{0.2s} \\\alpha=81.68rev/s^2$

and finally we substitute $I$ and $\alpha$ into the torque equation $\tau=I \alpha$ : $\tau=(0.271kg/m^2)(81.68rev(s^2)\\\tau=22.13Nm$

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4 years ago

There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

$V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3$

Similarly

$V_2=\frac{4}{24}\pi d_2^3$

Dividing the above two equations, we get

$\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3$

∴ The larger one hold 163.296 cm³ of paint

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3 years ago

Which waves can u see

NemiM [27]

light waves. the other waves like gamma rays or infrared waves or radio waves are all not visible to the eye. light is the only thing out of those you can see.

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4 years ago

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A solenoid with 385 turns per meter and a diameter of 17.0 cm has a magnetic flux through its core of magnitude 1.28 × 10-4 (T·m

shepuryov [24]

Answer:

(a)The current passes through the solenoid is 11.7 A.

(b) The new current will be one-fourth of the initial current.

Explanation:

Given that,

The number of turns per meter = 385

Diameter of solenoid = 17.0 cm = $17\times 10^{-2}$ m

Magnetic flux through core of solenoid $\phi$ = $1.28\times 10^{-4}$ Tm²

(a)

Magnetic field B= $\mu_0nI$

$\mu_0= 4\pi \times 10^{-7}$ T/amp m

Cross section area of the solenoid A= $\pi \frac{d^2}{4}$

$=\pi\frac{ (17\times 10^{-2})^2}{4}$ m²

The angle between magnetic field and cross section of the solenoid is $\theta =0^\circ$

The magnetic flux through a area A with magnetic fie;d B is

$\phi = BA cos\theta$

$\Rightarrow \phi =( \mu_0nI)(\pi \frac{d^2}4)cos \theta$

$\Rightarrow I =\frac{\phi}{(\mu_0\pi n \frac{d^2}4cos\theta)}$

$=\frac{4\phi}{(\mu_0n)(\pi d^2)cos\theta}$

$=\frac{1.28\times 10^{-4}\times 4}{(4\pi \times10^{-7}\times385 )\times(\pi\times17\times 10^{-2})^2cos 0^\circ}$

=11.7 A

The current of the solenoid is 11.7 A.

(b)

$I =\frac{4\phi}{(\mu_0\pi n d^2cos\theta)}$

From the above equation it is clear that, the current is inversely proportional to the square of the diameter of a solenoid.

$I\propto \frac1{d^2}$ .

Consider d' be the new diameter of the solenoid .

Since the new diameter of the solenoid is double of the initial diameter.

That is d'= 2d.

$\frac{I}{I'}= \frac{(d')^2}{d^2}$

$\Rightarrow \frac{I}{I'}=\frac{(2d)^2}{d^2}$

$\Rightarrow \frac{I}{I'}=4$

⇒I=4I'

$\Rightarrow I'=\frac{I}{4}$

The new current will be one-fourth of the initial current.

7 0

3 years ago