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3 years ago

7

HELP PLEASEEEEEEEEEEEEEEEE

1 answer:

prisoha [69]3 years ago

5 0

Solid has vibrating molecules that barely move to keep it's shape

liquid moves at an average speed and keeps it's volume but not it's shape

gases move quickly and all over the place so they don't have a shape or volume

plasma is the quickest moving and is like a gas

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If a liquid is heated and the temperature at which it boils is measured, the _____ property is being measured.

igor_vitrenko [27]

Answer: atomic think so

Explanation:

5 0

3 years ago

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The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 s, what are the (a) di

nlexa [21]

Answer and Explanation:

Let:

$x(t)=Acos(\omega t+ \phi)$

The equation representing a simple harmonic motion, where:

$x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time$

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

$v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)$

Also, you may know these fundamental formulas:

$f=\frac{\omega}{2 \pi} \\\\T=\frac{2 \pi}{\omega}$

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

$x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m$

(b)

$v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s$

(c)

$a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2$

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

$\phi = \frac{\pi}{5}$

(e)

$f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz$

(f)

$T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s$

8 0

3 years ago

Read 2 more answers

A girl pulls her younger brother on a sled along a flat sidewalk (the total mass of the sled is 30kg). A frictional force of 50N

harkovskaia [24]

Answer:

6.13 s

219 N

Explanation:

Newton's law in the x direction:

∑F = ma

150 cos 30° N − 50 N = (30 kg) a

a = 2.66 m/s²

Δx = v₀ t + ½ at²

(50 m) = (0 m/s) t + ½ (2.66 m/s²) t²

t = 6.13 s

Newton's law in the y direction:

∑F = ma

Fn + 150 sin 30° N − (30 kg) (9.8 m/s²) = 0

Fn = 219 N

4 0

3 years ago

A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 4.5 cm. The

ANTONII [103]

Answer:

Explanation:

a) ωp = 2π radians / 1.7 s = <u>3.7 rad/s</u>

b) ωs = 3.7 rad/s(9.5 cm / 4.5 cm) = 7.8 rad/s

v = (ωs)R = 7.8(65) = 507 cm/s or <u>5.1 m/s</u>

c) ωs = 3.5 m/s / 0.65 m = 5.38 rad/s

ωp = 5.38(4.5 cm / 9.5 cm) = 2.55 rad/s

t = θ/ω = 2π / 2.55 = 2.463... <u>2.5 s</u>

4 0

3 years ago

The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a

zlopas [31]

Answer:

$s = 9.6\times 10^{-12}kg/s$

Explanation:

Given:

Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel =0.10 cm

let the Solute Diffusion rate of new channel = s

now equating the diffusion rate per unit volume for both the channels

$\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}$

thus,

$s = 9.6\times 10^{-12}kg/s$

7 0

3 years ago