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Alekssandra [29.7K]

3 years ago

15

"In a pure substance all the particles must be identical; therefore pure substances are composed only of elements." Do you agree

with this statement? Why or why not?

1 answer:

kenny6666 [7]3 years ago

6 0

I do not agree with the statement.
The "substance" can be a compound. It's "pure"
as long as there's nothing else in it but its name.

'Pure' water is 100% H₂O with nothing else in it.
'Pure' table salt is 100% NaCl with nothing else in it.
'Pure' carbon dioxide is 100% CO₂ with nothing else in it.

These example substances are all compounds, not elements.

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Fig. 2.1 shows the extension-load graph for a spring.

trapecia [35]

Answer:

(1) Hooke's law

(2) a) Extension is directly proportional to the applied load

b) The starting point of the graph is the origin (0, 0) or absence of load, no extension

Explanation:

(1) The law obeyed by the spring is known as Hooke's law which states that the extension or compression, x, of a spring proportional to the applied force, F

F = -k × x

Where;

k = The spring constant

(2) Given that the law mathematically is F = -k × x

The two features of the graph that show that the law is obeyed are;

a) The extension increases as the load is increased

b) The extension is zero when the there is no applied load.

3 0

3 years ago

A 645-turn coil with a 20.250 m 2 area is spun in Earth’s 5.00×10 −5 T magnetic field, producing a 1.25-V maximum emf. A

Dmitriy789 [7]

Answer:

$\omega = 1.914\ rad/s$

Explanation:

Given,

Number of turns, N = 645 N

Area, A = 20.25 m²

Earth Magnetic field, B = 5 x 10⁻⁵ T

Maximum Emf = 1.25 V.

Angular velocity, ω = ?

Using Induced Emf formula

$\varepsilon = NAB\omega$

$\omega= \dfrac{\varepsilon}{NAB}$

$\omega= \dfrac{1.25}{645\times 20.25\times 5\times 10^{-5}}$

$\omega = 1.914\ rad/s$

Angular velocity of the coil = $\omega = 1.914\ rad/s$

5 0

4 years ago

what is the period of a sound wave whose wavelength is 20.0m? use values from the book and show all of your work

Lera25 [3.4K]

The wavelength of a sound wave is related to its frequency by the relationship:
$f= \frac{v}{\lambda}$
where
f is the frequency
v is the speed of the wave
$\lambda$ is the wavelength

The wave in our problem has wavelength of $\lambda=20.0 m$ and speed of $v=343 m/s$ (this is the speed of sound in air), therefore its frequency is
$f= \frac{343 m/s}{20.0 m}=17.15 Hz$

And the period of the wave is equal to the reciprocal of its frequency:
$T= \frac{1}{f}= \frac{1}{17.15 Hz}=0.058 s$

5 0

4 years ago

a 10 uF capacitpr is connected to a 12 V battery till its fully charged. it is then disconnected and to an uncharged 20 uF capac

likoan [24]

Answer:

Explanation:

The charge on 10μF capacitor = 10 x 12 x 10⁻⁶ = 120 μC

when it is connected with 20μF capacitor both acquires common potential whose value is

= 120 x 10⁻⁶ /( 10 +20) x 10⁻⁶ = 4 V.

Energy stored in 20μF capacitor =1/2 x 20 x 10⁻⁶ x 4 x 4 = 160 x 10⁻⁶ J.

6 0

3 years ago

An 8 g bullet leaves the muzzle of a rifle with

Elena-2011 [213]

Answer: 1872 N

Explanation:

This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:

$V^{2}=V_{o}^{2} + 2ad$ (1)

$F=ma$ (2)

Where:

$V=611.9 m/s$ is the bullet's final speed (when it leaves the muzzle)

$V_{o}=0$ is the bullet's initial speed (at rest)

$a$ is the bullet's acceleration

$d=0.8 m$ is the distance traveled by the bullet before leaving the muzzle

$F$ is the force

$m=8 g \frac{1 kg}{1000 g}=0.008 kg$ is the mass of the bullet

Knowing this, let's begin by isolating $a$ from (1):

$a=\frac{V^{2}}{2d}$ (3)

$a=\frac{(611.9 m/s)^{2}}{2(0.8 m)}$ (4)

$a=234013.5063 m/s^{2} \approx 2.34(10)^{5} m/s^{2}$ (5)

Substituting (5) in (2):

$F=(0.008 kg)(2.34(10)^{5} m/s^{2})$ (6)

Finally:

$F=1872 N$

4 0

3 years ago