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2 years ago

Determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3

Physics

1 answer:

nlexa [21]2 years ago

7 0

Explanation:

Relative density=density of liquid÷density of water

1.25=x ÷ 1000

x=1.25 × 1000

x=1250kgm^-3

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Electromagnetic waves do require a medium to travel.<br><br> True<br> False

pshichka [43]

Answer:

The Answer is false

Explanation:

Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.

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3 years ago

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Bob is pushing a box across the floor at a constant speed of 1.5 m/s, applying a horizontal force whose magnitude is 60 n. alice

earnstyle [38]

120n

since the speed is doubled, her force is doubled

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3 years ago

RHOOLIOTTO<br> How much mass would be needed to produce 2.7 x 1016 J?

Radda [10]

E = mc^2
m = e/c^2
m = 2.7*10^16/(300000^2)
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3 years ago

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

$r=\dfrac{d}{2}$

$r=\dfrac{15\times9.461\times10^{15}}{2}$

$r=7.09\times10^{16}\ m$

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

$M=\dfrac{v^2r}{G}$

Put the value into the formula

$M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}$

$M=3.44\times10^{37}\ Kg$

Hence, The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

5 0

3 years ago

Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of

EleoNora [17]

Answer:

tex]\lambda_{Be}[/tex] = 22.78 nm

Explanation:

Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)

$E_{n}$ = k e² / 2a₀ (1 /n²)

ao = h'² / k m e² h' = h/2πi

For another atom with a single electron in the last layer

a₀ ’= h’² / k m (Ze)²

a₀ ’= a₀ / Z²

Therefore, when replacing in the equation

$E_{n}$ = - Z² Eo/n²

E₀ = 13,606 eV

The transition occurs when the electron stops from one level to another

$E_{n}$ - $E_{m}$ = Z² E₀ (1 / n² - 1 / m²) = Z² ΔE

Let's relate this expression to the wavelength

c = λ f

E = h f

E = h c /λ

h c / λ = Z² ΔE

λ = 1 / Z² (hc / ΔE)

λ = 1 / Z² λ_hydrogen

Let's apply this last equation to our case

Lithium Z = 3

$E_{n}$ = - 9 Eo / n²

40.5 10-9 = 1/9 λ_hydrogen

Beryllium Z = 4

λ = 1/16 λ_hydrogen

Let's write our two equations is and solve

40.5 10-9 = 1/9 λ_hydrogen

tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen

40.5 10⁻⁹ = 1/9 (16 $\lambda_{Be}$ )

tex]\lambda_{Be}[/tex] = 40.5 9/16

tex]\lambda_{Be}[/tex] = 22.78 nm

6 0

3 years ago

Determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3​

Determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3