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3 years ago

The atomic mass unit is

1 answer:

3 0

The atomic mass unit is the atomic number of an element plus the number of neutrons in an element. Elements such as carbon, copper and lithium on the periodic table. Atomic Mass = Atomic number + Number of Neutrons.

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The block in the drawing has dimensions L0×2L0×3L0, where L0 =0.5 m. The block has a thermal conductivity of 250 J/(s·m·C˚). In

stira [4]

Answer:

I am sorry but what drawing

3 0

3 years ago

What is the wavelength of a monochromatic light beam, where the photon energy is 2.70 × 10^−19 J? (h = 6.63 ×10^−34 J⋅s, c = 3.0

SOVA2 [1]

Answer:

Wavelength = 736.67 nm

Explanation:

Given

Energy of the photon = 2.70 × 10⁻¹⁹ J

Considering:

$Energy=h\times frequency$

where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s

The relation between frequency and wavelength is shown below as:

c = frequency × Wavelength

Where, c is the speed of light having value = 3×10⁸ m/s

So, Frequency is:

Frequency = c / Wavelength

So, Formula for energy:

$Energy=h\times \frac {c}{\lambda}$

Energy = 2.70 × 10⁻¹⁹ J

c = 3×10⁸ m/s

h = 6.63 x 10⁻³⁴ J.s

Thus, applying in the formula:

$2.70\times 10^{-19}=6.63\times 10^{-34}\times \frac {3\times 10^8}{\lambda}$

Wavelength = 736.67 × 10⁻⁹ m

1 nm = 10⁻⁹ m

So,

Wavelength = 736.67 nm

8 0

3 years ago

What is the potential difference (voltage) if the resistance of 20 ohms produces a current of 10 amperes?

JulijaS [17]

Answer:

200 volts

Explanation:

Ohm's Law: V = IR

I = 10 A

R = 20 ohms

V = (10)(20) = 200 volts

8 0

3 years ago

Par 1/2

BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0

3 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 years ago