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3 years ago

Diffraction supports the: A. wave theory of light. B. particle theory of light.

Physics

2 answers:

Artemon [7]3 years ago

8 0

The answer is a hope its helps you

yanalaym [24]3 years ago

7 0

Answer:

Explanation:

Diffraction support particle theory of light.

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When nonmetals bond with other atoms, what usually happens?

Serga [27]

They share electrons because it is a covalent bond

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3 years ago

A stationary 500 kg tank fires a 20 kg miegile at 100 m/s. What is the velocity of the tank after the missile is fired? Assume t

dedylja [7]

Answer:

v₁ = 4 [m/s].

Explanation:

This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

$P=m*v$

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

$(m_{1}*v_{1})=(m_{2}*v_{2})$

where:

m₁ = mass of the tank = 500 [kg]

v₁ = velocity of the tank after firing the missile [m/s]

m₂ = mass of the missile = 20 [kg]

v₂ = velocity of the missile after firing = 100 [m/s]

$(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]$

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3 years ago

Why do we need to take care of the surface water/groundwater?

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Answer:

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3 years ago

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A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp

Anna71 [15]

$\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}$

Explanation:

Natural length of a spring is $0.5\text{ }m$ . The spring is streched by $0.02\text{ }m$ . The resultant energy of the spring is $0.5\text{ }J$ .

The potential energy of an ideal spring with spring constant $k$ and elongation $x$ is given by $\dfrac{1}{2}kx^{2}$ .

So, in the current problem, the natural length of the spring is not required to find the spring constant $k$ .

$\text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}$

∴ The spring constant of the spring = $2500\text{ }\frac{kg}{s^{2}}$

4 0

4 years ago

Find the center of mass (in cm) of a one-meter long rod, made of 50 cm of copper (density 8.92 g/cm3) and 50 cm of iron (density

Gennadij [26K]

Answer

given,

copper rod length = 50 cm

density of the copper = 8.92 g/cm³

iron rod length = 50 cm

density of iron = 7.86 g/cm³

mass of iron = density × volume

m_i = 7.86 × A × l/2

m_c = 8.96 × A × l/2

taking the intersection of copper and iron rod be starting point cooper side is taken as positive side and iron side length is taken to be -ve side.

center of mass

= $\dfrac{m_i\times \dfrac{-l}{4}+m_c\times \dfrac{l}{4}}{m_i+m_c}$

= $\dfrac{7.86\times A \times \dfrac{l}{2}\times \dfrac{-l}{4}+8.96\times A \times \dfrac{l}{2}\times \dfrac{l}{4}}{7.86\times A \times \dfrac{l}{2}+8.96\times A \times \dfrac{l}{2}}$

= $\dfrac{7.86\times \dfrac{-l}{4}+8.92\times \dfrac{l}{4}}{7.86+8.92}$

= $\dfrac{1.06\dfrac{l}{4}}{16.78}$

= 0.015793 m

= 1.579 m (+ve)

center of mass shift to cooper because cooper is heavy.

4 0

3 years ago