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4 years ago

4800 g to kg? With the work pls

Physics

1 answer:

Zigmanuir [339]4 years ago

3 0

Answer:

[See Below]

Explanation:

✦ Formula = $Mass / 1000$

✧ $4800 / 1000 = 4.8$

So $4800$ grams is equal to $4.8$ kilograms.

~<em>Hope this helps Mate. If you need anything feel free to message me. </em>

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Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

NemiM [27]

Answer:

their final velocity is 0.091 m/s

Explanation:

Given;

mass of the first train, m₁ = 138,000 kg

mass of the second train, m₂ = 123,000 kg

initial velocity of the first train, u₁ = 0.288 m/s

initial velocity of the seocnd train, u₂ = -0.131 m/s (opposite direction to the first)

Let their common final velocity after been coupled = v

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(138,000 x 0.288) + (-0.131 x 123,000) = v(138,000 + 123,000)

39,744 - 16,113 = v(261,000)

23,631 = v(261,000)

v = 23,631 / 261,000

v = 0.091 m/s

Therefore, their final velocity is 0.091 m/s

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3 years ago

Which of the following is a TRUE statement about the differences between electromagnetic and mechanical waves?

Butoxors [25]

' D ' is the only correct statement on the list.

Electromagnetic waves are capable of traveling through

many substances, and they're also capable of traveling

through vacuum. Mechanical waves can't travel in vacuum.

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4 years ago

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alexgriva [62]

Answer:

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Explanation:

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3 years ago

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Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

Harrizon [31]

Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$

Mass of the electron, $m_e=9.11\times 10^{-31}\ kg$

Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

The wavelength of the electron is given by :

$\lambda_e=\dfrac{h}{m_ev}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}$

$\lambda_e=1.09\times 10^{-10}\ m$

The wavelength of the proton is given by :

$\lambda_p=\dfrac{h}{m_p v}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}$

$\lambda_p=5.96\times 10^{-14}\ m$

(b) Kinetic energy, $K=1.71\times 10^{-15}\ J$

The relation between the kinetic energy and the wavelength is given by :

$\lambda_e=\dfrac{h}{\sqrt{2m_eK}}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}$

$\lambda_e=1.18\times 10^{-11}\ m$

$\lambda_p=\dfrac{h}{\sqrt{2m_pK}}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}$

$\lambda_p=2.77\times 10^{-13}\ m$

Hence, this is the required solution.

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3 years ago

Two application of heat energy

densk [106]

Answer:

Heat is very important in our daily life in warming the house, cooking, heating the water, and drying the washed clothes. The heat has many usages in the industry as making and processing the food and manufacture of the glass, the paper, the textile, and etc.

Explanation:

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3 years ago