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anyanavicka [17]

3 years ago

14

If an electron is accelerated from rest through a potential difference of 1200V find its approximate velocity at the end of this

process. (qe = 1.6 x 10^-19 C; me = 9.1 x 10^-11kg)

1 answer:

kolbaska11 [484]3 years ago

8 0

Answer: 2.1 × 10^7 m/s

Explanation:

Please see the attachments below

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If the mass of a material is 50 grams and the volume of the material is 5 cm^3, what would the density of the material be?

kotykmax [81]

D = M / V

D = 50 / 5

D = 10 g/cm^3

5 0

3 years ago

Read 2 more answers

In a belly-flop diving contest, the winner is the diver who makes the biggest splash upon hitting the water. the size

ad-work [718]

The second diver have to leap to make a competitive splash by 4.08 m high.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

PE = mgh

where, g = 9.81 m/s²

Given is the diver jumps from a 3.00-m platform. one diver has a mass of 136 kg and simply steps off the platform. another diver has a mass of 100 kg and leaps upward from the platform.

The potential energy of the first diver must be equal to the second diver.

P.E₁ = P.E₂

m₁gh₁ = m₂gh₂

Substitute the vales, we have

136 x 3 = 100 x h₂

h₂ = ₂4.08 m

Thus, the second diver need to leap by 4.08 m high.

Learn more about potential energy.

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2 years ago

Celine has raised the temperature of water to 100°C in order to cook pasta for her lasagna. What physical property is Celine dem

LuckyWell [14K]

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5 0

3 years ago

Which of the following is an example if harmonic motion?

german

Answer:

"A pendulum swinging back and forth" is an example of harmonic motion

X = Xo cos ω t

Explains the back and forth motion of the pendulum

3 0

2 years ago

An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is

Lady_Fox [76]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}$

$\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

<h3>☯ <u>By using formula of Lens</u> </h3>

$\\$

$\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}$

$\\$

$\dashrightarrow\:\: \sf{ v = 30 \ cm}$

$\\$

<h3>☯ <u>Now, Finding the magnification </u></h3>

$\\$

$\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}$

$\\$

$\dashrightarrow\:\: \sf{m = -2}$

$\\$

<h3>☯ <u>Hence</u>, $\\$ </h3>

$\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}$

$\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}$

3 0

3 years ago