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Murrr4er [49]
3 years ago
10

If u have a box with mass of 10kg and u push with a force of 200 N what is the acceleration of the box ?

Physics
1 answer:
AnnZ [28]3 years ago
5 0

Answer:

20m/s²

Explanation:

force = mass*acceleration

acceleration = force ÷ mass

acceleration = 200N ÷ 10kg

acceleration = 20m/s²

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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
2 years ago
A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just
Alex73 [517]

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

E = \frac {\sigma }{2\epsilon _{0}}

E = \frac{9\times 10^{-6}}{2\times 8.854\times 10^{-12}}

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7 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

KE = PE

\frac{1}{2} mv ^2 = \frac{1}{2} k A^2

Here,

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Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

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(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

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Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

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v = \frac{2\pi}{T} * \sqrt{A^2-x^2}

v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

v = 0.2049m/s

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