Physics, asked by sgsaliuzair01septemb, 2 months ago

if length is 10cm and width is 5cm and force is 50 Newton find pressure ​

Answers

Answered by Anonymous
3

Explanation:

Have Given :-

force = 50 N

length = 10cm

width = 5cm

 \sf \: Area = Length × Width

 \sf \: Area = 10 \times 5

 \sf \: Area = 50 \:  {cm}^{2}

We know that :-

 \sf \: Pressure = \frac{Force}{Area}

 \sf \: Pressure =  \frac{ \cancel{50}}{ \cancel{50}}

 \sf \: Pressure = 1 \: Pascal

Answered by prince100001kidiwani
1

Answer:

Answer:

sinα - αcosα

Step-by-step explanation:

Given that,

\lim_{x\to\alpha}\;\;|\frac{x\sin\alpha-\alpha\sin x}{x-\alpha}|lim

x→α

x−α

xsinα−αsinx

Let x - α = h

Thus when x ⇒ α , h ⇒ 0

\lim_{h\to0}\;\;|\frac{(h+\alpha)\sin\alpha-\alpha\sin(h+\alpha)}{h}|lim

h→0

h

(h+α)sinα−αsin(h+α)

Now we will evaluate left hand limit and right and limit separately

For LHL:-

\begin{gathered}\begin{gathered}\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cos(-h)]}{-h}+\sin\alpha-\frac{\alpha\sin(-h)\cos\alpha}{-h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{-h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{-h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha\end{gathered} \end{gathered}

h→0

lim

−h

αsinα[1−cos(−h)]

+sinα−

−h

αsin(−h)cosα

=

h→0

lim

−h

αsinα(1−cosh)

+sinα−

h

αsinhcosα

=αsinα[

h→0

lim

−h

(1−cosh)

]+sinα−αcosα[

h→0

lim

h

sinh

]

=0+sinα−αcosα

=sinα−αcosα

For RHL:-

\begin{gathered}\begin{gathered}\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cosh]}{h}+\sin\alpha-\frac{\alpha\sinh\cos\alpha}{h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha\end{gathered} \end{gathered}

h→0

lim

h

αsinα[1−cosh]

+sinα−

h

αsinhcosα

=

h→0

lim

h

αsinα(1−cosh)

+sinα−

h

αsinhcosα

=αsinα[

h→0

lim

h

(1−cosh)

]+sinα−αcosα[

h→0

lim

h

sinh

]

=0+sinα−αcosα

=sinα−αcosα

Thus limit is (sinα - αcosα).

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