can we increase elastic limit of material
Answers
Explanation:
Answer:
This means the material deforms irreversibly and does not return to its original shape and size, even when the load is removed. When stress is gradually increased beyond the elastic limit, the material undergoes plastic deformation. ... We can graph the relationship between stress and strain on a stress-strain diagram.
Explanation:
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Explanation:
Answer:
Given :-
A student starting from his house walks at a speed of 2 ½ km/h and reaches his school 6 minutes late.
Next day starting at the same time he increases his speed by 1 km/h and reaches 6 minutes early.
To Find :-
What is the distance between the school and his house.
Solution :-
Let,
\mapsto↦ The time be x
\mapsto↦ The distance be y
Now,
\implies \sf Speed_{(Student)} =\: 2\dfrac{1}{2}\: km/h⟹Speed(Student)=221km/h
\implies \sf Speed_{(Student)} =\: \dfrac{4 + 1}{2}\: km/h⟹Speed(Student)=24+1km/h
\implies \sf \bold{\green{Speed_{(Student)} =\: \dfrac{5}{2}\: km/h}}⟹Speed(Student)=25km/h
Now, let's find the time :
\implies \sf Time =\: (x + 6)\: minutes⟹Time=(x+6)minutes
\begin{gathered}\implies \sf Time =\: \bigg(x + \dfrac{\cancel{6}}{\cancel{6}0}\bigg)\: hours\: \: \bigg\lgroup \sf\bold{1\: minutes =\: \dfrac{1}{60}\: hours}\bigg\rgroup\\\end{gathered}⟹Time=(x+606)hours⎩⎪⎪⎪⎧1minutes=601hours⎭⎪⎪⎪⎫
\begin{gathered}\implies \sf \bold{\green{Time =\: \bigg(x + \dfrac{1}{10}\bigg)\: hours}}\\\end{gathered}⟹Time=(x+101)hours
Now, as we know that :
\clubsuit♣ Speed Formula :
\begin{gathered}\mapsto \sf\boxed{\bold{\pink{Speed =\: \dfrac{Distance}{Time}}}}\\\end{gathered}↦Speed=TimeDistance
According to the question by using the formula we get,
\begin{gathered}\implies \sf \dfrac{5}{2} =\: \dfrac{y}{\bigg(x + \dfrac{1}{10}\bigg)}\\\end{gathered}⟹25=(x+101)y
\begin{gathered}\implies \sf \dfrac{5}{2} =\: \dfrac{y}{\bigg(\dfrac{10x + 1}{10}\bigg)}\\\end{gathered}⟹25=(1010x+1)y
\implies \sf \dfrac{5}{2} =\: \dfrac{y}{1} \times \bigg(\dfrac{10}{10x + 1}\bigg)⟹25=1y×(10x+110)
\implies \sf \dfrac{5}{2} =\: \dfrac{10y}{10x + 1}⟹25=10x+110y
By doing cross multiplication we get,
\implies \sf 5(10x + 1) =\: 2(10y)⟹5(10x+1)=2(10y)
\implies \sf 50x + 5 =\: 20y⟹50x+5=20y
\implies \sf 50x - 20y =\: - 5⟹50x−20y=−5
\begin{gathered}\implies \sf\bold{\purple{50x - 20y =\: - 5\: ------\: (Equation\: No\: 1)}}\\\end{gathered}⟹50x−20y=−5−−−−−−(EquationNo1)
Again,
\mapsto↦ Next day starting at the same time he increases his speed by 1 km/h and reaches 6 minutes early.
\implies \sf \dfrac{5}{2} + 1 =\: \dfrac{y}{\bigg(x - \dfrac{1}{10}\bigg)}⟹25+1=(x−101)