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Answers posted by sreemathi.v

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answered Oct 9, 2014
Toolbox:Length of the latus rectum is $\large\frac{2b^2}{a}$Length of the minor axis is $2b$$e=\larg...
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answered Oct 9, 2014
Toolbox:The radius of the circle is $\sqrt{h^2+k^2-c}$ where $h$ and $k$ are the centres.Answer : $x...
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answered Oct 8, 2014
Toolbox:The radius of a circle is always perpendicular to the tangent.The perpendicular distance fro...
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answered Oct 8, 2014
Toolbox:General equation of the circle is $x^2+y^2+2hx+2ky+c=0$Answer : $x^2+y^2+2x-4y+20=0$Let the ...
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answered Oct 8, 2014
Toolbox:The centre of a circle is the midpoint of the diameter.Answer : $(1,2)$Equation of the given...
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answered Oct 8, 2014
Toolbox:The radius and a tangent to a circle are always perpendicular to each other.Length of the pe...
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answered Oct 8, 2014
Toolbox:If the slopes of two lines are equal,then the lines are parallel.The distance between the pa...
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answered Sep 2, 2014
Answer : 3600N.cmWe must work out the spring constant first (we will work in cm)$F=kx$So $1200 =k(2)...
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answered Sep 2, 2014
Answer : 12 JoulesWe know that F = 40 Nwhen $x = 60 cm = 0.60$ m.Since $F = kx$$k = F/x = (40 N)/(0....
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answered Sep 2, 2014
Answer : 525J$100 =\int_0^{0.5} kxdx$$\quad\;\;\;= \large\frac{1}{8}$$k$$\Rightarrow k=800$$W=\int\l...
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answered Sep 2, 2014
Answer : 24JFirst we need to find the spring constant.Let $x$ denote the distance the spring is stre...
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answered Sep 2, 2014
Answer : $\large\frac{27}{10}$From Hooke's law : $F=kx$ where $x$ is the distance the sprin is stret...
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answered Sep 2, 2014
Answer : $\large\frac{1}{18}$From Hooke's law : $F=kx$ where $x$ is the distance the spring is stret...
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answered Sep 1, 2014
Answer : 160 N.cmWe have $F(35-30)=50N=k(35-30)cm\large\frac{N}{cm}$$=k (5cm)$So $k= \large\frac{50}...
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answered Sep 1, 2014
Answer : 1.98JA force of 40N is required to stretch the spring $30cm-20cm=10cm=0.10m$ from its natur...
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answered Sep 1, 2014
Answer : 625J $W=\int\limits_a^b\overrightarrow{F}. \overrightarrow{ds}$ $\;\;\;\;=\int...
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answered Sep 1, 2014
Answer : 7.0J$W=\int\limits_2^3 3x^2dx+\int\limits_3^0 4dy$$\;\;\;\;=3\int\limits_2^3 x^2dx+4\int\li...
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answered Sep 1, 2014
Answer : -6JWe use the general definition of work (for a two-dimensional problem),$W=\int\limits_{x_...
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answered Sep 1, 2014
Answer : $\large\frac{\pi}{4}$The displacement variable is $x$ (not s) we can use the formula $dW = ...
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answered Sep 1, 2014
Answer : $\large\frac{1}{2}\normalsize m\large\frac{v^2}{t_1^2}\normalsize t^2$Work done =$F.s=ma\bi...
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answered Aug 28, 2014
Answer : $\large\frac{1}{2}$$ Cx_1^2$Work done = $\int\limits_{x_1}^{x_2} Fdx $$\qquad\qquad=\int\li...
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answered Aug 28, 2014
Answer : 135 JWork done =$\int\limits_{x_1}^{x_2}Fdx$$\qquad\qquad=\int\limits_0^5(7-2x+3x^2)dx$$\qq...
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answered Aug 28, 2014
Answer : 650000 ft-lbs We divide into two parts , the work needed to lift the coal and the ...
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answered Aug 28, 2014
Answer : 2250g JThe work done against gravity in lifting one box =$15g\times 1.5J=22.5g\;J$The work ...
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answered Aug 28, 2014
Answer : 12500 JAmount of Work is $W = F ⋅ d$$\qquad\qquad\qquad\qquad$=500 N.25 m$\qquad\qquad\qqua
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answered Aug 28, 2014
Answer : 18JDisplacement $S=r_2-r_1$$\qquad\qquad\quad\;\;=(\hat{i}+3\hat{j})-(4\hat{i}-5\hat{j})$$\...
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answered Aug 28, 2014
Answer : 11.8JTo lift a 3kg object at constant speed,an upward force equal in magnitude to its weigh...
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answered Aug 28, 2014
Answer : 20JWork is force times displacement through which the force acts.Here,force is in the same ...
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answered Aug 28, 2014
Answer : 17.6JFrom the free body diagram https://clay6.com/mpaimg/work2.png$T-mg =ma$$\Rightarrow T=...
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