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Gravitational potential of a ring of radius R and mass M on the axis at a distance x from the center is given by v(x) = − ((GM)/(√(R^2 + x^2 ))) Nm/kg Using the above expression find the gravitational potential of the disc of mass M and radius R on the axis at a distance x from the center of the disc.

$$\mathrm{Gravitational}\:\mathrm{potential}\:\mathrm{of}\:\mathrm{a}\:\mathrm{ring}\:\mathrm{of} \\ $$$$\mathrm{radius}\:{R}\:\mathrm{and}\:\mathrm{mass}\:{M}\:\mathrm{on}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{distance}\:{x}\:\mathrm{from}\:\mathrm{the}\:\mathrm{center}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${v}\left({x}\right)\:=\:−\:\frac{{GM}}{\sqrt{{R}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} }}\:\mathrm{Nm}/\mathrm{kg} \\ $$$$\mathrm{Using}\:\mathrm{the}\:\mathrm{above}\:\mathrm{expression}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{gravitational}\:\mathrm{potential}\:\mathrm{of}\:\mathrm{the}\:\mathrm{disc}\:\mathrm{of} \\ $$$$\mathrm{mass}\:{M}\:\mathrm{and}\:\mathrm{radius}\:{R}\:\mathrm{on}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{distance}\:{x}\:\mathrm{from}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{disc}. \\ $$

Question Number 15051 Answers: 1 Comments: 0

Solve simultaneously x + y + z = 6 ............ equation (i) x^3 + y^3 + z^3 = 92 .......... equation (ii) x − y = z ........... equation (iii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$ \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:............\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:=\:\mathrm{92}\:\:\:\:\:\:\:\:\:..........\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{x}\:−\:\mathrm{y}\:=\:\mathrm{z}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...........\:\mathrm{equation}\:\left(\mathrm{iii}\right) \\ $$

Question Number 15045 Answers: 1 Comments: 0

In a curve the x and y co-ordinate is function of t is given by the equation x = cos t and y = sin t, then find the length of the curve for t = 0 to t = (π/2).

$$\mathrm{In}\:\mathrm{a}\:\mathrm{curve}\:\mathrm{the}\:{x}\:\mathrm{and}\:{y}\:\mathrm{co}-\mathrm{ordinate}\:\mathrm{is} \\ $$$$\mathrm{function}\:\mathrm{of}\:{t}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{the}\:\mathrm{equation} \\ $$$${x}\:=\:\mathrm{cos}\:{t}\:\mathrm{and}\:{y}\:=\:\mathrm{sin}\:{t},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{for}\:{t}\:=\:\mathrm{0}\:\mathrm{to}\:{t}\:=\:\frac{\pi}{\mathrm{2}}. \\ $$

Question Number 15039 Answers: 2 Comments: 0

The acceleration of an object is given by a(t) = cos(πt) ms^(−2) and its velocity at time t = 0 is (1/(2π)) m/s at origin. Its velocity at t = (3/2) s is? The object′s position at t = (3/2) s is?

$$\mathrm{The}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{an}\:\mathrm{object}\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{by}\:{a}\left({t}\right)\:=\:\mathrm{cos}\left(\pi{t}\right)\:\mathrm{ms}^{−\mathrm{2}} \:\mathrm{and}\:\mathrm{its}\:\mathrm{velocity} \\ $$$$\mathrm{at}\:\mathrm{time}\:{t}\:=\:\mathrm{0}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\mathrm{m}/\mathrm{s}\:\mathrm{at}\:\mathrm{origin}.\:\mathrm{Its} \\ $$$$\mathrm{velocity}\:\mathrm{at}\:{t}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{s}\:\mathrm{is}? \\ $$$$\mathrm{The}\:\mathrm{object}'\mathrm{s}\:\mathrm{position}\:\mathrm{at}\:{t}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{s}\:\mathrm{is}? \\ $$

Question Number 15036 Answers: 1 Comments: 2

Rotating a curve y = (√x) about the x- axis produces a “head light” as shown below. What is the area of disc at any x?

$$\mathrm{Rotating}\:\mathrm{a}\:\mathrm{curve}\:{y}\:=\:\sqrt{{x}}\:\mathrm{about}\:\mathrm{the}\:{x}- \\ $$$$\mathrm{axis}\:\mathrm{produces}\:\mathrm{a}\:``\mathrm{head}\:\mathrm{light}''\:\mathrm{as}\:\mathrm{shown} \\ $$$$\mathrm{below}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{disc}\:\mathrm{at}\:\mathrm{any}\:{x}? \\ $$

Question Number 15019 Answers: 0 Comments: 0

1.00 Mol of a monoatomic gas initially at 3.00 × 10^2 K and occupying 2.00 × 10^(−3 ) m^3 is heated to 3.25 × 10^2 K and the final volume is 4.00 × 10^(−3) m^3 . Assuming ideal behaviour , Calculate the entropy change for the system.

$$\mathrm{1}.\mathrm{00}\:\mathrm{Mol}\:\mathrm{of}\:\mathrm{a}\:\mathrm{monoatomic}\:\mathrm{gas}\:\mathrm{initially}\:\mathrm{at}\:\mathrm{3}.\mathrm{00}\:×\:\mathrm{10}^{\mathrm{2}} \mathrm{K}\:\mathrm{and}\:\mathrm{occupying}\:\: \\ $$$$\mathrm{2}.\mathrm{00}\:×\:\mathrm{10}^{−\mathrm{3}\:} \mathrm{m}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{heated}\:\mathrm{to}\:\mathrm{3}.\mathrm{25}\:×\:\mathrm{10}^{\mathrm{2}} \mathrm{K}\:\mathrm{and}\:\mathrm{the}\:\mathrm{final}\:\mathrm{volume}\:\mathrm{is}\:\mathrm{4}.\mathrm{00}\:×\:\mathrm{10}^{−\mathrm{3}} \mathrm{m}^{\mathrm{3}} .\: \\ $$$$\mathrm{Assuming}\:\mathrm{ideal}\:\mathrm{behaviour}\:,\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{entropy}\:\mathrm{change}\:\mathrm{for}\:\mathrm{the}\:\mathrm{system}. \\ $$

Question Number 15017 Answers: 0 Comments: 0

Calculate the heat neccessary to raise the temperature of 5.00 mol of butane from 290K to 593K at a constant pressure. where Cp(19.41 + 0.233T)J/mol/K

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{heat}\:\mathrm{neccessary}\:\mathrm{to}\:\mathrm{raise}\:\mathrm{the}\:\mathrm{temperature}\:\mathrm{of}\:\mathrm{5}.\mathrm{00}\:\mathrm{mol}\:\mathrm{of}\:\mathrm{butane} \\ $$$$\mathrm{from}\:\mathrm{290K}\:\mathrm{to}\:\mathrm{593K}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{pressure}.\:\mathrm{where}\:\mathrm{Cp}\left(\mathrm{19}.\mathrm{41}\:+\:\mathrm{0}.\mathrm{233T}\right)\mathrm{J}/\mathrm{mol}/\mathrm{K} \\ $$

Question Number 15006 Answers: 1 Comments: 8

Question Number 14988 Answers: 0 Comments: 2

Solve on Z_4 ax+b=[0]_4 a,b∈Z_4 ax^2 +bx+c=[0]_4 a,b,c∈Z_4

$${Solve}\:{on}\:\mathbb{Z}_{\mathrm{4}} \: \\ $$$${ax}+{b}=\left[\mathrm{0}\right]_{\mathrm{4}} \:\:{a},{b}\in\mathbb{Z}_{\mathrm{4}} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\left[\mathrm{0}\right]_{\mathrm{4}} \:\:{a},{b},{c}\in\mathbb{Z}_{\mathrm{4}} \\ $$

Question Number 14977 Answers: 2 Comments: 2

Find the real roots of the equation cos^4 x + sin^7 x = 1 in the interval [−π, π].

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cos}^{\mathrm{4}} \:{x}\:+\:\mathrm{sin}^{\mathrm{7}} \:{x}\:=\:\mathrm{1}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\left[−\pi,\:\pi\right]. \\ $$

Question Number 14999 Answers: 1 Comments: 0

A 20 kg box is released from the top of an inclined plane that is 5 m long and makes an angle of 20° to the horizontal. A 60N friction force impedes the motion of the box . How long will it take to reach the bottom of the box.

$$\mathrm{A}\:\mathrm{20}\:\mathrm{kg}\:\mathrm{box}\:\mathrm{is}\:\mathrm{released}\:\mathrm{from}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{an}\:\mathrm{inclined}\:\mathrm{plane}\:\mathrm{that}\:\mathrm{is}\:\mathrm{5}\:\mathrm{m}\:\mathrm{long}\:\mathrm{and} \\ $$$$\mathrm{makes}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{20}°\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{A}\:\mathrm{60N}\:\mathrm{friction}\:\mathrm{force}\:\mathrm{impedes}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{box}\:.\:\mathrm{How}\:\mathrm{long}\:\mathrm{will}\:\mathrm{it}\:\mathrm{take}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{bottom}\:\mathrm{of}\:\mathrm{the}\:\mathrm{box}. \\ $$

Question Number 14971 Answers: 0 Comments: 0

Question Number 14970 Answers: 0 Comments: 0

Question Number 14965 Answers: 2 Comments: 2

Question Number 14991 Answers: 2 Comments: 0

Question Number 14964 Answers: 0 Comments: 0

proof that ∀ x,y ∈N ∃ a,b,c ∈N ∍ (4/(x^2 +y^2 ))=(1/a) + (1/b) + (1/c)

$$\mathrm{proof}\:\mathrm{that}\: \\ $$$$\forall\:{x},{y}\:\in\mathbb{N}\:\:\exists\:{a},{b},{c}\:\in\mathbb{N}\:\backepsilon\:\frac{\mathrm{4}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}} \\ $$

Question Number 14963 Answers: 1 Comments: 0

A resistor R is connected in series with a parallel combination of two resistors of 24 and 8 ohms . The total power disipated in the circuit is 64 watt when the applied voltage is 24 volt.Find R

$$\mathrm{A}\:\mathrm{resistor}\:\mathrm{R}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:\mathrm{with}\:\mathrm{a}\:\mathrm{parallel}\:\mathrm{combination}\:\mathrm{of}\:\mathrm{two}\:\mathrm{resistors} \\ $$$$\mathrm{of}\:\mathrm{24}\:\mathrm{and}\:\mathrm{8}\:\mathrm{ohms}\:.\:\mathrm{The}\:\mathrm{total}\:\mathrm{power}\:\mathrm{disipated}\:\mathrm{in}\:\mathrm{the}\:\mathrm{circuit}\:\mathrm{is}\:\mathrm{64}\:\mathrm{watt}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{applied}\:\mathrm{voltage}\:\mathrm{is}\:\mathrm{24}\:\mathrm{volt}.\mathrm{Find}\:\mathrm{R} \\ $$

Question Number 14962 Answers: 1 Comments: 0

Two 30 ohms resistor are connected in parallel, what should be the resistance to be connected in series with this parallel combination such that the power in each 30 ohms is (1/4) th of total power.

$$\mathrm{Two}\:\mathrm{30}\:\mathrm{ohms}\:\mathrm{resistor}\:\mathrm{are}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{parallel},\:\mathrm{what}\:\mathrm{should}\:\mathrm{be}\:\mathrm{the}\:\mathrm{resistance} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:\mathrm{with}\:\mathrm{this}\:\mathrm{parallel}\:\mathrm{combination}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{power} \\ $$$$\mathrm{in}\:\mathrm{each}\:\mathrm{30}\:\mathrm{ohms}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{th}\:\mathrm{of}\:\mathrm{total}\:\mathrm{power}. \\ $$

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