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Question Number 17017 Answers: 1 Comments: 0

solve the simultaenous equation x+y=3 (2^x /x)=(2^y /y) find xand y.show ur workings....

$s o l v e t h e s i m u l t a e n o u s e q u a t i o n$ $x + y = 3$ $\frac{2^{x}}{x} = \frac{2^{y}}{y}$ $f i n d x a n d y . s h o w u r w o r k i n g s . . . .$

Question Number 19167 Answers: 1 Comments: 1

Two particles A and B move with constant velocities v_1 and v_2 along two mutually perpendicular straight lines towards the intersection point O. At moment t = 0, the particles were located at distances d_1 and d_2 from O respectively. Find the time, when they are nearest and also this shortest distance.

$Two particles A and B move with$ $constant velocities v_{1} and v_{2} along two$ $mutually perpendicular straight lines$ $towards the intersection point O . At$ $moment t = 0, the particles were$ $located at distances d_{1} and d_{2} from O$ $respectively . Find the time, when they$ $are nearest and also this shortest$ $distance .$

Question Number 16859 Answers: 0 Comments: 0

In dealing with motion of projectile in air, we ignore effect of air resistance on motion. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

$In dealing with motion of projectile in$ $air, we ignore effect of air resistance$ $on motion . What would the trajectory$ $look like if air resistance is included ?$ $Sketch such a trajectory and explain$ $why you have drawn it that way .$

Question Number 16691 Answers: 1 Comments: 0

The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of horizontal motion of projectile, giving it a constant horizontal acceleration equal to g. Under the same conditions of projection, the new range will be (g = acceleration due to gravity) [Answer: R + 4H]

$The horizontal range of a projectile$ $is R and the maximum height attained$ $by it is H . A strong wind now begins to$ $blow in the direction of horizontal$ $motion of projectile, giving it a constant$ $horizontal acceleration equal to g .$ $Under the same conditions of projection,$ $the new range will be$ $(g = acceleration due to gravity)$ $[Answer : R + 4 H]$

Question Number 16682 Answers: 1 Comments: 0

Question Number 16663 Answers: 0 Comments: 0

Question Number 16647 Answers: 0 Comments: 4

A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height (d/2). Neglecting subsequent motion and air resistance its velocity V varies with height h above the ground is

$A ball is dropped vertically from a$ $height d above the ground . It hits the$ $ground and bounces up vertically to a$ $height \frac{d}{2} . Neglecting subsequent$ $motion and air resistance its velocity$ $V varies with height h above the$ $ground is$

Question Number 16645 Answers: 1 Comments: 1

A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x_1 (t) after time ′t′ and that of the second body by x_2 (t) after the same time interval. Which of the following graphs correctly describes (x_1 − x_2 ) as a function of time ′t′?

$A body is at rest at x = 0 . At t = 0, it$ $starts moving in the positive x - direction$ $with a constant acceleration . At the$ $same instant another body passes$ $through x = 0 moving in the positive$ $x - direction with a constant speed . The$ $position of the first body is given by$ $x_{1} (t) after time^{'} t^{'} and that of the$ $second body by x_{2} (t) after the same$ $time interval . Which of the following$ $graphs correctly describes (x_{1} - x_{2}) as$ $a function of time^{'} t^{'} ?$

Question Number 16635 Answers: 0 Comments: 0

Solve for x, y, z (√x) + (√y) + (√z) = 4 x^2 + y^2 + z^2 = 40.125 e^(xyz) = 4.771

$Solve for x, y, z$ $\sqrt{x} + \sqrt{y} + \sqrt{z} = 4$ $x^{2} + y^{2} + z^{2} = 40.125$ $e^{xyz} = 4.771$

Question Number 16634 Answers: 1 Comments: 2

Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as function of time will be

$Consider a rubber ball freely falling$ $from a height h = 4.9 m onto a$ $horizontal elastic plate . Assume that$ $the duration of collision is negligible$ $and the collision with the plate is$ $totally elastic . Then the velocity as a$ $function of time and the height as$ $function of time will be$

Question Number 16623 Answers: 1 Comments: 0

A ball is thrown up in a lift with a velocity u relative to the lift. If it returns to the lift in time t, then acceleration of the lift is [Answer: ((2u − gt)/t) upwards]

$A ball is thrown up in a lift with a$ $velocity u relative to the lift . If it$ $returns to the lift in time t, then$ $acceleration of the lift is [Answer :$ $\frac{2 u - g t}{t} upwards]$

Question Number 16621 Answers: 1 Comments: 1

A body moves in a straight line with a velocity whose square decreases linearly with displacement between two points A and B as shown. Determine the acceleration of the particle. [Answer: (8/3) ms^(−2) ]

$A body moves in a straight line with a$ $velocity whose square decreases$ $linearly with displacement between$ $two points A and B as shown .$ $Determine the acceleration of the$ $particle . [Answer : \frac{8}{3} {ms}^{- 2}]$

Question Number 16506 Answers: 0 Comments: 0

A train 1 moves from east to west (clockwise) and another train 2 moves from west to east (anticlockwise) on the equator with equal speeds relative to ground. The ratio of their centripetal acceleration (a_1 /a_2 ) relative to centre of earth is (1) > 1 (2) < 1

$A train 1 moves from east to west$ $(clockwise) and another train 2 moves$ $from west to east (anticlockwise) on$ $the equator with equal speeds relative$ $to ground . The ratio of their$ $centripetal acceleration \frac{a_{1}}{a_{2}} relative to$ $centre of earth is$ $(1) > 1$ $(2) < 1$

Question Number 16505 Answers: 1 Comments: 0

What is the maximum magnitude of change in velocity of a motorcycle moving with a uniform speed v_0 in a circular path of length l = (π/3)R and radius R? Treat motorcycle as a particle (1) ∣Δv^→ ∣ = (v_0 /2) (2) ∣Δv^→ ∣ = v_0

$What is the maximum magnitude of$ $change in velocity of a motorcycle$ $moving with a uniform speed v_{0} in a$ $circular path of length l = \frac{π}{3} R and$ $radius R ? Treat motorcycle as a$ $particle$ $(1) ∣ Δ \vec{v} ∣ = \frac{v_{0}}{2}$ $(2) ∣ Δ \vec{v} ∣ = v_{0}$

Question Number 16504 Answers: 2 Comments: 0

An object moves in a circular path with a constant speed in the xy plane with the centre at the origin. When the object is at x = −2 m, its velocity is −(4 m/s)j^∧ . Then objects velocity at y = 2 m is (1) 4 m/s i^∧ (2) (−4 m/s) i^∧ Using this data, find objects acceleration when it is at y = 2 m (1) 8 m/s^2 i^∧ (2) −8 m/s^2 j^∧

$An object moves in a circular path with$ $a constant speed in the xy plane with$ $the centre at the origin . When the$ $object is at x = - 2 m, its velocity is$ $- (4 m / s) \overset{\land}{j} . Then objects velocity at$ $y = 2 m is$ $(1) 4 m / s \overset{\land}{i}$ $(2) (- 4 m / s) \overset{\land}{i}$ $Using this data, find objects acceleration$ $when it is at y = 2 m$ $(1) 8 m / s^{2} \overset{\land}{i}$ $(2) - 8 m / s^{2} \overset{\land}{j}$

Question Number 16499 Answers: 1 Comments: 0

A particle is moving in parabolic path x^2 = y, with constant speed u. Find the acceleration of the particle when it crossess origin. Also find the radius of curvature at origin.

$A particle is moving in parabolic path$ $x^{2} = y, with constant speed u . Find the$ $acceleration of the particle when it$ $crossess origin . Also find the radius of$ $curvature at origin .$

Question Number 16491 Answers: 0 Comments: 0

Men are running in a line along a road with velocity 9 km/hr behind one another at equal distances of 20 m. Cyclists are also riding along the same line in the same direction at 18 km/hr at equal intervals of 30 m. The speed with which an observer must travel along the road in opposite direction of so that whenever he meets a runner he also meets a cyclist is (1) 9 km/h (2) 12 km/h (3) 18 km/h (4) 6 km/h

$Men are running in a line along a road$ $with velocity 9 km / hr behind one$ $another at equal distances of 20 m .$ $Cyclists are also riding along the same$ $line in the same direction at 18 km / hr$ $at equal intervals of 30 m . The speed$ $with which an observer must travel$ $along the road in opposite direction of$ $so that whenever he meets a runner he$ $also meets a cyclist is$ $(1) 9 km / h$ $(2) 12 km / h$ $(3) 18 km / h$ $(4) 6 km / h$

Question Number 16497 Answers: 1 Comments: 0

Two particles are revolving on two coplanar circles with constant angular velocities ω_1 and ω_2 respectively. Their time periods are T_1 and T_2 then prove that the time taken by second particle to complete one revolution more than the first particle, T, is given by T = ((T_1 T_2 )/(T_1 − T_2 ))

$Two particles are revolving on two$ $coplanar circles with constant angular$ $velocities ω_{1} and ω_{2} respectively . Their$ $time periods are T_{1} and T_{2} then prove$ $that the time taken by second particle$ $to complete one revolution more than$ $the first particle, T, is given by$ $T = \frac{T_{1} T_{2}}{T_{1} - T_{2}}$

Question Number 16466 Answers: 1 Comments: 0

Question Number 16455 Answers: 2 Comments: 0

The speed of a projectile when it is at its greatest height is (√(2/5)) times its speed at half the maximum height. What is its angle of projection?

$The speed of a projectile when it is at$ $its greatest height is \sqrt{\frac{2}{5}} times its$ $speed at half the maximum height .$ $What is its angle of projection ?$

Question Number 16310 Answers: 1 Comments: 3

Question Number 16281 Answers: 0 Comments: 0

A plane moves in windy weather due east while the pilot points the plane somewhat south of east. The wind is blowing at 50 km/hr directed 30° east of north, while the plane moves at 200 km/hr relative to the wind. What is the velocity of the plane relative to the ground and what is the direction in which the pilot points the plane?

$A plane moves in windy weather due$ $east while the pilot points the plane$ $somewhat south of east . The wind is$ $blowing at 50 km / hr directed 30 ° east$ $of north, while the plane moves at 200$ $km / hr relative to the wind . What is$ $the velocity of the plane relative to the$ $ground and what is the direction in$ $which the pilot points the plane ?$

Question Number 16271 Answers: 1 Comments: 0

The unemployment rate among workers under 25 in a state went from 8.2% to 7.5% in one year. Assume an average of 1340200 workers and estimate the decrease in the number unemployed.

$T h e u n e m p l o y m e n t r a t e a m o n g w o r k e r s$ $u n d e r 25 i n a s t a t e w e n t f r o m 8 . 2 %$ $t o 7 . 5 % i n o n e y e a r . A s s u m e a n a v e r a g e$ $o f 1340200 w o r k e r s a n d e s t i m a t e$ $t h e d e c r e a s e i n t h e n u m b e r u n e m p l o y e d .$

Question Number 16616 Answers: 1 Comments: 1

A particle starts from the origin with velocity (√(44)) ms^(−1) on a straight horizontal road. Its acceleration varies with displacement as shown. The velocity of the particle as it passes through the position x = 0.2 km is [Answer: 18 ms^(−1) ]

$A particle starts from the origin with$ $velocity \sqrt{44} {ms}^{- 1} on a straight$ $horizontal road . Its acceleration varies$ $with displacement as shown . The$ $velocity of the particle as it passes$ $through the position x = 0.2 km is$ $[Answer : 18 {ms}^{- 1}]$

Question Number 16156 Answers: 0 Comments: 2

A body in a uniform horizontal circular motion possesses a variable velocity. Does it mean that the K.E. of the body is also variable?

$A body in a uniform horizontal$ $circular motion possesses a variable$ $velocity . Does it mean that the K . E .$ $of the body is also variable ?$

Question Number 16155 Answers: 1 Comments: 0

A body of mass m is projected with a speed v making an angle θ with the vertical. What is the change in momentum of the body along the Y- axis; between the starting point and the highest point of its path?

$A body of mass m is projected with a$ $speed v making an angle θ with the$ $vertical . What is the change in$ $momentum of the body along the Y -$ $axis; between the starting point and the$ $highest point of its path ?$

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