Motion in a Plane

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New answer posted•a year ago

Motion in a Plane Class 11th

4.28 Can you associate vectors with

(a) The length of a wire bent into a loop

(b) A plane area

(c) A sphere?

Explain.

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

4.28

(a) We cannot associate the length of a wire bent into a loop with a vector.

(b) We can associate a plane area with a vector.

(c) We cannot associate the volume of a sphere with a vector.

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New answer posted•a year ago

Motion in a Plane Class 11th

4.27 A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?

13 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

4.27 No in the both the cases.

A physical quantity which is having both direction and magnitude is not necessarily a vector. For instance, in spite of having direction and magnitude, current is a scalar quantity. The basic necessity for a physical quantity to fall in a vector category is that it ough

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New answer posted•a year ago

Motion in a Plane Class 11th

4.26 A vector has magnitude and direction. Does it have a location in space? Can it vary with time? Will two equal vectors a and b at different locations in space necessarily have identical physical effects? Give examples in support of your answer.

32 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

4.26 Yes and No

A vector in space has no distinct location. The reason behind this is that a vector stays unchanged when it displaces in a way that its direction and magnitude do not change.

A vector changes with time. For instance, the velocity vector of a ball moving with a specific speed fluctuates

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New answer posted•a year ago

Motion in a Plane Class 11th

velocity of a freely falling body is v= $\sqrt[]{2 g h}$

And $λ = \frac{h}{m v} = \frac{h}{m \sqrt[]{2 g h}}$

$λ = h$ -1

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10-3 nm
(c) 1.2 x 10-6 nm (d). 1.2 x 10 nm

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

velocity of a freely falling body is v= $\sqrt[]{2 g h}$

And $? = \frac{h}{m v} = \frac{h}{m \sqrt[]{2 g h}}$

$? = h$ -1

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10-3 nm
(c) 1.2 x 10-6 nm (d). 1.2 x 10 nm

Upvote

New answer posted•a year ago

Motion in a Plane Class 11th

Explanation- velocity of a freely falling body is v= $\sqrt[]{2 g h}$

And $λ = \frac{h}{m v} = \frac{h}{m \sqrt[]{2 g h}}$

$λ = h$ -1

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10-3 nm
(c) 1.2 x 10-6 nm (d). 1.2 x 10 nm

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

(d)

We can write

OS + PS > OP …….(i)

OS > PS – OP …….(ii)

$|\overset{?}{a} - \overset{?}{b}|$ $>$ $|\overset{?}{a}|$ + $|\overset{?}{b}|$ ….(iii)

The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus of both sides as:

$||\overset{?}{a} - \overset{?}{b}||$ > $||\overset{?}{a}| - |\overset{?}{b}||$ ….(iv)

If the two vectors

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New answer posted•a year ago

Motion in a Plane Class 11th

**Q.4.21 A particle starts from the origin at t = 0 s with a velocity of 10.0 ? m/s and moves in the x-y plane with a constant acceleration of (8.0 ? + 2.0 ?) m s^-2.

(a) At what time is the x- coordinates of the particle 16 m? What is the y-coordinate of the particle at that time?

(b)** What is the speed of the particle at the time?

12 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Ans.4.21:

(a) Velocity , $\overset{?}{v}$ = 10.0 ? m/s

Acceleration, $\overset{?}{a}$ = (8.0 ? + 2.0 ?) m s-2

We know $\overset{?}{a}$ = $\frac{d \overset{?}{v}}{d t}$ = 8.0 ? + 2.0 ?

$\overset{?}{d v}$ = (8.0 ? + 2.0 ?)dt

Integrating both sides we get $\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$ ,

Where, $\overset{?}{u} =$ velocity vector of the particle at t =0

$\overset{?}{v} =$ velocity vector of the particle at time t

But $\overset{?}{v}$ = $\frac{\underset{d r}{?}}{d t}$

$\overset{?}{d r}$ = $\overset{?}{v}$ dt

= (8.0t ? + 2.0t ? + $\overset{?}{u}$ )dt

Integrating both sides with the condition at t = 0, r =0 and at t =t, r = r

$\overset{?}{r} =$ $\overset{?}{u}$ t + ½ $\times$ 8.0 t2 ? + ½ $\times$ 2.0 $\times$ t2 ? = $\overset{?}{u}$ t + 4.0 t2 ? + t2 ?

Substituting the value of $\overset{?}{u}$ , we get

$\overset{?}{r} =$ ( 10.0 ?)t + 4.0 t2 ? + t2 ? . This equation can be expressed as

x ? + y ? = 4.0 t2 ? + ( 10.0t + t2) ?

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of ? and ?, we get

x = 4.0 t2 and y = 10.0t + t2

t = $\sqrt{x / 4}$

(a) When x = 16m, t = 2 s, y = 24m

(b) Velocity of the particle

$\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$

At t = 2 s,

$\overset{?}{v}$ (t) = 8.0 $\times$ 2 ? + 2.0 $\times 2$ ? + 10 $\times$ ? = 16 ? + 14 ?

The magnitude of $\overset{?}{v}$ (t) is given by

$|\overset{?}{v}|$ = ( 162 + 142)1/2 = 21.26 m/s

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New answer posted•a year ago

Motion in a Plane Class 11th

4.25 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?

326 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

4.25

Height of the aircraft from the ground = 3400m

Let A and B be the positions of the aircraft, making an angle of AOB = 30 $°$ . The perpendicular OC and OC is the height of the aircraft. Angles AOC = angle BOC = 15 $°$

In AOC, AC = OC $\tan$
$15 °$ = 3400 $\tan ? 15 °$ = 911.03m

AB = AC + CB = 2AC = 1

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New answer posted•a year ago

Motion in a Plane Class 11th