Tags Motion in a Plane

Motion in a Plane

Get insights from 112 questions on Motion in a Plane, answered by students, alumni, and experts. You may also ask and answer any question you like about Motion in a Plane

Follow Ask Question

112

Questions

Discussions

Active Users

Followers

New answer posted

2 months ago

Motion in a Plane Class 11th

A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution of its speed is 8 AU per second?

Given : 1 ly = 9.46 * 10¹⁵m

1 AU = 1.5 * 10¹¹m

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

$R = \frac{l}{θ}$

$\Rightarrow T i m e = \frac{4 * 2 π R}{V} = \frac{4 * 2 π x}{V} (\frac{l}{θ})$

Time $= \frac{4 * 2 π * 4.4 * 9.64 * 1 0^{15}}{8 * 1.5 * 1 0^{11} * \frac{4}{3600} * \frac{π}{180}}$

$\Rightarrow T i m e = 4.5 * 1 0^{10} s o c s$

0 0

New answer posted

2 months ago

Motion in a Plane Class 11th

The magnitude of vectors $\vec{O} A, \vec{O} B a n d \vec{O} C$ in the given figure are equal. The direction of $\vec{O} A + \vec{O} B - \vec{O} C$ with x-axis will be:-

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

Let's say $| \vec{O A} | = | \vec{O B} | = | \vec{O C} | = l$

$\vec{O A} = l c o s 30 ° \hat{i} + l s i n 30 ° \hat{j}$

$\vec{O A} + \vec{O B} - \vec{O C}$

= $[l c o s 30 ° \hat{i} + l s i n 30 ° \hat{j}] + [l c o s 60 ° \hat{i} + l s i n 60 ° (- \hat{j})] - [l c o s 45 ° (- \hat{i}) + l s i n 45 ° (+ \hat{j})]$

$= l (\frac{\sqrt[]{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt[]{2}}) \hat{i} + l [\frac{1}{2} - \frac{\sqrt[]{3}}{2} - \frac{1}{\sqrt[]{2}}] \hat{j}$

$t a n θ = \frac{l [\frac{1}{2} - \frac{\sqrt[]{3}}{2} - \frac{\sqrt[]{2}}{2}]}{l [\frac{\sqrt[]{3}}{2} + \frac{1}{2} + \frac{\sqrt[]{2}}{2}]} = \frac{1 - \sqrt[]{3} - \sqrt[]{2}}{1 + \sqrt[]{3} + \sqrt[]{2}}$

0 0

New answer posted

2 months ago

Motion in a Plane Class 11th

Given below are two statements. One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A: Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. If A and B reached the maximum height h1 and h2 respectively, then $R = 4 \sqrt[]{h_{1} h_{2}}$

Reason R : Product of said heights.

$h_{1} h_{2} = (\frac{u^{2} s i n^{2} θ}{2 g}) . (\frac{u^{2} c o s^{2} θ}{2 g})$

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

First angle, 1 = $R = \frac{u^{2} s i n 2 ?}{g}$ $\frac{^{}}{}$

Another angle (2 = 90 - ) for which range will be

Same as that of 1 =

$R^{1} = \frac{u^{2} s i n 2 (90 ? ?)}{g} = \frac{u^{2}}{g} s i n 2 ? = R$

at $?_{1} = ?, ? ? ? ? ? h_{1} = \frac{u^{2} s i n^{2} ?}{2 g}$ $_{}_{} \frac{^{}^{}}{}$

$& ? ? ? ?_{2} = 90 ? ?, h_{2} = \frac{u^{2} s i n^{2} ?_{2}}{2 g} = \frac{u^{2} c o s^{2} ?}{2 g}$

$? h_{1} h_{2} = \frac{1}{4^{2}} {[\frac{u^{2} s i n 2 ?}{g}]}^{2}$

$h_{1} h_{2} = \frac{1}{4^{2}} R^{2}$

$R = 4 \sqrt[]{h_{1} h_{2}}$

So, Both statement is true & Reason is correct explanation for statement 1.

0 0

New answer posted

2 months ago

Motion in a Plane Class 11th

Which one of the following relations is true for two unit vector $\hat{A} a n d \hat{B}$ making an angle θ to each other?

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$| \hat{A} + \hat{B} | = \sqrt[]{A^{2} + B^{2} + 2 A B c o s θ}$ $^{}^{}$

$= \sqrt[]{1 + 1 + 2 c o s θ}$

$= \sqrt[]{2 (1 + c o s θ)}$

= $\sqrt[]{2 * 2 c o s^{2} \frac{θ}{2}}$ $^{} \frac{}{}$

$| \hat{A} + \hat{B} | = 2 c o s \frac{θ}{2}$ -(1)

$| \hat{A} - \hat{B} | = \sqrt[]{A^{2} + B^{2} - 2 A B c o s θ}$

$= \sqrt[]{1 + 1 - 2 c o s θ}$

$= 2 s i n \frac{θ}{2}$ -(2)

(2) $\div$ (1)

$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \frac{2 s i n \frac{θ}{2}}{2 c o s \frac{θ}{2}}$

$\Rightarrow | \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | t a n \frac{θ}{2}$

0 0

New answer posted

2 months ago

Motion in a Plane Class 11th

$\vec{A}$ is a vector quantity such that $| \vec{A} |$ = non-zero constant. Which of the following expression is true for $\vec{A}$ ?

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$\vec{A} * \vec{A} = A A s i n θ \hat{n}$

$= A A s i n 0 ° \hat{n}$

= 0 [since Angle between the vectors are zero degree]

$\vec{A} * \vec{A} = 0$

0 0

New answer posted

2 months ago

Motion in a Plane Class 11th

A projectile is launched at an angle 'α' with the horizontal with a velocity 20 ms^-¹. After 10 s, it inclination with horizontal is β. The value of tanβ will be: (g = 10ms^-²).

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

We know that horizontal speed will remain constant

20 cos = v cos ……… (i)

Along y-axis

$U_{y} = 20 s i n α, V_{y} = v s i n β$

$V_{y} = V_{y} + a_{y} t$

$V s i n β = 20 s i n α - 10 * 10$ ……. (ii)

$\Rightarrow (i i) \div (i)$

$\Rightarrow t a n β = t a n α - 5 s e c α$

0 0

New answer posted

2 months ago

Motion in a Plane Class 11th

A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60m, then the magnitude of displacement would be:

(Given $c o s 135 °$ = -0.7)

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

d = R

$60 = R [\frac{3 π}{4}]$

$R = \frac{60 * 4}{3 π} = \frac{80}{π} m$

Displacement = $\sqrt[]{R^{2} + R^{2} - 2 R^{2} c o s 135}$ $^{}^{}^{}$

= $\sqrt[]{2 R^{2} - 2 R^{2} (- 0.7)}$ $^{}^{}$

= $\sqrt[]{3.4 R^{2}}$ $^{}$

= $\sqrt[]{3.4 {(\frac{80}{4})}^{2}}$ ${\frac{}{}}^{}$

47 m

0 0

New answer posted

2 months ago

Motion in a Plane Class 11th

The sum of two forces $\vec{P}$ and $\vec{Q}$ is $\vec{R}$ such that $| \vec{R} | = | \vec{P} |$ . The angle $θ$ (in degrees) that the resultant of $2 \vec{P}$ and $\vec{Q}$ will make with $\vec{Q}$ is,

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

$| \overset{?}{P} + \overset{?}{Q} | = | \overset{?}{P} |$

^{$P^{2} + Q^{2} + 2 P Q c o s ? θ = P^{2}$}

$\Rightarrow Q + 2 P c o s ? θ = 0$

$\Rightarrow c o s ? θ = - \frac{Q}{2 P}$

$t a n ? α = \frac{2 P s i n ? θ}{2 P c o s ? θ + Q} = \infty$

$? [2 P c o s ? θ + Q = 0]$

α = 90^{?}

0 0

New answer posted

2 months ago

Motion in a Plane Class 11th

Two inclined planes are placed as shown in figure. A block in Projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is t $t (\sqrt[]{2} + 1) s$ . The value of t is__________.

(use g = 10 m/s²)

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

From conservation of Energy

$\frac{1}{2} m v_{0}^{2} = m g h$

$v_{0} = 10 \sqrt[]{2}$

For A to B

$a = - g s i n 45 ° = \frac{- 10}{\sqrt[]{2}}$

T = t₁ + t₂

$\begin{array}{l} = 2 \sqrt[]{2} + 2 \\ = 2 (\sqrt[]{2} + 1) \end{array}$

0 0

New answer posted

2 months ago

Motion in a Plane Class 11th

An object is projected in the air with initial velocity u at an angle q. The projectile motion is such that the horizontal range R, in maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be_______degree.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$R = \frac{u^{2} s i n (2 * 45 °)}{g} = \frac{u^{2}}{g}$

$\frac{R}{2} = \frac{u^{2}}{2 g} = \frac{u^{2} s i n 20}{g}$

$s i n 2 θ = \frac{1}{2}$

$\Rightarrow 2 θ = 30 ° \Rightarrow θ = 15 °$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Motion in a Plane

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience