{tex}\eqalign{ & {\rm{Here,}}\vec a = 2\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over i} + 3\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over j} + \hat k \cr & {\rm{ }}\vec b = \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over i} - \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over j} + 2\hat k \cr & {\rm{ \vec a \times \vec b = }}\left| {\matrix{ {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over i} } & {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over j} } & {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over k} } \cr 2 & 3 & 1 \cr 1 & { - 1} & 2 \cr } } \right| \cr & = \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over i} \left[ {3 \times 2 - 1 \times ( - 1)} \right] - \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over j} \left[ {2 \times 2 - 1 \times (1)} \right] + \hat k\left[ {2 \times ( - 1) - 3 \times 1} \right] \cr & = 7\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over i} - 3\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over j} - 5\hat k \cr} {/tex}

Force =Mass* Accln

N=Kg.m/s²

^so

T=( Kg.m/N)^^0.5

x-t curve gives the position time graph which tells the posistion of the body at any instant. The derviative or slope of the tangent  to the curve on x-t curve gives velocity at that instant. On v-t cuve similarily if tangent is drawn at a point acceleration at that instant  will be the slope of the tangent drawn at that point.

x-t curve cannot give acceleration graph however the slopes at two points on the graph gives speed at two different times. From this avg acceleration can be calculated.

For acceleration graph,  acceleration at all instants needs to be claculated.

Slope at t₁ and t₂ be v₁ and v₂ 

Then acceleration between these two periods is (v₂ -v₁₎ /(t₂-t₁)

U=5

a=2

S=20

Apply second eq of motion . 

Distance travelled in last second of journey = 24.5 m 

Applying,

s_{​​​​​​n} = {tex}u+{1\over 2}a(2n-1){/tex}

as initial velocity u = 0 and body is under freely Fall so a = g = 9.8 

{tex}24.1 = 0 +{9.8\over 2}(2n-1){/tex}

{tex}=> 5 = 2n-1{/tex}

=> n = 3

It means body was in air for 3 seconds.

Total distance covered in 3 seconds

{tex}s = {1\over 2}gt^2{/tex}

{tex}=> s = {1\over 2}\times 9.8\times 9{/tex}

=> s = 44.1 m

Ans= 44 m

Astronomers estimate the distance of nearby objects in space by using a method called stellar parallax, or trigonometric parallax. Simply put, they measure a star's apparent movement against the background of more distant stars as Earth revolves around the sun. ... They also measure small angles in arcseconds.

3x

3x

3x

Let initial velocity = u

Acceleration = a

Using equation

{tex}s = ut+{1\over 2}at^2{/tex}

For first 2 sec

{tex}20 = 2u+{1\over 2}4a{/tex}

{tex}20 = 2u+2a{/tex}

{tex}10=u+a \ ....... (1){/tex}

Now distance covered in 2 to 5 sec is given 40m

{tex}40 = 5u+{1\over 2}25a-20{/tex}

=> 120 = 10u +25 a

=> 24 = 2u+5a ....(2)

Solving (1) and (2), we get

{tex}a= {4\over 3}\ m/s^2{/tex}

And u {tex}={26\over 3}\ m/s{/tex}

check {tex} {d^2}y\over dt^2{/tex},  {tex} {d^2}z \over dt^2{/tex},and  {tex} {d^2}x \over dt^2{/tex} i.e check for magnitude of acceleration in three directions

if any two of them is non zero and third one is zero then force is two dimesnional

if three of them are non zeros its three dimensional

if all are zero then no force exists

check {tex} {d^2}y\over dt^2{/tex},  {tex} {d^2}z \over dt^2{/tex},and  {tex} {d^2}x \over dt^2{/tex} i.e check for magnitude of acceleration in three directions

if any two of them is non zero and third one is zero then force is two dimesnional

if three of them are non zeros its three dimensional

if all are zero then no force exists

check {tex} {d^2}y\over dt^2{/tex},  {tex} {d^2}z \over dt^2{/tex},and  {tex} {d^2}x \over dt^2{/tex} i.e check for magnitude of acceleration in three directions

if any two of them is non zero and third one is zero then force is two dimesnional

if three of them are non zeros its three dimensional

if only one is non zero and all others zero its one dimensional

if all are zero then no force exists

33.3 m

First case: 

Initial velocity u = 30 km/h {tex}= {25\over 3}\ m/s{/tex}

As it stops so Final velocity v = 0

time taken t  = 8 m

Acceleration of car = a 

We know,

{tex}a = {v-u\over t}{/tex}

={tex}-{25\over 24}\ m/s^2{/tex}

Second case : 

As same breaks applied so acceleration is same.

initial velocity u = 60 km/h {tex}={50\over 3}\ m/s{/tex}

As it comes to rest so final velocity v = 0 m/s

Distances covered before rest = s

We know,

{tex}v^2-u^2= 2as{/tex}

=> {tex}(0)^2-({50\over 3})^2= 2\times {-25\over 24}\times s{/tex}

{tex}=> s =-{2500\over 9}\times {-24\over 25}\times {1\over 2}{/tex}

= 33.3 m

Area of photo=1.75 cm²=1.75×10^-4m²

Area of image of photo=1.55m²

Magnification (areal)=1.55/1.75×10^-4=0.885×10^-4

Magnification(linear)=√areal magnification

                                      =√0.885×10⁴

                                       =0.94×10²

Let h be the height of the tower and t the time taken 

u=0 and g= 10m/s²

Using S= ut +1/2 gt²

h= 0+5t²  -----(1)

distance travelled in last 2 seconds is 40 m

again using S= ut +1/2 gt²

h-40= 0+5(t-2)² -----(2)

subracting 2 from 1

h-h+40= 5(t² - (t-2)²  

8= (t² - (t-2)² 

8= 4t-4

t=3

substituting t in equation 1

h= 5x 3 x 3= 45 m

Height of tower = 45 m

Centrifugal force is defined as, “The apparent force, equal and opposite to the centripetal force, drawing a rotating body away from the center of rotation, caused by the inertia of the body

A force, arising from the body's inertia, which appears to act on a body moving in a circular path and is directed away from the centre around which the body is moving.

Speed of the man, v_m = 4 km/h

Width of the river = 1 km

Time taken to cross the river = {tex}Width\ of \ the \ river \over Speed \ of \ the \ river {/tex}{tex}= { 1\over 4} h ={ 1 × 60 \over 4 }= 15\ min{/tex}

v²-u²=2as (use this formula)

{tex}s = {v^2 \over 2f}{/tex} 

s₂=v*t

s₃={tex} {v^2 \over f}{/tex}  ,s₃=2*s 

total distansce=5s

s+{tex}t* {\sqrt{s*2f}}{/tex}+2s=5s

{tex}t* {\sqrt{s*2f}}=2s{/tex}

{tex}s=ft^2/2{/tex}

Force = 500- 100t

Impulse( as a function of time) = {tex}dF\over dt{/tex}

                                                  = {tex}d(500-100t)\over dt{/tex}

                                                  = -100

Weber is SI unit of magnetic flux is Weber.

Dimensional formula = {tex}ML^2T^{−2}A^{−2}{/tex}

{tex}ML^2T^{-2}A^{-2}{/tex}

By using tangent table we can find this value

{tex}\begin{array}{l} \theta = {\tan ^{ - 1}}(\frac{{40}}{{30}})\\ \theta = {\tan ^{ - 1}}(1.33)\\ \theta = {53^ \circ }7' \end{array}{/tex}