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Slav-nsk [51]
1 year ago
5

Given three vectors A = 24i + 33j, B = 55i - 12j and C = 2i + 43j (a) Find the magnitude of each vector. (b) Write an expression

for the vector difference A - C. (c) Find the magnitude and direction of the vector difference A-B. (d) In a vector diagram show vector A + B, and A - B, and also show that your diagram agrees qualitatively with your answer.

Physics
1 answer:
slega [8]1 year ago
6 0

Answer:

(a) , .  and .

(b)\vec A - \vec C=22 \hat i -10 \hat j.

(c)|\vec A - \vec B|=63.13 and the direction \theta = 124.56°.

Explanation:

Given that,

,

and

\vec {C}=2 \hat i +43 \hat j

(a) The magnitude of a vector is the square root of the sum of the square of all the components of the vector, i.e. for a ,.

So, the magnitude of the is

|\vec A|=\sqrt {24^2+ 33^2}

\Rightarrow |\vec A|=\sqrt {1665}

.

The magnitude of the is

|\vec B|=\sqrt {55^2+ (-12)^2}

\Rightarrow |\vec B|=\sqrt {3169}

.

And, the magnitude of the is

|\vec C|=\sqrt {2^2+ 43^2}

\Rightarrow |\vec C|=\sqrt {1853}

.

(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is

\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)

\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j

\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j

(c) The expression of is

\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)

\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j

\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)

The magnitude of is

|\vec A - \vec B|=\sqrt {(-31)^2+55^2}

\Rightarrow |\vec A - \vec B|=\sqrt {3986}

\Rightarrow |\vec A - \vec B|=63.13

Now, if a vector \vec V= -\alpha \hat i +\beta \hat j in 3rd quadrant having direction \theta with respect to \hat i direction, than

in the anti-clockwise direction.

Here, from equation (i), for the vector \vec A - \vec C, \alpha=31 and \beta=45.

\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)

180°-55.44° [as \pi radian= 180°]

124.56° in the anti-clockwise direction.

(d) Vector diagrams for \vec A +\vec B and \vec A - \vec B has been shown  

in the figure(b) and figure(c) recpectively.

Vector \vec A - \vec B is in 3rd quadrant as calculated in part (c).

While Vector \vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)

\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j, which is in 1st quadrant as both the components are position has been shown in figure(b).

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The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati
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Answer:

The  tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, m=33.2\ kg

Coefficient of static friction is, \mu = 0.214

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Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are mg\cos \theta and normal N. Now, there is no motion in the direction perpendicular to the inclined plane. So,

N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N

Consider the direction along the inclined plane.

The forces acting along the plane are mg\sin \theta and frictional force, f, down the plane and tension, T, up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N

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Electric charge is uniformly distributed inside a nonconducting sphere of radius 0.30 m. The electric field at a point P, which
krok68 [10]

Answer:

E_{max}=41666.66\ N/C

Explanation:

Given that,

The radius of sphere, r = 0.3 m

Distance from the center of the sphere to the point P, x = 0.5 m

Electric field at point P, E_P=15000\ N/C (radially outward)

The maximum electric field is at the surface of the sphere. We know that the electric field is inversely proportional to the distance. So,

\dfrac{E_{max}}{E_p}=\dfrac{0.5^2}{0.3^2}

\dfrac{E_{max}}{15000}=\dfrac{0.5^2}{0.3^2}

{E_{max}}=\dfrac{0.5^2}{0.3^2}\times 15000

E_{max}=41666.66\ N/C

So, the magnitude of the electric field due to this sphere is 41666.66 N/C. Hence, this is the required solution.

6 0
1 year ago
An object is dropped from rest into a pit, and accelerates due to gravity at roughly 10 m/s2. It hits the ground in 5 seconds. A
vitfil [10]

Answer:

Second pit is 375 m deeper compared to first pit.

Explanation:

We have equation of motion s = ut + 0.5at²

First object hits the ground after 5 seconds,

          Initial velocity, u = 0 m/s

         Acceleration, a = 10 m/s²

         Time, t = 5 s

    Substituting,

                  s = ut + 0.5 at²

                 s = 0 x 5 + 0.5 x 10 x 5²

                    s = 125 m

           Depth of pit 1 = 125 m

Second object hits the ground after 10 seconds,

          Initial velocity, u = 0 m/s

         Acceleration, a = 10 m/s²

         Time, t = 10 s

    Substituting,

                  s = ut + 0.5 at²

                 s = 0 x 10 + 0.5 x 10 x 10²

                    s = 500 m

           Depth of pit 2 = 500 m

Difference in depths = 500 - 125 = 375 m

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1 year ago
A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c
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Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

  • The plane wall is initially with constant properties is initially at a uniform temperature, To.
  • Suddenly the surface x=L is exposed to convection process such that T∞>To.
  • The other surface x=0 is maintained at To
  • Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

  1. There is only conduction in 1-D.
  2. The system bears constant properties.
  3. The volumetric heat generation is uniform

From the given data, the condition are as follows

<u>Initial Condition</u>

At t≤0

T(x,0)=T_o

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

<u>Boundary Conditions</u>

<u>At x=0</u>

<u />T(0,t)=T_o<u />

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

<u>At x=L</u>

<u />-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]<u />

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

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2 years ago
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