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Slav-nsk [51]
2 years 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]2 years 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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A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in
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<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
  • \boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}
  • \boxed{a = acceleration \ (constant)}
  • \boxed{d = distance \ travelled}

The formula we follow for this problem are as follows:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (in m/s²)
  • u = initial velocity  
  • v = final velocity
  • d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

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Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

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Arte-miy333 [17]

Answer:

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Explanation:

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The charge Q on the resistor will be;

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Q= 6×10^-6 ×33.3

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Q= 1.99×10^-4Coulombs

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2 years ago
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