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svlad2 [7]
2 years ago
6

To visit your favorite ice cream shop, you must travel 490 m west on Main Street and then 920 m south on Division Street. Suppos

e you take 49 s to complete the 490-m displacement and 75 s to complete the 920-m displacement.
a. What is the magnitude of your average velocity during this 121-second period of time?
b. What is the direction of your average velocity during this 121-second period of time?
c. What is your average speed during the trip?
Physics
1 answer:
topjm [15]2 years ago
5 0

Answer:

a) The magnitude of your average velocity during the 121 s is 8.61 m/s.

b) The direction of the average velocity is 61.9° south of west.

c) Your average speed during the trip is 11.7 m/s

Explanation:

Hi there!

a) The average velocity (a.v) is calculated as the displacement divided by the time it took to do such a displacement.

The displacement is calculated as the distance between the initial position and the final position:

Displacement = Δ(x,y) = final position - initial position

Let's consider that your initial position is the origin of our frame of reference and let's also consider that west and south are positive directions (+x and +y respectively). Then the displacement vector will be:

Δ(x,y) = final positon - initial position

Δ(x,y) = (490, 920) m - (0, 0) m = (490, 920) m

The average velocity will be:

a.v = Δ(x,y) / t

a.v = (490, 920) m / 121 s

a.v = (4.05, 7.60) m/s

The magnitude of the average velocity is calculated as follows:

 

The magnitude of your average velocity during the 121 s is 8.61 m/s.

b) To find the direction of the average velocity, we have to use trigonometric rules of right triangles. Notice that the x and y-components of the average velocity (vx and vy) together with the average velocity vector (v), with magnitude 8.61 m/s, form a triangle (see figure).

Also, notice that v is the hypotenuse of the triangle and that vx is the side adjacent to the angle θ while vy is the side opposite to θ.

Using trigonometry, we can calculate the value of the angle θ:

cos θ = adjacent side / hypotenuse

cos θ = vx / v

cos θ = 4.05 m/s / 8.61 m/s

θ = 61.9°

The direction of the average velocity is 61.9° south of west.

c) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it took to cover that distance (t). In total, you traveled (490 m + 920 m) 1410 m in 121 s, then the average speed will be:

a.s = d/t

a.s = 1410 m / 121 s

a.s = 11.7 m/s

Your average speed during the trip is 11.7 m/s

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Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The value is F_j  =  550\ N

Explanation:

From the question we are told that

   The length of the stretcher is  d =  2.0 \  m

    The weight of the stretcher is W  =  100 \  N

    The weight for Wayne is  W_w =  800 \ N

     The distance of  center of gravity for Wayne from Chris is c_w = 75 cm  =  0.75 \ m

Generally taking moment about the first end where Chris is

         F_j *  d              => upward moment

Here F_j is the force applied by Jamie

Generally  taking moment about the second end where Jamie is

      W *  ( \frac{d}{2} ) +  W_w * (d - c_w)      => downward moment

Generally at equilibrium , the upward moment is equal to the downward moment

     F_j *  d = W *  ( \frac{d}{2} ) +  W_w * (d - c_w)

=>   F_j *  2  = 100 *  ( \frac{ 2}{2} ) +  800 * (2 - 0.75)

=>    F_j  =  550\ N

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2 years ago
A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc
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Answer:

E=88200\ J

Explanation:

Given:

  • mass of car, m=1500\ kg
  • distance of skidding after the application of brakes, d=15\ m
  • coefficient of kinetic friction, \mu_k=0.4

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

f=\mu_k.N

where N = normal reaction by ground on the car

f=0.4\ties 1500\times 9.8

f=5880\ N

<em>Now from the work-energy equivalence:</em>

E=f.d

E=5880\times 15

E=88200\ J is the dissipated energy.

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Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

v^{2}=u^{2}+2as

0^{2}=35^{2}- 2 \times 9.8 \times h

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

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The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

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Answer:

Explanation:

a )

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So rate which energy is coming out of coal per second

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efficiency = (800 / 2025) x 100

= 39.5 % .

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