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Archy [21]
2 years ago
7

Jamison walks to a friends house. He walks 1000 meters north, then realizes he walked to far so he turns around and walks 200 me

ters south. The entire walk takes him 12 minutes. What is his average velocity in meters per second.
Physics
1 answer:
olasank [31]2 years ago
4 0

We will find that his average velocity is 1.11 m/s north-wise.

<h3>How to get the average velocity?</h3>

We define average velocity as the quotient between the displacement and the time it takes to make that displacement.

We know that he starts at the point 0m, where this measures distance in the north direction.

Then he moves 1000 meters north, so the new position is 1000m.

Then he moves 200 meters south, so the new position is:

1000m - 200m = 800m.

The displacement is equal to the difference between the final position (800m) and the initial position (0m).

D = 800m - 0m = 800m.

And we know that it takes 12 minutes to walk that distance. Because we want the average velocity in meters per second, we write:

12 minutes = 12*(60 seconds) = 720s

Then the average velocity is just:

AV = (800m)/(720s) = 1.11 m/s

And the direction of this average velocity is north-wise.

If you want to learn more about average velocity, you can read:

brainly.com/question/4931057

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5. A cable is attached 32.0 m from the base of a flagpole that is about to
soldi70 [24.7K]

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, F_v = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

F_v + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

∴ W × d/2 ×cos(60.0°) ≈  4513.093·d  

W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N

R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373

R ≈ 30,382.373 N

Taking moment about the point of attachment of the cable to the ground, we have;

W × ((d/2) × cos(60.0°) + 32) = R × 32

∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281

d = 2 × 43.71281 ≈ 87.43

The length of the flagpole, d ≈ 87.43 m

7 0
3 years ago
With good tires and breaks, a car traveling 50 mi/hr on dry pavement can travel 400 ft when the driver reacts to something he se
Bogdan [553]

Answer:6.72 m/s^2

Explanation:

Given

initial velocity u=50 mi/hr\approx 73.33 ft/s

Distance traveled before stopping s=400 ft

using equation of

v^2-u^2=2as

where v=final velocity

u=initial Velocity

a=acceleration

s=displacement

v=0 as car stops after travelling 400 ft

0-73.33^2=2\times a\times 400

a=-6.72 m/s^2

negative sign indicates it is deceleration

3 0
3 years ago
Now, let’s see what happens when the cannon is high above the ground. click on the wheel of the cannon, and drag it upward as fa
Setler79 [48]
The equation for range is:

R = v₀²sin(2θ)/g
To find the maximum R, differentiate the equation and equate to zero. The solution is as follows:

dR/dθ = (v₀²/g)(sin 2θ)
dR/dθ = (v₀²/g)(cos 2θ)(2) = 0
cos 2θ = 0
2θ = cos⁻¹ 0 = 90
θ = 90/2
<em>θ = 45°</em>
5 0
3 years ago
What velocity will it take to slingshot a planet with 9.8m/s gravity
Yuri [45]
It depends on how close you get to it. Remember that its gravity decreases as you get farther from it.
3 0
3 years ago
Two students try to move a heavy box. One pushes with the force of the 80N while the other pulls with a force of 40N in the same
alekssr [168]

Answer:

<em>Good Luck!</em>

Explanation:

Force applied by first student (F1) = 80 N

Force applied by the second student (F2) = 40 N

Displacement (d) = 10 m

Work done by first student (W1) = F1d = 80*10 = 800 J

Work done by second student (W2) = F2d = 40*10 = 400 J

Hence, the work done by the pushing student is 800 J and that by pulling student is

<h2> <u><em>400 J</em></u></h2>
5 0
3 years ago
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