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klemol [59]
3 years ago
15

It’s a sunny Saturday afternoon and you are walking around the lake by your house, enjoying the last few days of summer. The sid

ewalk surrounding the perimeter of the circular lake is crowded with walkers and runners. You then notice a runner approaching you wearing a T-shirt with writing on it. You read the first two lines, but are unable to read the third line before he passes. You wonder, ”Hmmm, if he continues around the lake, I bet I’ll see him again but I should anticipate the time when we’ll pass again.”
You look at your watch and it is 5:07pm. You estimate your walking speed at 3 m/s and the runner’s speed to be about 14 m/s. You also estimate that the diameter of the lake is about 2
miles. At what time should you expect to read the last line of the t-shirt?
Physics
1 answer:
Citrus2011 [14]3 years ago
7 0

The anticipated time when he will appear again is 5:17 pm

The given parameters;

your speed, V_a = 3 m/s

the runner's speed, V_b = 14 m/s

the diameter of the lake, d = 2 miles = 3218.69 meters

  • Let the anticipated time when he will appear again = t

The circumference of the lake is calculated as;

C = \pi d\\\\C = 3.142 \times 3218.69 = 10,113.12 \ m

Apply concept of relative velocity to determine the time, in which he will appear again.

By the time he appears again;

the distance you moved + distance he moved = circumference of the circle

V_at + V_bt = 10, 113.12\\\\(V_a + V_b)t = 10,113.12\\\\(3 + 14) t = 10,113.12\\\\17t = 10,113.12\\\\t = \frac{10,113.12}{17} \\\\t = 594.89 \ s = 9.92 \ \min \ \approx 10 \ \min

t\  \approx \ \ 5:17 \ pm

Thus, the anticipated time when he will appear again is 5:17 pm

Learn more here: brainly.com/question/20662992

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Figure 3 shows a bicycle of mass 15 kg resting in a vertical position, with the front and back
Vinil7 [7]

Explanation:

There are three forces on the bicycle:

Reaction force Rp pushing up at P,

Reaction force Rq pushing up at Q,

Weight force mg pulling down at O.

There are four equations you can write: sum of the forces in the y direction, sum of the moments at P, sum of the moments at Q, and sum of the moments at O.

Sum of the forces in the y direction:

Rp + Rq − (15)(9.8) = 0

Rp + Rq − 147 = 0

Sum of the moments at P:

(15)(9.8)(0.30) − Rq(1) = 0

44.1 − Rq = 0

Sum of the moments at Q:

Rp(1) − (15)(9.8)(0.70) = 0

Rp − 102.9 = 0

Sum of the moments at O:

Rp(0.30) − Rq(0.70) = 0

0.3 Rp − 0.7 Rq = 0

Any combination of these equations will work.

3 0
3 years ago
How much force would you need to accelerate a 528 kg motorcycle at 3 m/s2
mart [117]

Answer:

<h2>1584 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 528 × 3

We have the final answer as

<h3>1584 N</h3>

Hope this helps you

4 0
3 years ago
A satellite has a mass of 5850 kg and is in a circular orbit 4.1 x10 to the 5th power m above the surface of a planet. The perio
koban [17]

Answer:

 W = 24.28 kN

Explanation:

given,

Mass of satellite = 5850 Kg

height , h = 4.1 x 10⁵ m

Radius of planet = 4.15 x 10⁶ m

Time period = 2 h

                    = 2 x 3600 = 7200 s

Time period of satellite

T = \dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}

R is the radius of planet

h is the height of satellite

T^2 = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{g}}

now calculation of acceleration due to gravity

g = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{T^2}}

g = \dfrac{4\pi^2}{(4.15\times 10^6)^2}\ {\dfrac{(4.15\times 10^6+4.1\times 10^5)^3}{(7200)^2}}

g = 4.15 m/s²

True weight of satellite

W = m g

W = 5850 x 4.15

W = 24277.5 N

 W = 24.28 kN

True weight of the satellite is   W = 24.28 kN

8 0
3 years ago
How to find net force
nexus9112 [7]
 net force is mass multiplied by acceleration. hope this helps
3 0
3 years ago
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Rocks that fall out of the sky and land on Earth are called _____________. Question 7 options: A.meteorites B.asteroids C.comets
kenny6666 [7]
They’re called meteorites
3 0
3 years ago
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