1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
A train at rest emits a sound at a frequency of 1057 Hz. An observer in a car travels away from the sound source at a speed of 2
il63 [147K]

Answer:

993.52 Hz

Explanation:

The frequency of sound emitted by the stationery train is 1057 Hz.

The car travels away from the train at 20.6 m/s.

The frequency the observer hears is given by the formula:

f_o = \frac{v - v_o}{v}f

where v = velocity of sound = 343 m/s

vo = velocity of observer

f = frequency from source

This phenomenon is known as Doppler's effect.

Therefore:

f_o = \frac{343 - 20.6}{343} * 1057\\ \\f_o = 322.4 / 343 * 1057\\\\f_o = 993.52 Hz

The frequency heard by the observer is 993.52 Hz.

4 0
4 years ago
A 5 kg ball of clay is moving with a speed of 25 m/s directly toward a 10 kg ball of clay which is at rest. The two clay balls c
Dafna11 [192]

Answer:

8.3m/s

Explanation:

Given parameters:

mass of clay ball  = 5kg

Speed of clay ball  = 25m/s

mass of clay ball at rest  = 10kg

speed of clay ball at rest  = 0m/s

Unknown:

Velocity after collision  = ?

Solution:

 Since the balls stick together, this is an inelastic collision:

   m1v1 + m2v2  = v(m1 + m2)

  5(25) + 10(0)  =  v (5 + 10)

         125 = 15v

           v  = 8.3m/s

7 0
3 years ago
A grasshopper jumps at a 65.0 degree angle at 5.42m/s. At what time does it reach its maximum height?
crimeas [40]
When the grasshoppers vertical velocity is exactly zero.
v = -g•t + v0.
v: vertical part of velocity. Is zero at maximum height.
g: 9.81
t: time you are looking for
v0: initial vertical velocity
Find the vertical part of the initial velocity, by using the angle at which the grasshopper jumps.
6 0
3 years ago
Read 2 more answers
The most reactive metals are the?
photoshop1234 [79]
I Know B for sure trust me had this quiz
7 0
3 years ago
Read 2 more answers
If you attach a 100 g mass to the spring whose data are shown in the graph, what will be the period of its oscillations?
ycow [4]

if for a force of 0.5 N we have a displacement of -0.02m we can calculate the elastic constant(k) with the formula F=-kx(F=0.5N x=-0.02)

k=F/x k=0.5/0.02=25N/m

now we can calculate the period by the formula

T=2π√(m/k)

in the mass we convert grams to kilograms so 100g=0.1kg

T=6.28√(0.1/25)⇒T=0.39seconds

8 0
3 years ago
Read 2 more answers
Other questions:
  • Suppose the sun were to suddenly disappear. what would happen to the orbital path of earth? it would stay the same, but the eart
    13·2 answers
  • Which volcanic hazard can block the sunlight and temporarily cool the Earth’s surface?
    13·2 answers
  • The temperature of an object is a measure of the average thermal energy of the molecule in the object true or false
    11·1 answer
  • Tightness of the sciatic nerve can result in lower back pain true or false
    6·1 answer
  • a 2.00 kg friction-less block is attached to an ideal spring with force constant 315 N/m.Initially, the spring is neither stretc
    11·1 answer
  • You can use one of your 5 senses to make _____ during an inquiry activity.
    14·2 answers
  • A woman pushes a 3.2kg textbook on a table as she walks forward 3.5 meters. How much work does the vertical component of the for
    11·1 answer
  • During a solar eclipse our tiny moon totally blocks the sun because of .....
    12·1 answer
  • If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average
    7·1 answer
  • •<br> 2.<br> particles penetrate through paper.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!