answer.
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
Ronch [10]
2 years ago
8

A water-skier is moving at a speed of 12.0 m/s . When she skis in the same direction as a traveling wave, she springs upward eve

ry 0.600 s because of the wave crests. When she skis in the direction opposite to that in which the wave moves, she springs upward every 0.500 s in response to the crests. The speed of the skier is greater than the speed of the wave. Determine (a) the speed and (b) the wavelength of the wave.
Physics
1 answer:
Kisachek [45]2 years ago
7 0

Answer:

Mechanical waves require a medium to transfer their energy.

What are examples of mechanical waves?

Explanation:

Mechanical waves require a medium to transfer their energy.

What are examples of mechanical waves?

You might be interested in
Suppose that air resistance cannot be ignored. For the position at which the person has jumped from the platform and the cord re
Kamila [148]

Answer:

The stretch cord stores potential energy as a result of stretching but due to kinetic energy, it will move back to its original state. Since air resistance is not being ignored in this case, it will experience a slight delay in stretching at first.

Explanation:

In case, where air resistance is being ignored the stretch cord will stretch as it normally does.              

  • Air resistance is a force that any object experiences as a result of its motion through the air.  

There are various factors that affect air resistance like speed, the density of air, area, the shape of an object etc. Meanwhile, the density of air changes with temperature or altitude. <em>Hence this force is not constant but is thought to be constant during short time frames.  </em>

6 0
2 years ago
In some amazing situations, people have survived falling large distances when the surface they land on is soft enough. During a
OleMash [197]

Answer:

-3413 ft/s2

Explanation:

We need to know the velocity with which he landed on the snow.

He 'dropped' from 512 feet. This is the displacement. His initial velocity is 0 and the acceleration of gravity is 32 ft/s2.

We use the equation of mition

v^2 = u^2 + 2as

v and u are the initial and final velocities, a is the acceleration and s is the displacement. Putting the appropriate values

v^2 = 0^2 + 2\times32\times512

v = \sqrt{2\times32\times512} = 128\sqrt{2}

This is the final velocity of the fall and becomes the initial velocity as he goes into the snow.

In this second motion, his final velocity is 0 because he stops after a displacement of 4.8 ft. We use the same equation of motion but with different values. This time, u=128\sqrt{2}, v = 0 and s = 4.8 ft.

0^2 = (128\sqrt{2})^2 + 2a\times4.8

a = -\dfrac{2\times128^2}{2\times4.8} = -3413

Note that this is negative because it was a deceleration, that is, his velocity was decreasing.

5 0
2 years ago
A frog jumps up at time t = 0.
valina [46]
Is there a picture that I can see
6 0
2 years ago
Read 2 more answers
At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is
Nastasia [14]
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
     = 314.37 lb/ft²
     = 314.37/144 = 2.18 lb/in²

Answer: 2.2 psi

5 0
2 years ago
A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho
Gnoma [55]

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)

80=5(50-x)

80=250-5x

5x=170

x=\frac{170}{5}

x=34 cm

4 0
2 years ago
Other questions:
  • A 2 kg bar of metal weighs about 4.4 pounds. Approximately how much does it weigh in newtons?
    10·2 answers
  • If a bowl of fruit is sitting on a table in a state of rest, what is the net force acting on it?
    15·2 answers
  • The pupil of an eagle’s eye has a diameter of 6.0 mm. Two field mice are separated by 0.010 m. From a distance of 204 m, the eag
    5·1 answer
  • Compare the density, weight, mass, and volume of a pound of gold to a pound of iron on the surface of Earth.
    11·1 answer
  • A child's toy is suspended from the ceiling by means of a string. The Earth pulls downward on the toy with its weight force of 8
    5·1 answer
  • A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle
    14·1 answer
  • All of these are examples of pseudopsychology EXCEPT:
    15·2 answers
  • Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and
    7·1 answer
  • One of the Lady Spartans was falling to the ground after
    15·1 answer
  • A 16 g ball at the end of a 1.4 m string is swung in a horizontal circle. It revolves once every 1.09 s. What is the magnitude o
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!