PHYSICS COLLEGE

According to the World Flying Disk Federation, the world distance record for a flying disk throw in the men’s 85-years-and-older category is held by Jack Roddick of Pennsylvania, who on July 13, 2007, at the age of 86, threw a flying disk for a distance of 54.0 m. If the flying disk was thrown horizontally with a speed of 13.0 m/s, how long did the flying disk remain aloft? (Jack Roddick was also a physics teacher! Read more about him at

Answers

Answer 1

Answer:

Answer:

t = 4.15 seconds

Explanation:

It is given that,

Distance traveled by a flying disk, d = 54 m

The speed at which it was thrown, v = 13 m/s

We need to find the time for which the flying disk remain aloft. Let the distance is d. We know that, speed is equal to the distance covered divided by time. So,

$t=(d)/(v)\n\nt=(54\ m)/(13\ m/s)\n\nt=4.15\ s$

Hence, for 4.15 seconds the flying disk remain aloft.

Related Questions

An earthquake on the ocean floor produced a giant wave called a tsunami. The tsunami traveled through the ocean and hit a remote island, causing a lot of damage. Is the water that hit the island the same water that was above the earthquake on the ocean floor?A No, the water from above the earthquake stayed in the same place and only the energy was transferred.B No, the energy in the wave pushed the water particles from above the earthquake in the opposite direction.C Yes, the water particles moved toward the island while the energy remained above the earthquake.
An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Determine the speed at this time and the maximum height at which it reaches.
The electric field just above the surface of the charged drum of a photocopying machine has a magnitude E of 2.5 × 105 N/C. What is the surface charge density on the drum, assuming that the drum is a conductor?
Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.
The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7 T · N/A and the speed of light is 2.99792 × 108 m/s. Find the magnitude of Em for the electromagnetic waves at the top of the atmosphere. Answer in units of N/C.

How do you perceive the importance of online safety security ethics and etiquette

Answers

Answer:

is called online etiquette, or netiquette for short.

...

Here are some tips to help you get started:

Create Complex Passwords.

Boost Your Network Security.

Use a Firewall.

Click Smart.

Be a Selective Sharer.

Protect Your Mobile Life.

Practice Safe Surfing & Shopping.

Keep up to date.

A helicopter is hovering above the ground. Jim reaches out of the copter (with a safety harness on) at 180 m above the ground. A package is launched upward, from a point on a roof 10 m above the ground. The initial velocity of the package is 50.5 m/s. Consider all quantities as positive in the upward direction. Does Jim Bond have a chance to catch the package? (calculate how high will it go)

Answers

Answer:

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

Explanation:

Hi there!

The equation of height and velocity of the package are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the package at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).

v = velocity of the package at a time t.

First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:

v = v0 + g · t

0 = 50.5 m/s - 9.8 m/s² · t

Solving for t:

-50.5 m/s / -9.81 m/s² = t

t = 5.15 s

Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:

h = h0 + v0 · t + 1/2 · g · t²

h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²

h = 140 m

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

While testing at 30 feet below the surface in Lake Minnetonka, with the sub stopped and in equilibrium, one of the students aboard the sub drops a hammer that goes through the hull of the submarine, and sticks out of the submarine handle first. When this happens, a seal forms immediately around the handle, so that no water enters the sub. What is the new equilibrium position for the sub?

Answers

Answer:

Explanation:

The equilibrium position of the sub is at the surface of the lake

When a body falls freely under gravity, then the work done by the gravity is ___________

Answers

Answer: positive

Explanation:

Gravity can be defined as the force with which the body is attracted towards the center of the earth, or towards any other body. If the force acting on the body is in the direction of displacement then the word done by the applicable force is positive. This causes the free fall of the ball under the influence of gravity is also positive.

Two violin strings are tuned to the same frequency 294 H. The tension in one string is then decreased by 2.0%. What will be the beat frequency heard when the two strings are played together?

Answers

The beat frequency heard when the two strings are played together is 2.95 Hz.

Given data:

The tuning frequency of the violin is, f = 294 Hz.

Decrement in the tension is, 2 %.

Since, tension is reduced at the rate of 2%. Then the new magnitude of tension on the string is,

T = (100 - 2 )/100

T = 0.98

Then the expression for the beat frequency heard when the two strings are played together is given as,

$f_(b)=f -(√(T * f))$

Solving as,

$f_(b)=294-(√(0.98 * 294))\n\nf_(b)=2.95\;\rm Hz$

Thus, we can conclude that the beat frequency heard when the two strings are played together is 2.95 Hz.

Learn more about the beat frequency here:

brainly.com/question/20347530

Answer:

Beat frequency together = 2.95 Hz (Approx)

Explanation:

Given:

Frequency (F) = 294 H

Decrease in tension = 2%

Find:

Beat frequency together

Computation:

Tension = (100 - 2) / 100

Tension (T) = 0.98

Beat frequency together = Frequency (F) - (√T × F)

Beat frequency together = 294 - (√0.98 × 294)

Beat frequency together = 2.95 Hz (Approx)

We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 129 ∘C. The gas expands and, in the process, absorbs an amount of heat equal to 1180 J and does an amount of work equal to 2160 J . What is the final temperature Tfinal of the gas? Use R = 8.3145 J/(mol⋅K) for the ideal gas constant.