You push a box 3.5 m across the floor with a constant force of 12.4 N. How much work do you do on the box?

Mr. Llama walked from his house to the bus stop. The bus stop is 2 miles from his house. He returned back to his house from the

Hoochie [10]

Answer:

Displacement of Mr. Llama: Option D. 0 miles.

Explanation:

The magnitude of the displacement of an object is equal to the distance between its final position and its initial position. In other words, as long as the initial and final positions of the object stay unchanged, the path that this object took will not affect its displacement.

For Mr. Llama:

Final position: Mr. Llama's house;
Initial position: Mr. Llama's house.

The distance between the final and initial position of Mr. Llama is equal to zero. As a result, the magnitude of Mr. Llama's displacement in the entire process will also be equal to zero.

7 0

4 years ago

I need help with the last question

Mariana [72]

The best object to show the Earth's true appearance is something round and smooth. Since the earth is only slightly oblate, it doesn't matter based on its size, and you would use a pool ball as the perfect scale model, because it is very round and smooth.

3 0

3 years ago

Read 2 more answers

A guitar string is supposed to have a fundamental frequency 256 Hz. It currently has a fundamental frequency 248 Hz when the str

umka2103 [35]

Answer:

The tension to bring the guitar string into tune is 372.95 Hz.

Explanation:

Given;

current frequency, f₁ = 248 Hz

current tension, T₁ = 350 N

fundamental frequency, f₂ = 256

The tension on the string to bring the guitar string into tune is calculated as;

$v = \sqrt{\frac{T}{\mu} } \\\\f\lambda = \sqrt{\frac{T}{\mu} } \\\\f^2\lambda^2 = \frac{T}{\mu} \\\\f^2 = \frac{T}{\mu \lambda^2}\\\\let \ {\mu \lambda^2} = k\\\\f^2 =\frac{T}{k} \\\\k = \frac{T}{f^2} \\\\\frac{T_1}{f_1^2} = \frac{T_2}{f_2^2}\\\\T_2 = \frac{T_1 f_2^2}{f_1^2} \\\\T_2 = \frac{350 \times 256^2}{248^2} \\\\T_2 = 372.95 \ Hz$

Therefore, the tension to bring the guitar string into tune is 372.95 Hz.

3 0

3 years ago

TP-10 Which of the following is considered a safe refueling practice?

erik [133]

Answer:

The correct option is;

Sending all passengers below while refueling

Explanation:

Safety tips for refueling a boat includes;

1) People that are not involved in the refueling of the boat should be made to leave the area

2) The engines, electronics, ignition system and open flames should be off

3) Ensure the boat is properly secured to the dock

4) Do not use the hands-free clips and try not to be distracted

5) Ensure not to completely fill the fuel tank as fuel has the tendency to expand as its temperature rises. The tank should have about 90% filling

7 0

3 years ago

The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto

igor_vitrenko [27]

a) The angular velocity of the electron is $4.12\cdot 10^{16} rad/s$

b) The number of revolutions per second is $6.54\cdot 10^{15}$ rev/s

c) The centripetal acceleration of the electron is $8.98\cdot 10^{22} m/s^2$

Explanation:

a)

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

$\omega = \frac{v}{r}$

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

$v=2.18 \cdot 10^6 m/s$

$r=5.29\cdot 10^{-11} m$

Therefore, the angular velocity is

$\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s$

b)

The period of revolution of the electron is given by

$T=\frac{2\pi}{\omega}$

where

$\omega = 4.12\cdot 10^{16}rad/s$ is the angular velocity

Substituting,

$T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s$

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

$f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15}$ rev/s