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3 years ago

14

How is potential energy unique?

1 answer:

ahrayia [7]3 years ago

3 0

Answer:

is uniquely defined.

Explanation:

The gravitational potential energy gained by an object being lifted is equal to the work done on the object by an applied force, which exactly cancels gravity. The applied force does positive work. When the object is being lifted, the gravitational force does negative work.

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Five different forces act on an object. Is it possible for the net force on the object to be zero?

bazaltina [42]

No because there must be an even # if their is an even amount one of the forces isn’t being cancelled

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2 years ago

What is this answer

Katyanochek1 [597]

Answer: an ectomorph is a body type that struggles to gain weight and muscle

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2 years ago

Read 2 more answers

A lawyer drives from her home, located 88 milesmiles east and 1818 milesmiles north of the town courthouse, to her office, lo

myrzilka [38]

Answer:

the distance between the lawyer's home and her office is 124 miles

Explanation:

given information:

first lets assume that

x-axis (west = positive, east = negative)

y-axis (north = positive, south = negative)

thus,

distance of the house = (-88,18)

distance of the office = (13, -54)

thus, the distance between the lawyer's home and her office

R = √(x₂ - x₁)² + (y₂ - y₁)²

= √(13 - (-88))² + (-54 -18)²

= 124 miles

7 0

3 years ago

You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli

FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

$f(x,t)=Acos(kx-\omega t)$

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

$\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}$

$k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}$

$T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m$

b) Hence, the wave function is:

$f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)$

c) for x=3m you have:

$f(3,t)=0.075cos(1.047*3-12.56t)$

d) the speed of the medium:

$\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)$

you can see the velocity of the medium for example for x = 0:

$v=\frac{df}{dt}=13.15cos(12.56t)$

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2 years ago

1. li took 2 } seconds for a car's vclocity to change from 20 m/s to 15 m/s. The mass of the car was 1370 kg. What force was req

Pani-rosa [81]

Answer:

Do you still need help??

Explanation:

3 0

2 years ago