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2 years ago

A student moves a box across the floor by exerting 56.7 N of force and doing 195 J of

Physics

1 answer:

Paha777 [63]2 years ago

8 0

Answer:

A. 3.4 m

Explanation:

Given the following data;

Force = 56.7N

Workdone = 195J

To find the distance

Workdone is given by the formula;

$Workdone = force * distance$

Making "distance" the subject of formula, we have;

$Distance = \frac {workdone}{force}$

Substituting into the equation, we have;

$Distance = \frac {195}{56.7}$

Distance = 3.4 meters.

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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:

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

7 0

2 years ago

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

wel

Answer:

Normal force = 8.75 N

Explanation:

given,

frictional force between the steel spatula and the Teflon frying pan=0.350 N

coefficient of friction between material =0.04

normal force = ?

using formula,

Frictional force = coefficient of friction × normal force

$normal\ force = \dfrac{Frictional\ force}{coefficient\ of\ friction}$

$normal\ force = \dfrac{0.350}{0.04}$

Normal force = 8.75 N

8 0

2 years ago

Balance the following chemical reactions: question attached hellpp

Mice21 [21]

1 Ti + 2 Cl2 —> 1 TiCl4

3 0

2 years ago

3. In 1989, Michel Menin of France walked on a tightrope suspended under a

Tamiku [17]

Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

Let the distance of Menin's pocket to the ground be x

Let the distance between Menin's pocket to the balloon be y

Hence, x=3150-y------1

Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

3070m

Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

6 0

3 years ago

A gravitational force of 2.54 x 10^-10 is exerted by two objects that are 0.25 m apart. If the first object has a mass of 0.35 k

Murrr4er [49]

Answer :

what is the answer options?

Explanation:

3 0

1 year ago