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Misha Larkins [42]

2 years ago

13

You have found that the net force exerted on the cylinder depends on several different independent variables in different ways.

how can you combine those separate dependencies into one equation relating the net force to the relevant independent variables

1 answer:

postnew [5]2 years ago

4 0

The net force on the cylinder depends on several independent variables such as mass, acceleration, velocity, time, etc, and it is given as F(net) = ma.

<h3>Net force exerted on the cylinder</h3>

The net force on the cylinder is calculated from Newton's second law of motion.

F(net) = ma

F(net) = m(v/t)

where;

m is mass of the cylinder
a is acceleration of the cylinder
v is velocity of the cylinder
t is time of motion of the cylinder

Thus, the net force on the cylinder depends on several independent variables such as mass, acceleration, velocity, time, etc, and it is given as F(net) = ma.

Learn more about net force here: brainly.com/question/11556949

#SPJ1

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A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav

Agata [3.3K]

Answer:

The tension in the string is $T = 1.49*10^{-6}N$ .

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For a string with tension $T$ and linear density $\mu_d$ carrying a transverse wave at speed $v$ it is true that

$v = \sqrt{\dfrac{T}{\mu_d} }$

solving for $T$ we get:

$T = \dfrac{v^2}{\mu_d}.$

Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is

$v =\dfrac{7.4*10^{-3}m}{0.88s} \\\\v = 8.4*10^{-3}s$

And it's linear density (mass per unit length) is

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Therefore, the tension in the cord is

$T = \dfrac{(8.4*10^{-3}m/s^2)^2}{47.3kg/m}.$

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Find the net force and acceleration. 15 points. Will give brainliest!

gladu [14]

Answer:

$\boxed{F_{net} = 28.7 \ N}$

$\boxed{a = 2.1 \ m/s^2}$

Explanation:

<u><em>Finding the net force:</em></u>

<u><em>Firstly , we'll find force of Friction:</em></u>

$F_{k} = (micro)_{k}mg$

Where $(micro)_{k}$ is the coefficient of friction and m = 13.6 kg

$F_{k} = (0.16)(13.6)(9.8)\\$

$F_{k} = 21.32 \ N$

<u><em>Now, Finding the net force:</em></u>

$F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\$

$F_{net} = 28.7 \ N$

<u><em>Finding Acceleration:</em></u>

$a = \frac{F_{net}}{m}$

$a = \frac{28.7}{13.6}$

$a = 2.1 \ m/s^2$

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3 years ago