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2 years ago

Determine the torque produced by a perpendicular force of 75 N at the end of a 0.2 m wrench.

Physics

2 answers:

nataly862011 [7]2 years ago

8 0

Answer:

Explanation:

the answer is C.)

trasher [3.6K]2 years ago

7 0

Answer:c

Explanation:

Because I took uty

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Which best describes the direction of heat?

Gekata [30.6K]

Answer:

MRCORRECT has answered the question

Explanation:

According to second law of thermodynamic,heat flows from higher to lower temperature region. Heat is the transfer of energy from an object at higher temperature to an object at a lower temperature. so energy flows from warmer to colder objects

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3 years ago

You ran a power plant what could you do to make this generator produce electricity

DIA [1.3K]

You could use a magnetic generator or you could use hydraulic power

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3 years ago

A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica

Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed $\omega =40rad/s$

mass of $m_2=0.8 kg$ dropped at r=0.3 m from center

let $\omega _2$ be the final angular velocity of cylinder

Conserving Angular momentum

$L_1=L_2$

$\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2$

$\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2$

$26.01\times 40=26.082\times \omega _2$

$\omega _2=39.88 rad/s$

3 0

3 years ago

Cmon Guys help your pal abigail

NemiM [27]

13-16 that is where they’re located at

8 0

3 years ago

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

3 years ago