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3 years ago

What is the definition of a reference point

Physics

1 answer:

riadik2000 [5.3K]3 years ago

8 0

A reference point would be something used to understand or judge or describe a certain topic in discussion.

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A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (

Anika [276]

Answer:

$a=9.8 rad/s^{2}$

Explanation:

Torque, $\tau$ is given by

$\tau=Fr$ where F is force and r is perpendicular distance

$R=0.5Lcos\theta$ where $\theta$ is the angle of inclination

Torque, $\tau$ can also be found by

$\tau=Ia$ where I is moment of inertia and a is angular acceleration

Therefore, Fr=Ia and F=mg where m is mass and g is acceleration due to gravity

Making a the subject, $a=\frac {Fr}{I}=\frac {mgr}{I}$ and already I is given as

$I=\frac {mL^{2}}{3} and r is 0.5Lcos\theta$ hence

$a=\frac {0.5mgLcos\theta}{1/3 mL^{2}}$

$a=\frac {3gcos\theta}{2L}$

Taking g as 9.81, $\theta$ is given as 37 and L is 1.2

$a=\frac {3*9.81cos37}{2*1.2}=9.7932679419$

$a=9.8 rad/s^{2}$

4 0

3 years ago

The smallest known galaxy, Segue 2, has an approximate radius of 1.05 × 1015 kilometers. Use the conversion factors 1 light-year

scoundrel [369]

( 1.05 x 10¹⁵ km ) x ( 1 LY / 9.5 x 10¹² km ) x ( 1 psc / 3.262 LY ) =

(1.05) / (9.5 x 3.262) x (km · LY · psc) / (km · LY) x (10¹⁵⁻¹²) =

(0.03388) x (psc) x (10³) =

33.88 parsecs

5 0

3 years ago

What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?

viktelen [127]

Answer: 27.21 V

Explanation:

The electric potential $V_{E}$ due to a point charge is expressed as:

$V_{E}=k\frac{q}{r}$

Where:

$k=9(10)^{9}\frac{Nm^{2}}{C^{2}}$ is the electric constant

$q=1.6(10)^{-19}C$ is the electric charge of the hydrogen nucleus, which is positive

$r=5.292(10)^{-11}m$ is the distance

Rewritting the equation with the known values:

$V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}$

Finally:

$V_{E}=27.21 V$

5 0

3 years ago

6. Velocity is a ____________. a) vector b) value c) arrow d) none of these

Olenka [21]

the speed of something in a given direction. so i think none of these

4 0

3 years ago

Read 2 more answers

Gulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells.

kumpel [21]

Answer:

v = 17.71 m / s

Explanation:

We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity

v² = v₀² - 2 g (y -y₀)

v² = 0 - 2g (y -y₀)

when it hits the stone the height is zero and part of the height of the seagull I

v² = 2g y₀

v = Ra (2g i)

let's calculate

v =√ (2 9.8 16)

v = 17.71 m / s

8 0

3 years ago