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2 years ago

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An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of mass M is attac

hed to the other end of the rod. The pendulum thus formed is hung by the hoop onto a revolving shaft. The coefficient of friction between the shaft and the hoop is µ. Find the equilibrium angle between the rod and the vertical.

1 answer:

Roman55 [17]2 years ago

8 0

Answer:

$arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})$

Explanation:

By the Law of Sines,

$sin \theta = \frac{sin \phi R}{ l + R}$

From Newton's Law,

$mg = N\sqrt{\mu^2+1}$

And the last equation again from Newton's Law,

$\mu N = mgsin\phi$

Then if we collect all equations together,

$\mu N = mgsin\phi = N\sqrt{\mu^2+1}sin\phi\\$

$sin\theta = \frac{\mu R}{ (l + R)\sqrt{\mu^2+1}}$

Thus,

$\theta = arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})$

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A red ball, initially at rest, is simultaneously hit by a blue ball traveling from west to east at 5 m/s and a green ball travel

andrezito [222]

Answer:

C. Between North and West

Explanation:

Since all have equal masses and the red ball and green ball are moving in south and east direction, the blue ball would most likely be moving between the north and West direction.

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2 years ago

A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with 1010 excess electrons

liraira [26]

When the metals touch together, half the charge of the charged metal flows to the other because the electrons all repel each other. Therefore this also means that each metal ball contains the same amount of electrons. Each ball has 5^10 electrons, this is equivalent to a total charge of:

Q1 = Q2 = (1.602 * 10^-19 coulombs / electron) 5^10 electrons = Q

Q = 1.564 * 10^-12 C

Now using the Coulombs law to find for the electric force:

F = k q1 q2 / r^2 = k (Q)^2 / r^2

where k is a contant = 9 * 10^9 N m^2 / C^2

r = the distance of the two metals = 0.2 m

So,

F = (9 * 10^9 N m^2 / C^2) (1.564 * 10^-12 C)^2 / (0.2 m)^2

F = 5.51 * 10^-13 N

Since the two metals repel therefore they are the one which exerts the force hence the magnitude must be negative:

<span>F = - 5.51 * 10^-13 N</span>

7 0

2 years ago

A bird is flying to the right when a gust of wind causes the bird to accelerate leftward at 0.5\,\dfrac{\text m}{\text s^2}0.5 s

Licemer1 [7]

The initial velocity of the bird before the gust of wind : <u>4 m/s</u>

<h3>Further explanation</h3>

An equation of uniformly accelerated motion

$\large {\boxed {\bold {x = xo + vo.t + \frac {1} {2} at ^ 2}}}$

v = vo + at

Vt² = vo² + 2a (x-xo)

x = distance on t

vo/vi = initial speed

vt/vf = speed on t /final speed

a = acceleration

Acceleration is a change in speed within a certain time interval

a = Δv /Δ t

$\displaystyle a=\frac{v2-v1}{t2-t1}$

Let complete the task :

A bird is flying to the right when a gust of wind causes the bird to accelerate leftward at 0.5 m/s² for 3 s. After the wind stops, the bird is flying to the right with a velocity of 2.5 m/s.

Assuming the acceleration from the wind is constant, what was the initial velocity of the bird before the gust of wind?

we can use formula :

vf = vi + a.t

vf = final velocity = 2.5 m/s

vi = asked

a = - 0.5 m/s²(leftward=negative)

t = 3 s

then :

$\displaystyle 2.5=vi-0.5.3\\\\vi=2.5+1.5\\\\vi=\boxed{\bold{4\frac{m}{s}}}$

<h3>Learn more</h3>

The car reach the end of the road

brainly.com/question/13750982

Keywords: uniformly accelerated motion, distance, velocity, acceleration

#LearnwithBrainly

7 0

2 years ago

When the 3.0 kg cylinder fell 500 m, the final temperature of the water was

miss Akunina [59]

Answer:

Change in temperature =7.14° C

Explanation:

Assuming all potential energy of cylinder is converted into heat energy to increase the temperature .

Work done by gravity in first case =m *g*h= 3*10*500=15000

Let quantity of water be 1 kg at 0°C.

Also specific heat capacity of water is =4200 Joule

So potential energy is converted to m*4200*T(1) =15000

T(1)=3.57

Work done by gravity in first case =m *g*h= 9*10*500=45000

Here also potential energy is converted into m*4200*T(2)=45000

T(2)=10.71

Here temperature is increasing.

Thus change in temperature =T(1) -T(2) =7.14° C

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2 years ago

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed

$\alpha = 0.745 rad/s2 \rightarrow$ The angular acceleration

$r = 4.65 m \rightarrow$ The distance

The relation between the linear velocity and angular velocity is

$v = r\omega$

Where,

r = Radius

$\omega =$ Angular velocity

At the same time we have that the centripetal acceleration is

$a_c = \frac{v^2}{r}$

$a_c = \frac{(r\omega)^2}{r}$

$a_c = \frac{r^2\omega^2}{r}$

$a_c = r \omega^2$

$a_c = (4.65 )(1.25 rad/s)^2$

$a_c = 7.265625 m/s^2$

Now the tangential acceleration is given as,

$a_t = \alpha r$

Here,

$\alpha =$ Angular acceleration

r = Radius

$\alpha = (0.745)(4.65)$

$\alpha = 3.46425 m/s^2$

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

$|a| = \sqrt{a_c^2+a_t^2}$

$|a| = \sqrt{(7.265625)^2+(3.46425)^2}$

$|a| = 8.049 m/s^2 \approx 8.05 m/s2$

Therefore the correct answer is C.

7 0

2 years ago