Ask question

Ask question

All categories

14 days ago

10

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:

ValentinkaMS [3.4K]14 days ago

8 0

Answer:

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

Explanation:

By applying the Law of Sines,

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

Based on Newton's Law,

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

And the final equation also derived from Newton's Law,

$\mu N = mgsin\phi$

Then by consolidating all the 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,

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

You might be interested in

Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula KE = mv2, where m is t

ValentinkaMS [3465]

Kinetic energy is represented as

KE = (0.5) m v²

In each scenario, v = the velocity of the bottle set at 4 m/s

with m = 0.125 kg

KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J

for m = 0.250 kg

KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J

if m = 0.375 kg

KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J

when m = 0.500 kg

KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J

6 0

1 month ago

Read 3 more answers

We fully submerge an irregular lump of material in a certain fluid. The fluid that would have been in the space now occupied by

Ostrovityanka [3204]

Answer:

the lump descends

Explanation:

The full articulation of the query is

Upon fully submerging a 3 kg lump of material in a certain fluid, the fluid that would have occupied the space now filled by the lump weighs 2 kg. (a) When released, does the lump float up, sink, or remain steady

(a)

$F_{b}$ = Buoyant force acting upward on the lump

$M$ = mass of irregular lump = 3 kg

$m$ = mass of fluid displaced = 2 kg

The upward buoyant force on the lump is given by the weight of the displaced fluid, thus

$F_{b} = mg = (2) (9.8) = 19.6 N$

the weight of the irregular lump of material is represented as

$W = mg\\W = (3) (9.8)\\W = 29.4 N$

Given that the weight of the lump downward exceeds the upward buoyant force, the lump will indeed descend

4 0

14 days ago

A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat

ValentinkaMS [3465]

The focal length of the lens while in water is noted to be 150 cm, whereas in air, it measures 60 cm. To derive these values, the formula incorporates the variations in the refractive index of glass compared to that of the surrounding medium.

5 0

24 days ago

Read 2 more answers

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Sav [3153]

Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

$\texttt{ }$

Additional explanation

The Ideal Gas Law that should be remembered is:

$\large {\boxed {PV = nRT} }$

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

$\texttt{ }$

Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

$\texttt{ }$

Next, we will determine the work performed on the gas:

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

Step B:

By utilizing the methodology mentioned earlier:

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

Next, we will ascertain the work completed on the gas:

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

Ultimately, we can calculate the total work done and heat supplied as follows:

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

Learn more

Minimum Coefficient of Static Friction:
The Pressure In A Sealed Plastic Container:
Effect of Earth’s Gravity on Objects:

$\texttt{ }$

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

29 days ago

9. A 2 liter bottle of Coke weighs about 2 kilograms. If that bottle were combined with an equal amount of anti-Coke, how many J

Maru [3345]

Answer:

The energy expected to be released is calculated to be 4182 Joules.

Explanation:

The total mass of coke is 2 kg, which is equivalent to 2000 g

1 calorie per gram corresponds to 4.184 Joules of energy

4.184 J/gC * 2000g results in 8368 J

1 food calorie approximates to 4186 J

By subtracting, we find 8368 - 4186

Hence, the total energy that will be released amounts to 4182 Joules.

3 0

1 month ago