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4 years ago

14

A uniform disk of unknown mass M and radius R = 10 cm is free to rotate about its axis. A light cord is wrapped around the rim o

n the disk and then tied to a small can of mass m = 50 gm. The cord does not slip as it unwinds on the disk. When released the can moves down with acceleration 3.27 m/s2. Take g = 9.81 m/s2. What is the angular acceleration of the disk?

1 answer:

Ne4ueva [31]4 years ago

8 0

Answer:

$32.7\ rad/sec^2$

Explanation:

We have given the acceleration of the cord is $3.27 m/sec^2$ and acceleration due to gravity is $9.81 m/sec^2$ there is a relation between the angular acceleration and the linear acceleration that is

Linear acceleration = angular acceleration × radius

We have given radius R = 10 cm =0.1 m

So $\alpha = \frac{a}{R}=\frac{3.27}{0.1}=32.7\ rad/sec^2$ here α is angular acceleration a is linear acceleration and R is radius

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A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication

koban [17]

Answer:

r = 4.24x10⁴ km.

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$

$r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}$

$r_{1} = 4.24 \cdot 10^{4} km$

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0

3 years ago

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generate

tigry1 [53]

Explanation:

It is given that,

Average power per unit mass generated by Lance, $\dfrac{P}{m}=6.5\ W/kg$

$P=6.5\times 75=487.5\ W$

(a) Distance to cover race, $d = 160\ km =160\times 10^3\ m$

Average speed of the person, v = 11 m/s

If t is the time taken to cover the race.

$t=\dfrac{d}{v}$

$t=\dfrac{160\times 10^3\ m}{11\ m/s}$

t = 14545.46 s

Let W is the work done. The relation between the work done and the power is given by :

$P=\dfrac{W}{t}$

$W=P\times t$

$W=487.5\times 14545.46$

W = 7090911.75 J

(b) Since, $1\ J=2.389\times 10^{-4}\ calories$

So, in 7090911.75 J, $W=7090911.75 \times 2.389\times 10^{-4}$

W = 1694.01 J

Hence, this is the required solution.

6 0

3 years ago

What is the DISPLACEMENT of the motorbike rider in the picture?

lutik1710 [3]

120m north east hope this helps

6 0

3 years ago

T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?

Debora [2.8K]

Answer:

$P_J=195N$

Explanation:

From the question we are told that

$Mass\ of T-joe\ M_J=65\\Velocity\ of T-joe\ V_J=3m/s\\Mass of\ T-Brud\ M_B=50kg\\Velocity\ of T-Brud\ V_B=3m/s\\$

Generally the equation for momentum is mathematically given by

$P=mv$

Therefore

T-Joe momentum $P_J$

$P_J=65*3$

$P_J=195N$

5 0

3 years ago

A spring with spring constant 15.5 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

Aloiza [94]

Answer:

A) 138.8g

B)73.97 cm/s

Explanation:

K = 15.5 Kn/m

A = 7 cm

N = 37 oscillations

tn = 20 seconds

A) In harmonic motion, we know that;

ω² = k/m and m = k/ω²

Also, angular frequency (ω) = 2π/T

Now, T is the time it takes to complete one oscillation.

So from the question, we can calculate T as;

T = 22/37.

Thus ;

ω = 2π/(22/37) = 10.5672

So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g

B) In simple harmonic motion, velocity is given as;

v(t) = vmax Sin (ωt + Φ)

It is from the derivative of;

v(t) = -Aω Sin (ωt + Φ)

So comparing the two equations of v(t), we can see that ;

vmax = Aω

Vmax = 7 x 10.5672 = 73.97 cm/s

6 0

3 years ago