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3 years ago

9

If mass1 of a student is 70 kg and mass of the Jupiter is 1.901 x 10^27 kg 1

1 answer:

tamaranim1 [39]3 years ago

6 0

Answer:

lol girly same

Explanation:

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A system consists of a disk of mass 2.33 kg and radius 50 cm upon which is mounted an annular cylinder of mass 2.20 kg with inne

Gekata [30.6K]

Answer:

Kinetic energy of the system = 2547.41 Joules.

Explanation:

Given:

Disk:

Mass of the disk (m) = $2.33$ kg

Radius of the disk (r) = $50$ cm = $\frac{50}{100} =0.5$ m

Cylinder:

Mass of the annular cylinder (M) = $2.20$ kg

Inner radius of the cylinder $(R_i)$ = $0.2$ m

Outer radius of the cylinder $(R_o)$ = $0.3$ m

The angular speed of the system $(\omega)$ = $15.1$ rev/s

Angular speed in in terms of Rad/sec = $15.1\times 2\pi =94.876$ rad/sec

Formula to be used:

Rotational Kinetic energy, $(KE)_r$ = $\frac{I\times \omega^2}{2}$

So, before that we have to work with the moment of inertia (MOI) of the system.

⇒ MOI of the system = MOI of the disk + MOI of the cylinder

⇒ MOI (system) = $\frac{mr^2}{2} +\frac{M(R_i+R_o)^2}{2}$

⇒ MOI (system) = $\frac{2.33\times (0.5)^2}{2} + \frac{2.20\times (0.2+0.3)^2}{2}$

⇒ MOI (system) = $0.566$ kg.m^2

Now

The rotational Kinetic energy.

⇒ $(KE)_r =\frac{I\omega^2}{2}$

Plugging the values.

⇒ $(KE)_r=\frac{0.566\times (94.876)^2}{2}$

⇒ $(KE)_r=2547.41$ Joules

Then

The kinetic energy of the rotational system is 2547.41 J.

5 0

3 years ago

Suppose you swing a ball of mass (m) in a vertical circle on a string of length (L). As you probably know from experience, there

KATRIN_1 [288]

Answer:

a) $\omega =\sqrt{\dfrac{g}{L}}$

b)N= 21.29 rpm

Explanation:

Given that

Mass of the ball =m

Length of string = L

Lets take angular speed = ω

The centripetal force on the ball

F = m ω² L

To complete the circle ,at the top condition the force due to gravity should be equal to the centripetal force

Gravity force = mg

F= mg

m ω² L = m g

ω² L = g

$\omega =\sqrt{\dfrac{g}{L}}$

When L= 2 m

Lets take g =10 m/s²

$\omega =\sqrt{\dfrac{g}{L}}$

$\omega =\sqrt{\dfrac{10}{2}}$

ω = 2.23 rad/s

To convert in rpm

$\omega =\dfrac{2\pi N}{60}$

N=Speed in rpm

$2.23 =\dfrac{2\pi N}{60}$

N= 21.29 rpm

8 0

3 years ago

Playing tetherball, a boy hits the 0.45 kg ball, moving toward him initially at 4.6 m/s, in the opposite direction at 2.5 m/s. H

saul85 [17]

Answer:

HI

Explanation:

8 0

3 years ago

Hi, why do waves with longer wavelengths travel further than waves with shorter wavelengths?

oee [108]

Answer:

ones longer than the other daa

Explanation:

6 0

3 years ago

Read 2 more answers

A long solenoid with 11.2 turns/cm and a radius of 9.74 cm carries a current of 24.2 mA. A current of 17.1 A exists in a straigh

sveticcg [70]

Answer with Explanation:

We are given that

Number of turns =n=11.2 /cm= $11.2\times 100=1120/m$

1 m=100 cm

I=24.2 mA= $24.2\times 10^{-3} A$

$1m A=10^{-3} A$

Radius of solenoid,r=9.74 cm= $9.74\times 10^{-2} m$

Current in straight conductor,I'=17.1 A

a.Magnetic field in solenoid= $B_s=\,u_0 nI=4\pi \times 10^{-7}\times 1120\times 24.2\times 10^{-3}=3.4\times 10^{-5} T$

Magnetic field in straight wire= $B'=\frac{\mu_0I'}{2\pi r'}=\frac{2\times 10^{-7}\times 17.1}{r'}=\frac{34.2\times 10^{-7}}{r'}$

Where $\frac{\mu_0}{4\pi}=10^{-7}$

$\theta=62.3^{\circ}$

$\frac{B'}{B}=tan 62.3^{\circ}$

$\frac{34.2\times 10^{-7}}{r'\times 3.4\times 10^{-5}}=1.9047$

$r'=\frac{34.2\times 10^{-7}}{3.4\times 10^{-5}\times 1.9047}=0.053m$

b.Magnitude of magnetic field= $\sqrt{B^2+B'^2}$

$\sqrt{(3.4\times 10^{-5})^2+(\frac{2\times 10^{-7}\times 17.1}{0.053})^2}= 7.29\times 10^{-5} T$

Hence,the magnitude of magnetic field = $7.29\times 10^{-5} T$

3 0

3 years ago