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2 years ago

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Two masses hang below a massless meter stick. Mass 1 is located at the 10cm mark with a weight of 15kg, while mass 2 is located

at the 60cm mark with a weight of 30kg. At what point in between the two masses must the string be attached in order to balance the system?

1 answer:

yan [13]2 years ago

7 0

Answer:43.33 cm mark

Explanation:

Given

mass 1 is located at the 10 cm mark with weight of 15 kg

mass 2 is located at 60 cm mark with weight of 30 kg

string should be attached between them to balance the system

so the distance between the the two masses is 50 cm

For system to be balance torque of both the weight must nullify each other

Let us suppose string is at a distance of x cm from 15 kg mass so 30 kg mass is at a distance of 50-x cm

Balancing torque

$15\times x-30\times (50-x)$

$x=\frac{100}{3}=33.33$

so string should be at a mark of 10+33.33=43.33 cm

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Two balls of unequal mass are hung from two springs that are not identical. The springs stretch the same distance as the two sys

baherus [9]

Answer:

a. Springs oscillate with the same frequency

Explanation:

As they both are in the same height at equilibrium, so

weight of ball must be balanced with spring force, that is

k×x=mg

k= stiffness constant of spring

x=distance stretched

g= acceleration due to gravity

so, we can write

k/m=g/x

as the g is a constant and they stretched to same distance x so the g/x term becomes constant and

$f\propto\sqrt{k/m}$

and k/m is same for both the springs so they will oscillate at the same frequency.

hence option a is correct.

3 0

2 years ago

A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintains that velocity for 10 s, and then decelerates at th

zhannawk [14.2K]

Answer:

Final speed of car = 12 m/s

Explanation:

We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.

a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s

v = ?

u = 0 m/s

a = 4.0 m/s²

t = 5 s

v = u + at = 0 + 4 x 5 = 20 m/s

b) Then maintains that velocity for 10 s

v = ?

u = 20 m/s

a = 0 m/s²

t = 10 s

v = u + at = 20 + 0 x 10 = 20 m/s

c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

v = ?

u = 20 m/s

a = -2.0 m/s²

t = 4 s

v = u + at = 20 + -2 x 4 = 12 m/s

Final speed of car = 12 m/s

3 0

1 year ago

A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta

yuradex [85]

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

5 0

1 year ago

4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that

Basile [38]

The strength of the magnetic field is $4.8\cdot 10^{-5} T$

Explanation:

According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:

$\epsilon = \frac{N\Delta \Phi}{\Delta t}$ (1)

where

N = 505 is the number of turns in the coil

$\Delta \Phi$ is the change in magnetic flux through the coil

$\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s$ is the time interval

$\epsilon = 0.166 V$

The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:

$\Delta \Phi = B A cos \theta$

where

B is the strength of the magnetic field

A is the area of the coil

$\theta=0^{\circ}$ is the angle between the normal to the coil and the field

The area of the coil can be written as

$A=\pi r^2$

where

$r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m$ is its radius

Substituting everything into eq.(1) and solving for B, we find:

$\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T$

Learn more about magnetic fields:

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#LearnwithBrainly

8 0

1 year ago

Read 2 more answers

A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences

NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, $q=3\times 10^{-6}\ C$

Mass of charge particle, $m=2\times 10^{-6}\ C$

Speed, $v=5\times 10^{6}\ m/s$

Acceleration, $a=3\times 10^{4}\ m/s^2$

We need to find the minimum magnetic field that would produce such an acceleration. So,

$ma=qvB\ sin\theta$

For minimum magnetic field,

$ma=qvB$

$B=\dfrac{ma}{qv}$

$B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}$

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0

2 years ago