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stellarik [79]
3 years ago
10

A thin hoop is hung on a wall, supported by a horizontal nail. The hoop's mass is M=2.0 kg and its radius is R=0.6 m. What is th

e period of small oscillations of the hoop? Type your answer below, accurate to two decimal places, and assuming it is in seconds. [Recall that the moment of inertia of a hoop around its center is Icm=MR2.]
Physics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

Explanation:

Given that,

Mass of the thin hoop

M = 2kg

Radius of the hoop

R = 0.6m

Moment of inertial of a hoop is

I = MR²

I = 2 × 0.6²

I = 0.72 kgm²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

I = 0.72 kgm²

M is the mass of the hoop

M = 2kg

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = r = 0.6m

Then, applying the formula

T = 2π √ (I / MgR)

T = 2π √ (0.72 / (2 × 9.8× 0.6)

T = 2π √ ( 0.72 / 11.76)

T = 2π √0.06122

T = 2π × 0.2474

T = 1.5547 seconds

T ≈ 1.55 seconds to 2d•p

Then, the period of oscillation is 1.55seconds

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A donkey starts from rest and accelerates for 6.7 s at a rate of 1.6 m/s².
vitfil [10]

The maximum speed of the donkey is 10.72m/s

The question is based on the principle of motion in one dimension and hence formulas of motion in one dimension can be applied.

It is given that donkey attains an acceleration of 1.6 m/s^2

The time taken to accelerate  to given speed is 6.7 seconds

We use the formula v=u + at to find the fastest speed

v is the final or maximum speed

u is the initial speed which in this case is 0 as the donkey is at rest

a is the acceleration of the donkey

t is  the time taken in seconds

v = u + at

v= 0 + 1.6 x 6.7

 = 10.72 m/s

Hence the donkey obtains the speed of 10.72 m/s

For further reference:

brainly.com/question/24478168?referrer=searchResults

#SPJ9

3 0
2 years ago
What can change of matter from one state to another state?
Anna35 [415]

Answer:

answer is Heating

Explanation:

take a solid and heat it it will become a liquid

8 0
3 years ago
A circuit is constructed with six resistors and two batteries as shown. the battery voltages are v1 = 18 v and v2 = 12 v. the po
VladimirAG [237]

Answer:

V4=9.197v

Explanation:

Given:

V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms

V4= I4R4 = V2/(R4 + R5)×R4

V4= 12×125 /(125 + 58)

V4=1500/183 =9.197v

5 0
3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
Read 2 more answers
What is the momentum of a photon having the same total energy as an electron with a kinetic energy of 100 keV?
statuscvo [17]

Answer:

The momentum of the photon is 1.707 x 10⁻²² kg.m/s

Explanation:

Given;

kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000  x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J

Kinetic energy is given as;

K.E = ¹/₂mv²

where;

v is speed of the electron

K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} =  (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} =  (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s

Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s

6 0
3 years ago
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