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3 years ago

14

Suppose, the same angular momentum is transferred to two rotating bodies of different moment of inertia , how will you compare t

he angular velocities of the two bodies as a result of angular momentum transfer.

1 answer:

Sonbull [250]3 years ago

4 0

Answer:

As per the law of conservation of angular momentum, the angular velocity will be higher for the body with a lower moment of inertia and vice versa.

Explanation:

Angular momentum L of a body is given by:

$L=I\times \omega$

Now when the same angular momentum is transferred to two different bodies with different moment of inertia, the body with a higher moment of inertia will have lower angular velocity and vice versa.

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A person exerts a tangential force of 37.7 N on the rim of a disk-shaped merry-go-round of radius 2.75 m and mass 144 kg. If the

Shkiper50 [21]

Answer:

ω = 0.467 rad/s

Explanation:

given,

tangential force exerted by the person = 37.7 N

radius of merry-go-round = 2.75 m

mass of merry-go-round = 144 Kg

angle = 33.2°

moment of inertia

$I = \dfrac{1}{2} m R^2$

$I = \dfrac{1}{2}\times 144 \times 2.75^2$

I = 544.5 kg.m²

torque = force x radius

τ = 37.7 x 2.75

τ = 103.675 N.m

angular acceleration

$\alpha= \dfrac{\tau}{I}$

$\alpha= \dfrac{103.675}{544.5}$

α = 0.190 rad/s²

now ,

distance = $33.2\times \dfrca{2\pi}{360}$

d = 0.579 rad

we know,

using equation of rotational motion

$d = \omega t + \dfrac{1}{2}\alpha t^2$

$0.579 = \dfrac{1}{2}\times 0.190\times t^2$

t = 2.46 s

angular speed

ω = α x t

ω = 0.19 x 2.46

ω = 0.467 rad/s

7 0

2 years ago

so I know you can solve this either by using Vox or Voy. I'm getting 3.08s when using Vox and 3.14s for Voy way. For Voy I'm usi

Tatiana [17]

The person's horizontal position is given by

$x=v_0\cos40^\circ t$

and the time it takes for him to travel 56.6 m is

$56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s$

so your first computed time is the correct one.

The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity $v_y$ at time $t$ is

$v_y=v_{0y}-gt$

which tells us that he would reach the ground at about $t=3.15\,\mathrm s$ . In this time, he would have traveled

$x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m$

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity $v_{0y}$ would have been a bit smaller than $\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ$ .

7 0

2 years ago

Calcium chloride is used to make pickles. The calcium atom transfers electrons to chlorine atoms, leading to the formation of ch

GarryVolchara [31]

Calcium chloride contains ionic bonds.

Pennies contain metallic bonds.

Hydrochloric acid contains covalent bonds.

You're welcome.

3 0

3 years ago

Read 2 more answers

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

Anarel [89]

Answer:

$R = 24.3 m$

Explanation:

As we know that the orbital speed is given as

$v = \sqrt{\frac{GM}{R + h}}$

here we know that

v = 5500 m/s

$R = 4.48 \times 10^6 m$

$h = 630 km$

now we have

$5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}$

$M = 2.34 \times 10^24 kg$

now acceleration due to gravity of planet is given as

$a = \frac{GM}{R^2}$

$a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}$

$a = 7.7 m/s^2$

now range of the projectile on the surface of planet is given as

$R = \frac{v^2 sin2\theta}{g}$

$R = \frac{14.6^2 sin(2\times 30.8)}{7.7}$

$R = 24.3 m$

3 0

3 years ago

Short Answer Questions:

Otrada [13]

According to Archimede's principle, a physical object experiences an upthrust due to a difference in pressure between upper and lower surfaces.

<h3>What is an upthrust?</h3>

An upthrust is also referred to as buoyancy and it can be defined as an upward force which is exerted by a fluid (liquid or gas), so as to oppose the weight of a partially or fully immersed physical object that is floating in it.

Based on scientific information, a physical object experiences an upthrust when it is immersed in a fluid due to a difference in height and pressure between upper (top) and lower (bottom) surfaces.

According to Archimede's principle, there is a higher pressure at the bottom of the physical object due to height, and a lower pressure at the top of the physical object.

Read more on upthrust here: brainly.com/question/24389514

#SPJ1

4 0

1 year ago