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2 years ago

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A uniform solid sphere of mass M and radius R is free to rotate about a horizontal axis through its center. A string is wrapped

around the sphere and is attached to an object of mass m. Assume that the string does not slip on the sphere. (Use the following as necessary: M, m, and g.) Find the acceleration of the object.

1 answer:

LenaWriter [7]2 years ago

6 0

Answer:

$a = \frac{mg}{m + \frac{2}{5}M}$

Explanation:

To calculate the Acceleration and the tension of the object, we start by considering the value of the Tension through its moment of Inertia and Acceleration based on the angular velocity

$\tau = I\alpha = Tension(T)*R$

And $a = \alpha R$

Replacing,

$T*R = I\alpha = (\frac{2}{5} MR^2)*\frac{a}{R})\\T*R = \frac{2}{5}MaR\\T = \frac{2}{5}Ma$

The following forces occur in the body,

$mg - T = ma$

By this way we have the acceleration

$mg - \frac{2}{5}Ma = ma$

$a(m + \frac{2}{5})M) = mg$

$a = \frac{mg}{m + \frac{2}{5}M}$

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An eagle is flying horizontally at a speed of 3.80 m/s when the fish in her talons wiggles loose and falls into the lake 3.90 m

Dovator [93]

Answer:

the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal

Explanation:

initial veetical speed V₀y=0

Horizontal speed Vx = Vx₀= 3.80m/s

Vertical drop height= 3.90m

Let Vy = vertical speed when it got to the water downward.

g= 9.81m/s² = acceleration due to gravity

From kinematics equation of motion for vertical drop

Vy²= V₀y² +2 gh

Vy²= 0 + ( 2× 9.8 × 3.90)

Vy= √76.518

Vy=8.747457

Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below

V= √Vy² + Vx²

V=√3.80² + 8.747457²

V=9.537m/s

The angle can also be calculated as

θ=tan⁻¹(Vy/Vx)

tan⁻¹( 8.747457/3.80)

=66.52⁰

the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal

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2 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 $4.2\; {\rm s}$ .

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10 months ago

Read 2 more answers

What is the definition of half-life?

kenny6666 [7]

The time for half the atoms in a radioactive substance to disintegrate

https://www.vocabulary.com/dictionary/half-life

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2 years ago

Why can't a hot air balloon go to higher layers?

Elena L [17]

If you go to high you’ll run out of oxygen and possibly be blown off due to high winds.

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2 years ago

How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?

Illusion [34]

Power is defined as the rate at which the body is doing work:
$P=\frac{W}{t}$
Work is defined as displacement done by the force times that displacement:
$W=F\cdot h$
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
$W=62N\cdot10m=620J$
of work.
Now we just divide that by 5s to get how much power is required:
$P=\frac{620J}{5s}=124W$

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2 years ago