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A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90-kgkg uniform rod of length 0.95 mm, and compute the torque the player applies to one end of it.

Answers

Answer 1

Answer:

Answer:

$\tau=22.13Nm$

Explanation:

information we have:

mass: $m=0.9kg$

lenght: $L=0.95m$

frequency: $f=2.6rev/s$

time: $t=0.2s$

and from the information we have we can calculate the angular velocity $\omega$ . which is defined as

$\omega=2\pi f$

$\omega=2\pi (2.6rev/s)\n\omega=16.336 rev/s$

----------------------------

Now, to calculate the torque

We use the formula

$\tau=I \alpha$

where $I$ is the moment of inertia and $\alpha$ is the angular acceleration

moment of inertia of a uniform rod about the end of it:

$I=(1)/(3)mL^2$

substituting known values:

$I=(1)/(3) (0.9kg)(0.95m)^2\nI=0.271kg/m^2$

for the torque we also need the acceleration $\alpha$ which is defined as:

$\alpha=(\omega)/(t)$

susbtituting known values:

$\alpha=(16.336rev/s)/(0.2s) \n\alpha=81.68rev/s^2$

and finally we substitute $I$ and $\alpha$ into the torque equation $\tau=I \alpha$ : $\tau=(0.271kg/m^2)(81.68rev(s^2)\n\tau=22.13Nm$

Answer 2

Answer:

Final answer:

To calculate the torque, we need to use the formula: Torque = Moment of Inertia * Angular Acceleration. By approximating the bat as a uniform rod and using its length and mass, we can find the moment of inertia. Then, using the given angular velocity, we can calculate the angular acceleration. Finally, we can determine the torque by multiplying the moment of inertia by the angular acceleration.

Explanation:

To compute the torque the player applies to one end of the bat, we need to use the formula:

Torque = Moment of Inertia * Angular Acceleration

Given that the bat is approximated as a uniform rod and we know its length and mass, we can calculate the moment of inertia. Then, using the given angular velocity, we can compute the angular acceleration. Finally, we can find the torque by multiplying the moment of inertia by the angular acceleration.

Learn more about Torque here:

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An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Determine the speed at this time and the maximum height at which it reaches.

Answers

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2* 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+(at^2)/(2)$

$s=0+(2* 6^2)/(2)$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=(144)/(2* 9.8)=7.34 m$

$s+s_0=36+7.34=43.34 m$

A worker is holding a filled gas cylinder still. Which two sentences are true about the energy of the filled gas cylinder?

Answers

Answer:

B and C

i think its right

Answer:

B and C

Explanation:

They are right and they are the only ones that make sense

The planet uranus is tilted nearly on its side so that its axis or rotation is only 8 degress abway from its orbit plane. if you lived at latitude 45 degrees on uranus for what fraction of the uranian year would answer?

Answers

The rotation of Uranus, like that of Venus, is retrograde and its axis of rotation is inclined almost ninety degrees above the plane of its orbit. During its orbital period of 84 years one of the poles is permanently illuminated by the Sun while the other remains in the shade. Exactly its rotation period is equivalent to 17 hours and 14 Earth minutes and its translation period is equivalent to 84 years, 7 days and 9 Earth hours.

Only a narrow band around the equator experiences a rapid cycle of day and night, but with the Sun very low on the horizon as in the polar regions of the Earth. On the other side of the orbit of Uranus, the orientation of the poles in the direction of the Sun is inverse. Each pole receives about 42 years of uninterrupted sunlight, followed by 42 years of darkness. Therefore an observer at latitude of 45 degrees in Uranus will probably experience a long winter night that is equivalent to one third of the year uranium.

Using an argument based on the general form of the Schrödinger equation, explain why if \psi (x) is a solution to the Schrödinger equation, then A\psi(x) must also be a solution if A is a constant.I saw an explanation for this from another posted question, but this person put the explanation in numerical/equation form. Is there any way someone can explain the answer to this question in words (NON numerical/equation form)?

Answers

From a mathematical point of view, the Schrödinger Equation is a LINEAR partial differential equation, as is a partial differential equation that is defined by a linear polynomial in the solution and its derivatives.

For a linear differential equation, if you got two different solutions $\psi$ and $\phi$ , then the linear combination $\alpha \psi + \beta \phi$ , where $\alpha$ and $\beta$ are scalars, is also a solution.

This also is valid for only one solution (think of the other solution as equal to zero, $\phi = 0$ ). So, as the Schrödinger Equation is a Linear partial differential equation, then if $\psi$ is a solution, then $A \psi$ must also be a solution.

This is extremely important for physicist, as let us know that the superposition principle is valid.

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave perspective, the antenna is a circular aperture through which the microwaves diffract. What is the diameter of the radar beam at a distance of 30 km

Answers

Answer:

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Hope this helps.

An unladen swallow that weighs 0.03 kg flies straight northeast a distance of 125 km in 4.0 hours. With the x x direction due east and the y y direction due north, what is the average momentum of the bird (in unit vector notation)?

Answers

The average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.

Total displacement

Since the unladen swallow that weighs 0.03 kg flies straight northeast (that is at a bearing of 45°) a distance of 125 km in 4.0 hours.

Its position vector after 4.0 hours is d = (125kmcos45)i + (125kmsin45)j = (125000 × 1/√2)i + (125000 × 1/√2)j

= (62500√2)i + (62500√2)j.

If the initial position of the swallow is d' = 0i + 0j, then its total displacement after 4 hours is, D = d - d'

= (62500√2)i + (625000√2)j - (0i + 0j)

= (62500√2)i + (62500√2)j m

Average velocity

The unladen swallow's average velocity, v = D/t where

D = total displacement = (62500√2)i + (62500√2)j m and
t = time = 4.0 hours = 4 × 60 min/hr × 60 s/min = 14400 s

So, v = [(62500√2)i + (62500√2)j m]/14400 s = (88388.35)i/14400 + (88388.35)j /1440

= 6.14i + 6.14j m/s

Average momentum

The average momentum of the unladen swallow is p = mv where

m = mass of unladen swallow = 0.03 kg and
v = average velocity = 6.14i + 6.14j m/s

So, p = mv

p = 0.03 kg × (6.14i + 6.14j m/s)

p = (0.1842i + 0.1842j) kgm/s

So, the average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.

Learn more about average momentum here:

brainly.com/question/25941500

Answer:

The average momentum of the bird is 0.26 kgm/s

Explanation:

The formula to be used here is that of momentum which is

momentum (in kgm/s) = mass (in kg) × velocity (in m/s)

The velocity of the bird is

velocity (in m/s) = distance (in meter) ÷ time (in seconds)

distance in meters = 125km × 1000 = 125,000 m

time in seconds = 4 hrs × 60 × 60 = 14,400 secs

velocity = 125000/14400

velocity = 8.68 m/s

momentum (p) = 0.03 × 8.68

p = 0.26 kgm/s

The average momentum of the bird is 0.26 kgm/s

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