All categories

2 years ago

• (-/1 Points) DETAILS OSCOLPHYS2016 9.5.WA.040.

Physics

1 answer:

Tanya [424]2 years ago

6 0

The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.

<h3>Force exerted outside the wheel</h3>

The force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.

∑τ = 0

Let the distance traveled by the load = 1.5 m
Let the radius of the wheel or position of the force = 0.45 m

∑τ = R(mg) - r(F)

rF = R(mg)

0.45F = 1.5(21,200 x 9.8)

F = 6.925 x 10⁵ N.

Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.

Learn more about angular momentum here: brainly.com/question/7538238

You might be interested in

Which theory best explains the present arrangement of continents oceans and landforms on earth? A the Pangaea theory B The conti

solong [7]

Plate Tectonic Theory

3 0

3 years ago

as earth rotates about it’s axis, it takes three hours for the united states to pass beneath a point above earth that is station

pickupchik [31]

interesting question.

how much fuel to hover over one place ?

The Foucault pendulum is a v v v long pendulum which can show the eart's rotation over time if the pend bob motion is tracked and recorded

6 0

3 years ago

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

$\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s$

The angle is changing at 0.0031792338 rad/s

6 0

3 years ago

A large truck is moving 22.0 m/s. If it’s momentum is 125,000 kg • m/s, what is the trucks mass

strojnjashka [21]

Answer:

p = mv

m = p/v = 125000/22 = 5682 kg

Explanation:

Direct application of the momentum equation

p = mv

where,

p: momentum

m: mass

v: object velocity

steps:

-------

1) check for units consistency ( SI or Imperial)

2) separate the variable you are looking for.

3) DONE! :DD

6 0

3 years ago

9. What is the magnitude of the velocity of a 25 kilogram mass that is moving with a

Ganezh [65]

Answer:

Explanation:

8 0

3 years ago