All categories

3 years ago

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

Physics

1 answer:

Tanya [424]3 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

The weight of a body kept at a distance of 6400 km from the centre of the

Vlad [161]

Answer:

2.5N

Explanation:

since force is inversely proportional to square of the distance apart. that means the ratio of distance apart. so u have 12800×12800÷(6400××6400)=4/1

so the weight will be reduced to 1/4 of 10N

6 0

4 years ago

Q. A mass of 300g is lifted to a<br> height of 10m<br> 205 by a person. Calculate his work done

sergiy2304 [10]

A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J

3 0

2 years ago

Which describes the changes in visible light moving from red to violet?

Alexxx [7]

The energy increases

5 0

3 years ago

A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio

Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft × 4 sorings

K = 100 lb/ft

Amplitude of the microscope $\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]$

where;

$\epsilon = 0$

$W_n = \sqrt { \frac{k}{m}}$

= $\sqrt { \frac{100*32.2}{200}}$

= 4.0124

replacing them into the above equation and making X the subject of the formula:

$X =$ $Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}$

$X =$ $0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}$

$X =$ $5.676*10^{-3} \ mm$

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

8 0

3 years ago

18. The orbits of planets being elliptical was one the planetary laws developed by

TEA [102]

It was a man named <span>Johannes Kepler. </span>

6 0

3 years ago