A closed system:

A soccer ball of diameter d and mass m rolls up a hill without slipping, reaching a maximum height of h above the base of the hi

stich3 [128]

Answer:

Convert the units to standard units.

r = d / 2 (cm)

r = r in cm [ 1 m / 100 cm ]

m = m in g * [ 1 kg / 1000 g] = m in kg

Write the general equation for this problem.

The KE at the bottom of the hill = the PE at the top.

1/2 I w^2 + 1/2 m v^2 = mgh

I for a hollow sphere = 2/3xmxr^2

1/2 (2/3 x mr^2 x w^2) + 1/2 x m x v^2 = mgh

Now we need to relate w with v

v = w x r

w = v / r put this result in for w

1/2 (2/3 x mxr^2 z v^2 / r^2) + 1/2 m x v^2 = mgh

1/2 (2/3 x m v^2) + 1/2 x m x v^2 = mgh

1/3 m x v^2 + 1/2 mv^2 = mgh

5/6 m x v^2 = mgh Notice that the m's are all the same on both sides. They cancel.

5/6 v^2 = gx h

Solve for v

v^2 = 6/5 x g x h

Find the Rotational Kinetic Energy (KE_r)

PE - KE_translational = KE_r

mgh - 1/2 x m x v^2 = KE_r

Find the rotational rate at the bottom w

KE_r = 1/2 I x w^2

I = 2/3 m R^2

7 0

3 years ago

In a charming 19th-century hotel, an old-style elevator is connected to a counterweight by a cable that passes over a rotating d

melisa1 [442]

Answer:

2.38732 rpm

1.22625 rad/s²

163.292°

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration = $\frac{1}{8}g$

d = Diameter of wheel = 2 m

r = Radius of wheel = $\frac{d}{2}=\frac{2}{2}=1\ m$

v = Speed of elevator = 25 cm/s

Angular speed is given by

$\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{0.25}{1}\\\Rightarrow \omega=0.25\ rad/s=0.25\times \frac{60}{2\pi}\\\Rightarrow \omega=2.38732\ rpm$

The angular speed of the wheel is 2.38732 rpm

Angular acceleration is given by

$\alpha=\frac{a}{r}\\\Rightarrow \alpha=\frac{\frac{1}{8}g}{r}\\\Rightarrow \alpha=\frac{\frac{1}{8}\times 9.81}{1}\\\Rightarrow \alpha=1.22625\ rad/s^2$

The angular acceleration of the wheel is 1.22625 rad/s²

Angular displacement is given by

$\theta=\frac{s}{r}\\\Rightarrow \theta=\frac{2.85}{1}\\\Rightarrow \theta=2.85\ rad=2.85\times \frac{360}{2\pi}\\ =163.292\ ^{\circ}$

The angle the disk turned when it has raised the elevator is 163.292°

4 0

3 years ago

What are saturated and non saturated vapour?

Elodia [21]

Answer:

A saturated vapour is one that is in balance with its own liquid. Space is considered to be unsaturated if it contains fewer vapours than the maximum amount it can retain at a given temperature. Saturated vapour pressure is unaffected by volume, although it rises as temperature rises.

Explanation:

8 0

3 years ago

Study the image of the moving car.

gayaneshka [121]

Answer:

While traveling downhill, the car’s potential is <u>increasing</u> and kinetic energy is <u>decreasing</u>

Explanation:

hope this helps!

6 0

2 years ago

Read 2 more answers

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

4 years ago