I need help answering this, I keep running into dead ends.

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.16

ycow [4]

We assign the variables: T as tension and x the angle of the string
The centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.

(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. 
(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. 
(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. 
(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.

This minimum v is obtained when T=0 and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.

4 0

3 years ago

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During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : The dry unit weight is 0.0616  $lb/in^3$ 

part b : The void ratio is 0.77

part c : Degree of Saturation is 0.43

part d : Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

The dry unit weight is 0.0616  $lb/in^3$ 

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

The void ratio is 0.77

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

Degree of Saturation is 0.43

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

3 years ago

The Astronomer Carl Sagan is famous. Why?

Zolol [24]

Mainly because he was Johnny Carson's advisor and consultant
on space, astronomy, and science in general, and he appeared
on The Tonight Show Starring Johnny Carson many times.

6 0

3 years ago

What is perfect machine

Fittoniya [83]

Answer:

Explanation:

A machine is which no part of the work done on the machine is wasted, is called an ideal or perfect machine

5 0

3 years ago

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5) Why does the first hill/drop on a roller coaster have to be the highest? (think about

prohojiy [21]

5) Due to the conservation of energy / due to the presence of friction

6) The work done is 4000 J

7) The distance travelled by the object is 4 m

8) The potential energy of the bird is 2450 J

Explanation:

5)

For a roller coaster in motion along the track, in absence of friction, the mechanical energy remains constant:

$E=U+K = const.$

where

U is the potential energy

K is the kinetic energy

At the beginning, the roller coaster is at rest, therefore its kinetic energy is zero and its mechanical energy is just equal to the potential energy:

$E=U=mgh$

where m is the mass of the roller coaster, g the acceleration of gravity, and h the height of the roller coaster above the ground at the first hill.

Since this amount of energy remains constant, the roller coaster cannot go to a higher hill: in fact, in that case it would reach a height $h'>h$ , but this is not possible, because it would mean that its mechanical energy would be $mgh' > E$ , larger than its mechanical energy, and since energy cannot be created (law of conservation of energy), the height of the next hills cannot be higher than the first one. Moreover, if we consider friction, then friction does work on the roller coaster in motion, and as a result, part of its mechanical energy is wasted (converted into thermal energy): therefore, the height of the following hills must be lower than the height of the first hill.

6)

The work done when lifting an object is equal to the gravitational potential energy gained by the object:

$W=Fh$

where

F is the weight of the object

h is the change in height of the object

For the object in this problem, we have

F = 200 N (weight)

h = 20 m (change in height)

Substituting, we find the work done:

$W=(200)(20)=4000 J$

7)

The work done when pushing an object in a direction parallel to the motion of the object is given by:

$W=Fd$

where

F is the magnitude of the force

d is the distance through which the object has travelled through

In this problem, we have:

W = 300 J (work done)

F = 75 N (magnitude of the force)

Solving the equation for d, we find the distance travelled:

$d=\frac{W}{F}=\frac{300}{75}=4 m$

8)

The potential energy of an object is the energy possessed by an object due to its position in a gravitational field. Near the Earth's surface, it is given by

$PE=mgh$