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Fudgin [204]
3 years ago
15

Shani poses as the statue of liberty for an art class. she holds a torch above her head with her right hand and a book against h

er side with her left hand. she holds the pose for ten minutes. at the end of the session, she mentions to her friend that it was a lot of work to pose and that her arms are really tired.which describes the work shani did while she posed?
Physics
1 answer:
kozerog [31]3 years ago
6 0
For Art class, Social Studies class, English class, American History class,
or Photography class, Shani has done an incredible amount of work. 
She'll tell her parents all about it at dinner, and her arms will really ache
when she gets up tomorrow morning.

But it's a different story in Physics class.  In Physics, there is a formal definition
for "work".  (That's so it can be measured, and numerically compared to all forms
of energy.) 

The scientific definition of work is 

                           (force exerted) times (distance moved) .

With this definition, if the force doesn't move through a distance,
then the work done is zero. 

Anything you do without moving, even if it's holding a small car over
your head for an hour while your muscles tremble and sweat pours
down, represents no "work" in the scientific sense.
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Is Mercury a heavier element than tin
Kryger [21]
Yes because mercury has more protons and electrons that tin. (30 more)
4 0
2 years ago
The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2

Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

3 0
3 years ago
A cannonball is shot up in the air with a vertical speed of 24 miles what is the cannon balls vertical speed just before it hits
meriva

Answer:

I think it will back towards the earth because earth gravitional field will attract to Wards

8 0
2 years ago
The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr
Dvinal [7]

Answer:

T_A_v_g=9.918192559$^{\circ}C

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

T(t)=10-5*sin(\frac{\pi}{12t})

The average temperature over a given interval can be calculated as:

T_a_v_g=\frac{T_o+T_f}{2}

Where:

T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C

T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C

Hence, the average temperature between noon and midnight is:

T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C

3 0
2 years ago
a uniform metre rule is pivoted at its centre and weights of 5 N and 12 N are hung at the 3 cm and 5 cm marks respectively. how
schepotkina [342]

In order to balance the stick on the pivot, the total "moments" must be equal on both sides.  A "moment" is (a weight) x (its distance from the center).

for the 5N weight: Moment = (5N) x (3 cm) = 15 N-cm

for the 12N weight: Moment = (12N) x (5 cm) = 60 N-cm

Sum of the moments trying to pull the stick down on that side = 75 N-cm

Whatever we hang on the other side has to provide a moment of 75 N-cm in the other direction.  We have a 25N weight. Where should we hang it ?  

(25N) x (distance from the pivot) = 75 N-cm

Distance from the pivot = (75 N-cm) / (25 N)

<em>Distance from the pivot =  3 cm </em>

8 0
3 years ago
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