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2 years ago

What is a benefit of a medically supervised rehabilitation program?

Physics

1 answer:

Marina86 [1]2 years ago

3 0

A. Doctors can safely monitor the physical demands of detox

A. Doctors can safely monitor the physical demands of detoxB. Doctors can provide accountability and emotional support

A. Doctors can safely monitor the physical demands of detoxB. Doctors can provide accountability and emotional supportC. Doctors can prescribe drugs to counteract the effects of alcohol

A. Doctors can safely monitor the physical demands of detoxB. Doctors can provide accountability and emotional supportC. Doctors can prescribe drugs to counteract the effects of alcoholD. Doctors can help shorten the time needed for detox and rehab

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The type of data collected through remote sensing includes _____.

Anit [1.1K]

Remote sensing is used in numerous fields, including geography<span> and most Earth Science disciplines.</span><span> It also has military, intelligence, commercial, economic, planning, and humanitarian applications.</span>

8 0

3 years ago

A piston–cylinder device initially contains 0.22 kg of steam at 200 kPa and 300°C. Now, the steam is cooled at constant pressure

wel

Explanation:

Defining compressibility factor (Z), which is also known as the gas deviation factor, is a correction factor which describes the deviation of a real gas from ideal gas behaviour.

Z = p/(density * Rs *T)

Where Rs is specific gas constant =R/M

Deriving the volume values from steam tables,

At 200kPa and 300°C, V1 = 1.31623 m3/kg

At 200kPa and 150°C, V2 = 0.95986 m3/kg

Change in volume, DV = m(V1 - V2)

= 0.2*(1.31623 - 0.95986)

= 0.07128 m3/kg

Also from steam tables,

The critical pressure, Pc = 22.06MPa

The critical temperature, Tc = 647.1K

Remember, gas constant, R = 0.4615 kPa.m3/kg.K

Therefore, reduced temperature at initial state, Tr = T1/Tc

= (360 + 273)/647.1

= 0.886

Also, reduced pressure at initial state, Pr = P1/Pr

= 200 x 10^-3/22.06

= 0.0091

But compressibility factor at initial state, Z1 = 0.9956

Ideal volume, Vi = m*R*T/P1

= (0.2*0.4615*(300+273))/200

= 0.2644m3

Compressibility volume, Vc = Z1*Vi

= 0.9956* 0.2644

= 0.2633m3

Reduced temperature at final state, Tr = T2/Tc

= (150+273)/647.1

= 0.65

Reduced pressure at final state, Pr = P2/Pc

= 200 x 10^-3/22.06

= 0.0091

Compressibility factor at final state, Z2 = 0.9897

3 0

3 years ago

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.25 m and finds that it makes 419

Zina [86]

Answer:

$g=2406.8m/s^{2}$

Explanation:

firstly we have to find the frequency and then the angular frequency.

The frequency will be:

$f=\frac{419}{60} \\f=6.98Hz$

The angular frequency is:

ω $=2\pi f$

ω $=2*\frac{22}{7} *6.98$

ω $=43.88rad /sec$

Now we can apply simple pendulum relationship

ω $=\sqrt{\frac{g}{l} }$

make g the subject of the formula

$g=w^{2} l\\g=43.88^{2} *1.25\\g=2406.8m/s^{2}$

6 0

3 years ago

Read 2 more answers

If the temperature is held constant during this process and the final pressure is 671 torr , what is the volume of the bulb that

MariettaO [177]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

The initial pressure of the gas is $P_1 = 1.50 \ atm$

The volume of the second bulb is $V_2 = 0.800 \ L$

The final pressure of the gas is $P_2 = 671 \ torr = \frac{671 }{ 760} = 0.883 \ atm$

Let the unknown volume be represented as $V_1 = V \ L$

Generally from Boyle's law we have that

$P_1 * V_1 = P_2 * V_2$

=> $1.50 * V = 0.883 * 0.800$

=> $V = 0.479 \ L$

4 0

3 years ago

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc

____ [38]

Answer:

Part a)

$a = 0.36 m/s^2$

Part b)

$a = -1.29 m/s^2$

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

$F_x = Fcos25$

$F_x = 80 cos25 = 72.5 N$

$F_y = Fsin25$

$F_y = 80 sin25 = 33.8 N$

now in vertical direction we can use force balance for the box to find the normal force on it

$F_n + F_y = mg$

$F_n = (25)(9.81) - 33.8$

$F_n = 211.45 N$

now kinetic friction on the box opposite to applied force due to rough floor is given as

$F_k = \mu F_n$

$F_k = (0.300)(211.45) = 63.44 N$

now the net force on the box in forward direction is given as

$F_{net} = F_x - F_k$

$F_{net} = 72.5 - 63.44 = 9.065 N$

now the acceleration of the box is given as

$a = \frac{F_{net}}{m}$

$a = \frac{9.065}{25} = 0.36 m/s^2$

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

$F_x = 72.5 N$

$F_y = 33.8 N$

now force balance perpendicular to inclined plane is given as

$F_n + F_y = mgcos\theta$

$F_n = (25)(9.81)cos10 - 33.8 = 207.7 N$

now the friction force opposite to the motion on the box is given as

$f_k = \mu F_n$

$f_k = (0.300)(207.7) = 62.3 N$

now the net pulling force along the inclined plane is given as

$F_{net} = F_x - F_k - mgsin10$

$F_{net} = 72.5 - 62.3 - (25)(9.81)sin10$

$F_{net} = -32.38 N$

now the box will decelerate and it is given as

$a = \frac{F_{net}}{m}$

$a = \frac{-32.38}{25} = -1.29 m/s^2$

7 0

3 years ago