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labwork [276]
3 years ago
15

If low CVP precipitates a suction alarm, rapid infusion of volume can remedy the situation after dropping the P-level.

Physics
1 answer:
motikmotik3 years ago
8 0

Answer:

d

Explanation:

You might be interested in
g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the
MrRa [10]

Answer:

Approximately -1.58\; \rm rad \cdot s^{-2}.

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

v^2 - u^2 = 2\, a\, x,

where

  • v is the final velocity of the moving object,
  • u is the initial velocity of the moving object,
  • a is the (linear) acceleration of the moving object, and
  • x is the (linear) displacement of the object while its velocity changed from u to v.

The angular analogue of that equation will be:

(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta, where

  • \omega(\text{final}) and \omega(\text{initial}) are the initial and final angular velocity of the rotating object,
  • \alpha is the angular acceleration of the moving object, and
  • \theta is the angular displacement of the object while its angular velocity changed from \omega(\text{initial}) to \omega(\text{final}).

For this object:

  • \omega(\text{final}) = 0\; \rm rad\cdot s^{-1}, whereas
  • \omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}.

The question is asking for an angular acceleration with the unit \rm rad \cdot s^{-1}. However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}.

Rearrange the equation (\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta and solve for \alpha:

\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}.

7 0
3 years ago
How to convert 7500kilometers to meters​
Kazeer [188]

Answer:

7500000 metres

Explanation:

1 kilometre = 1000 metres

so 7500 kilometres would equal to 7500000 metres

7 0
3 years ago
A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 25 degrees hill.
zloy xaker [14]

Answer:

P=28.085\,hp

Explanation:

Given that:

  • mass of 1 skier, m=80kg
  • inclination of hill, \theta=25^{\circ}
  • length of inclined slope, l=220m
  • time taken to reach the top of hill, t=2.3 min= 138 s
  • coefficient of friction, \mu=0.15

<em>Now, force normal to the inclined plane:</em>

F_N=m.g.cos\theta

F_N=80\times 9.8\times cos25^{\circ}

F_N=710.54\,N

<em>Frictional force:</em>

f=\mu.F_N

f=0.15\times 710.54

f=106.58\,N

<em>The component of weight along the inclined plane:</em>

W_l=m.g.sin\theta

W_l=80\times 9.8\times sin25^{\circ}

W_l=331.33\,N

<em>Now the total force required along the inclination to move at the top of hill:</em>

F=f+W_l

F=106.58+331.33

F=437.91\,N

<em>Hence the work done:</em>

W=F.l

W=437.91\times 220

W=96340.80\,J

<em>Now power:</em>

P=\frac{W}{t}

P=\frac{96340.80}{138}

P=698.12\,W

<u>So, power required for 30 such bodies:</u>

P=30\times 698.12

P=20943.65\,W

P=\frac{20943.65}{745.7}

P=28.085\,hp

8 0
3 years ago
An elevator motor in a high-rise building can do 3500 kJ of work in 5 min. Find the power developed by the motor. Explain if you
Ilya [14]

Answer:

P = 11666.6 W

Explanation:

Given that,

Work done by the motor, W = 3500 kJ

Time, t = 5 min = 300 s

We need to find the power developed by the motor. Power developed is given by :

P=\dfrac{E}{t}\\\\P=\dfrac{3500\times 10^3}{300}\\\\P=11666.7\ W

So, the required power is 11666.6 W.

4 0
3 years ago
You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone
umka21 [38]

Answer:

0.37 m

Explanation:

Given :

Window height, h_1 = 1.27 m

The flowerpot falls 0.84 m off the window height, i.e.

h_2 = (1.27 x 0.84 ) m in a time span of $t=\frac{8}{30}$   seconds.

Assuming that the speed of the pot just above the window is v then,

h_2=ut+\frac{1}{2}gt^2

$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$

$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$

$v= 2.69$ m/s

Initially the pot was dropped from rest. So,  u = 0.

If it has fallen from a height of h above the window then,

$h = \frac{v^2}{2g}$

$h = \frac{(2.69)^2}{2 \times 9.81}$

h = 0.37 m

3 0
3 years ago
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