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3 years ago

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A collapsible plastic bag contains a glucose solution. The acceleration of gravity is 9.8 m/s2 . h Glucose solution If the avera

ge gauge pressure in the vein is 14800 Pa, what must be the minimum height of the bag in order to infuse glucose into the vein? Assume that the speciﬁc gravity of the solution is 1.03. Answer in units of m.

1 answer:

aliina [53]3 years ago

8 0

Answer:

The height of the bag is 1.466 m.

Explanation:

Given that,

Acceleration of gravity $g=9.8\ m/s^2$

Pressure = 14800 Pa

Specific gravity = 1.03

We need to calculate the density

Using formula of specific gravity

$\rho_{s}=\dfrac{\rho}{\rho_{w}}$

$rho=\rho_{s}\times{\rho_{w}}$

Where, $\rho$ = density of solution

$\rho_{w}$ = density of water

Put the value in to the formula

$\rho=1.03\times1000$

$\rho=1030\ kg/m^3$

We need to calculate the height

Using formula of pressure

$P=\rho gh$

$h=\dfrac{P}{\rho g}$

Where, P = pressure

g = acceleration due to gravity

h = height

Put the value into the formula

$h = \dfrac{14800}{1030\times9.8}$

$h=1.466\ m$

Hence, The height of the bag is 1.466 m.

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If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the

PolarNik [594]

Answer:

a. v₁ = 16.2 m/s

b. μ = 0.251

Explanation:

Given:

θ = 15 ° , r = 100 m , v₂ = 15.0 km / h

a.

To determine v₁ to take a 100 m radius curve banked at 15 °

tan θ = v₁² / r * g

v₁ = √ r * g * tan θ

v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s

b.

To determine μ friction needed for a frightened

v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg

v₂ = 4.2 m/s

fk = μ * m * g

a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²

a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²

F₁ = m * a₁ , F₂ = m * a₂

fk = F₁ - F₂ ⇒ μ * m * g = m * ( a₁ - a₂)

μ * g = a₁ - a₂ ⇒ μ = a₁ - a₂ / g

μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)

μ = 0.251

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4 years ago

A wall, acted upon by a force of 20 N, does not move. The work done on the wall in this process is

jenyasd209 [6]

Answer:

0 (Zero)

Explanation:

Definition of work done states that ,

Work done on an object is the force applied on that object in the direction of Displacement of the object .

means that ,

W = F.dS (Dot product of Force and displacement)

When the wall is acted upon by the force of 20 N , Wall does not move ,

Thus , displacement of the wall is zero.

So, W = 20×0 = 0.

Thus, Work done on the wall is zero.

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A turbine blade rotates with angular velocity ω(t) = 8.00 rad/s - 0.20 rad/s3 t 2. what is the angular acceleration of the blade

Irina18 [472]

Given: <span>ω(t) = 8.00 - 0.2*t^2

Differentiating both sides with respect to time we get:
angular acceleration = </span>α(t) = -0.2 * 2 *t

At t = 6.8 sec,
angular acceleration = α(t) = -0.2 * 2 *6.8 = -2.72 rad/s^2

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4 years ago

It took 2.5 hours for the bus to travel the distance between two cities at a velocity of 75.0 miles/hr. How many miles lie betwe

makvit [3.9K]

Velocity of the bus = 75 miles/hr
Time taken by the bus to travel the distance between two cities = 2.5 hours
Let us assume the distance between the two cities = x
Then
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Then Distance = Velocity * Time taken
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The distance between the two cities is 187.5 miles.I hope this answer is what you were looking for. This process is applicable for similar problems. In future you can always use this method to determine the answer.

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3 years ago

Read 2 more answers

To make a pendulum, a 210 g ball is attached to one end of a string that has a length of 1.2 m and negligible mass. (The other e

lakkis [162]

Answer:

a) $v \approx 1.233\,\frac{m}{s}$ , b) $v \approx 1.426\,\frac{m}{s}$ , c) $\theta_{2} \approx 22.61^{\textdegree}$

Explanation:

a) The speed of the ball is determined by applying the Principle of Energy Conservation:

$U_{g,1} + K_{1} = U_{g,2} + K_{2}$

The speed of the ball when the string makes an angle of 12° with the vertical is:

$K_{2} = U_{g,1} - U_{g,2} + K_{1}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot L \cdot [(1-\cos \theta_{1})-(1 -\cos \theta_{2})]$

$v = \sqrt{2\cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1})}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)\cdot (\cos 12^{\textdegree}-\cos 24^{\textdegree})}$

$v \approx 1.233\,\frac{m}{s}$

b) The maximum speed of the ball is:

$v = \sqrt{2\cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1})}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)\cdot (\cos 0^{\textdegree}-\cos 24^{\textdegree})}$

$v \approx 1.426\,\frac{m}{s}$

c) The angle between the string and the vertical when the speed of the ball is one-third its maximum value is obtained by proving different values of $\theta_{2}$ . The solution is approximately:

$\theta_{2} \approx 22.61^{\textdegree}$

5 0

3 years ago