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2 years ago

What is the mass of 2000 ml of an intravenous glucose solution with a density of 1.15 g/ml?

Physics

1 answer:

Scorpion4ik [409]2 years ago

5 0

According to the following formula, the answer is 2,300 g or 2.3 kg:

Volume (m)/Mass (m) Equals Density (p) (V)

Here, the density is 1.15 g/mL, allowing the formula described above to result in a mass of 2.00 L:

p=m/V

1.15 g/mL is equal to x g/2.00 L or x g/2,000 mL.

2,000 mL of x g = 1.15 g of g/mL

2.3 kg or 2,300 g for x g.

<h3>How many grams of glucose are in a 1000ml bag of glucose 5?</h3>

Its active ingredient is glucose. This medication includes 50 g of glucose per 1000 ml (equivalent to 55 g glucose monohydrate). 50 mg of glucose is present in 1 ml (equivalent to 55 mg glucose monohydrate). A transparent, nearly colourless solution of glucose in water is what is used in glucose intravenous infusion (BP) at 5% weight-to-volume.

Patients who are dehydrated or who have low blood sugar levels get glucose intravenously. Other medications may be diluted with glucose intravenous infusion before being injected into the body. Other diseases and disorders not covered above may also be treated with it.

learn more about glucose intravenous infusion refer

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A piano string having a mass per unit length of 5.00 g/m is under a tension of 1350 N. Determine the speed of transverse waves i

padilas [110]

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

$v=\sqrt{\dfrac{T}{\mu}}$

Where, $\mu$ = mass per unit length

T = tension

Put the value into the formula

$v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}$

$v =519.61\ m/s$

Hence, The speed of transverse waves in this string is 519.61 m/s.

6 0

3 years ago

Point A and B located at 4 meters and 9 meters from a source of the sound. If IA and IB are intensity at point A and point B, th

elena-14-01-66 [18.8K]

The intensity ratio at point A and B will be 81:16.

<u>Explanation:</u>

Sound waves are known to get faded with increase in the distance. This is because, the intensity of the sound is inversely proportional to the square of the distance of the source from the observer. So, if an observer is standing greater distance from the source of the sound, he/she will find difficulty in hearing the sound.

So, as the distance between the source and observer increases, the intensity of the sound wave decreases.

$I = \frac{1}{r^{2} }$

As here two points A and B are located at 4 m and 9 m distance from the source, then the intensity of sound at A and B will be inversely proportional to their respective square of the distance as shown below.

$I_{A} =\frac{1}{r_{A}^{2} } = \frac{1}{4 \times 4}=\frac{1}{16}$

Similarly,

$I_{B} =\frac{1}{r_{B}^{2} } = \frac{1}{9 \times 9}=\frac{1}{81}$

So, the ratio of intensity at point A and B will be

$\frac{I_{A} }{I_{B} } = \frac{81}{16} =81:16$

Thus, the intensity ratio at point A and B will be 81:16.

7 0

3 years ago

A 75 kg bungee jumper leaps from a bridge. When he is 30 meters above the water, and moving at a speed of 20 m/s, the bungee cor

jasenka [17]

Answer:

k = 52.2 N / m

Explanation:

For this exercise we are going to use the conservation of mechanical energy.

Starting point. When it is 30 m high

Em₀ = K + U = ½ m v² + m g h

Final point. Right when you hit the water

$Em_{f}$ = K_{e} = ½ k x²

in this case the distance the bungee is stretched is 30 m

x = h

as they indicate that there are no losses, energy is conserved

Em₀ = Em_{f}

½ m v² + m g h = ½ k h²

k = $\frac{m (v^{2} + 2 g h)}{h^{2} }$

let's calculate

k = $\frac{75 \ ( 20^{2} + 2 \ 9.8 \ 30)}{30^{2} }$

k = 52.2 N / m

3 0

3 years ago

What is the english system of measurement called?

Artemon [7]

The answer is imperial units

8 0

3 years ago

A canoe floats in a lake. Inside the canoe is a 25 kg steel solid ball. If the ball is thrown into the lake, does the level of t

Rus_ich [418]

Answer:

Explanation:

volume of ball bearing = 4/3 π r³

= 4/3 x 3.14 x 1.5³

= 14.13 cc.

if D be the density of steel , weight of the ball bearing

= 14.13 x D x g

For the first case , water will be displaced to keep it floating

weight of displaced water will be equal to weight of steel

weight of displaced water = 14.13 Dg

mass of displaced water = 14.13 D

volume of displaced water = mass / density of water

= 14.13 D / d ; (where d is density of water) .

Now when the steel ball bearing is dipped in water , it will displace water equal to its volume only and it will be drowned

its volume = 14 .13 cc

14.13 D / d > 14.13 ( because D/d is more than one , since D > d )

volume of water displaced in first case is greater

water level will go up higher in first case .

Hence in the second case water level will go down .

Same will happen in case of 25 kg steel .

4 0

3 years ago