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lorasvet [3.4K]
3 years ago
11

What are the typical units for measuring blood pressure?

Physics
1 answer:
VARVARA [1.3K]3 years ago
5 0

Answer:

Blood pressure is measured in units of millimeters of mercury (mmHg).

Explanation:The readings are always given in pairs, with the upper (systolic) value first, followed by the lower (diastolic) value. Hope This Helps :)

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What is kinetic energy?
Lemur [1.5K]

Answer:

A, the energy an object has due to its motion.

Explanation:

Kinetic energy is the energy created by motion.

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Below is a single of DNA, What is the complementary base pair for RNA?<br> AGC, CGT, ATA, GAT
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It’s GAT because I just learned about this
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Hydrogen/hydronium ion concentration
Valentin [98]

Answer:

Pure water is considered to neutral and the hydronium ion concentration is 1.0 x 10-7 mol/L which is equal to the hydroxide ion concentration. So the pH is the -log of the [hydronium ion].

Explanation:

Pure water is considered to neutral and the hydronium ion concentration is 1.0 x 10-7 mol/L which is equal to the hydroxide ion concentration. So the pH is the -log of the [hydronium ion].

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Group 7A of the periodic table contains the?
olga2289 [7]

The correct answer is A. Most reactive non metals.

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3 years ago
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Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass
kirza4 [7]

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

5 0
3 years ago
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