3 years ago

What are the requirements for an inertial frame of reference?

1 answer:

3 0

As simply put as I believe is possible, it's the immediate space-time co-oridinate in which the observer is not in motion. All measurements by the observer are in relation to that 'place'

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Blood contains positive and negative ions and therefore is aconductor. A blood vessel, therefore, can be viewed as anelectrical

CaHeK987 [17]

Answer:

<h2>Magnetic field required for the given induced EMF is 1.41 T</h2>

Explanation:

Potential difference across the blood vessel is given as

$E = vBd$

here we know that the speed is given as

$v = 14.8 cm/s$

$d = 4.80 mm$

$E = 1 mV$

now we have

$1 \times 10^{-3} = (14.8 \times 10^{-2})B(4.80 \times 10^{-3})$

$B = 1.41 T$

Now volume flow rate of the blood is given as

$Q = Av$

$Q = \frac{\pi d^2v}{4}$

from above equation we have

$v = \frac{E}{Bd}$

Now we have

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2 years ago

Some coastal regions of the world have cooler summers and warmer winters than inland regions at the same latitude. What accounts

maks197457 [2]

Water holds in heat very well. Keep the temperature more steady and average. The areas around the water will also have a less variant change in temperature as a result. This property of water is known as high specific heat.

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3 years ago

A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturi

Darina [25.2K]

Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

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2 years ago

Please help with this!!!!!

34kurt

(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K) I hoped that helped

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2 years ago

How does energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface? a. Organisms in the deep se

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that would be the energy from the core. because of the fact that ht eearths core is producing 45mill killawhats and giggawats

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2 years ago

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