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3 years ago

Why is the use of international units important in medicine?

Physics

1 answer:

Sidana [21]3 years ago

7 0

Answer:IU (international unit): An international unit (IU) is an internationally accepted amount of a substance.

This type of measure is used for the fat-soluble vitamins (such as vitamins AD and E) and certain hormones, enzymes, and biologicals (such as vaccines).

Explanation:

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A motorcyclist accelerates from rest to 10 mi/hr. what is the change in velocity

gulaghasi [49]

The change in velocity is 10 mi/h (4.47 m/s)

Explanation:

The change in velocity of the motorcyclist is given by

$\Delta v = v-u$

where

v is the final velocity

u is the initial velocity

In this problem, we have

u = 0 (the motorbike starts from rest)

v = 10 mi/h

Therefore, the change in velocity is

$\Delta v = 10 -0 = 10 mi/h$

And keeping in mind that

1 mile = 1609 m

1 h = 3600 s

We can convert it into m/s:

$\Delta v = 10 \frac{mi}{h} \cdot \frac{1609 m/mi}{3600 s/h}=4.47 m/s$

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

6 0

3 years ago

A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the

liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0

3 years ago

Please select the word from the list that best fits the definition

alexdok [17]

Answer:

yuuuhhhhhhh

Explanation:

it maxed Imao

4 0

3 years ago

Read 2 more answers

A circuit consists of one 330 ohm resistor in series with two other resistors (470 ohms and 220 ohms) connected in parallel. Wha

umka21 [38]

Answer:

power drain on an ideal battery, P = 0.017 W

Given:

$R_{1} = 330\ohm$

$R_{2} = 470\ohm$

$R_{3} = 220\ohm$

Since, $R_{2} = 470\ohm$ and $R_{3} = 220\ohm$ are in parallel and this combination is in series with $R_{1} = 330\ohm$ , so,

Equivalent resistance of the circuit is given by:

$R_{eq} = \frac{R_{2}R_{3}}{R_{2} + R_{3}} + R_{1}$

$R_{eq} = \frac{470\times 220}{470 + 220} + 330$

$R_{eq} = 149.85 + 330 = 479.85 \ohm$

power drain on an ideal battery, P = $\frac{V^{2}}{R_{eq}}$

P = $\frac{2.9^{2}}{479.85}$

P = 0.017 W

4 0

4 years ago

Two airplanes leave an airport at the same time.The velocity of the first airplane is 700 m/h at a heading of 31.3 the velocity

MArishka [77]

Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity

$v_{12} = v_1 - v_2$

here

speed of first plane is 700 mi/h at 31.3 degree

$v_1 = 700 cos31.3\hat i + 700 sin31.3\hat j$

$v_1 = 598.12\hat i + 363.7\hat j$

speed of second plane is 570 mi/h at 134 degree

$v_2 = 570 cos134 \hat i + 570 sin134 \hat j$

$v_2 = -396\hat i + 410\hat j$

now the relative velocity is given as

$v_{12} = (598.12 + 396)\hat i + (363.7 - 410)\hat j$

$v_{12} =994.12\hat i -46.3 \hat j$

now the distance between them is given as

$d = v* t$

$d = (994.12 \hat i - 46.3 \hat j)* 3$

$d = 2982.36\hat i - 138.9\hat j$

so the magnitude of the distance is given as

$d = \sqrt{2982.36^2 + 138.9^2}$

$d = 2985.6$ miles

so the distance between them is 2985.6 miles

5 0

3 years ago

Read 2 more answers