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Natasha2012 [34]

2 years ago

6

An electric field of 3x10^16 n/c is needed to create a spark in the air. If the charged particles in the field are separated by

a distance of 1mm how many coulombs of charge are needed for a spark (round to the nearest whole number)

1 answer:

meriva2 years ago

6 0

The answer of this problem is positive 3

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A client with hypertension who weighs 72.4 kg is receiving an infusion of nitroprusside (Nipride) 50 mg in D5W 250 ml at 75 ml/h

Mkey [24]

To solve this problem it is necessary to simply apply the concepts related to cross-multiply and proportion between units.

Let's start first by relating the amount of dose needed to be supplied per hour, in other words,

The infusion of 250ml should be supplied at a rate of 75ml / hour, so what amount x of mg hour should be supplied with 50Mg.

$\frac{x}{75ml/hour} \rightarrow \frac{50mg}{250ml}$

$x \rightarrow \frac{50mg*75ml/hour}{250ml}$

$x \rightarrow \frac{3750mg}{250hour}$

$x \rightarrow 15\frac{mg}{hour}$

Converting to mcg units we know that 1mg is equal to 1000mcg and that 1 hour contains 60 min, therefore

$x \rightarrow 15\frac{mg}{hour}$

$x \rightarrow 15\frac{mg}{hour}(\frac{1000mcg}{1mg})(\frac{1hour}{60min})$

$x \rightarrow 250mcg/min$

The dose should be distributed per kilogram of the patient so if the patient weighs 72.4kg,

$Dose = \frac{250mcg/min}{72.4kg}$

$Dose = 3.5 \frac{mcg/min}{kg}$

Therefore the client will receive 3.5mcg/kg/min.

8 0

2 years ago

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

2 years ago

Discuss five occasions when people dance

Xelga [282]

Answer:

Party, Birthday, Weddings, Nightclub, Just for fun

7 0

2 years ago

When you do work on an object some of your energy is ____ to that object

HACTEHA [7]

Maybe the word could be converted?

8 0

2 years ago

which of the following is the correct si unit to use in measuring the mass of a boulder a. tons b. kilograms c. ounces d. pounds

Salsk061 [2.6K]

B. kilograms iuyiuiuhikj

7 0

2 years ago

Read 2 more answers