A 2.00 kg cart on a frictionless track is pulled by force of 3.00 N. What is the acceleration of the cart?

How much work is done when a very large force is applied to an object but does not move the object?

Ierofanga [76]

Work = (force) x (distance)

If the distance is zero (the object doesn't move), then it
doesn't matter how great the force is. The work is still zero.

4 0

3 years ago

Read 2 more answers

Give two examples of where a permanent magnet might used. <3

N76 [4]

Two examples of where you can find permanent magnets are
1.compass
2.speakers

5 0

3 years ago

Please help me, I beg you

MA_775_DIABLO [31]

Answer:

40g

Explanation:

Solubility of Copper sulfate at 90°=60g

Solubility of potassium bromide at 90°=100g

100g-60g=40g

8 0

3 years ago

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

3 years ago

The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian

Llana [10]

Complete question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v = c* \sqrt{1 - \frac{1}{n^2} }$

Explanation:

From the question we are told that

a = nb

The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

$h = t \sqrt{1 - \frac{v^2}{c^2} }$

Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)

So t = a = nb

and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)

i.e h = b

So

$b = nb [\sqrt{1 - \frac{v^2}{c^2} } ]$

$[\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}$

$v^2 =c^2 [1- \frac{1}{n^2} ]$

$v^2 =c^2 [\frac{n^2 -1}{n^2} ]$

$v = c* \sqrt{1 - \frac{1}{n^2} }$

6 0

3 years ago