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3 years ago

What is work done in physics?

Physics

2 answers:

Eddi Din [679]3 years ago

6 0

Answer:

work done is the product of force and distance moved in the direction of a force

Kisachek [45]3 years ago

6 0

Answer:

work is said to be done when an energy or force is used.

Explanation:

when you carry a cement and you take a step, work is done.

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A 3.00 kg block moving 2.09 m/s

Talja [164]

Answer:11.64kgm/s

Explanation:

4 0

3 years ago

Explain the terms boiling and boiling point.

Volgvan

The particular temperature at which vaporisation occurs is known as the boiling point of liquid. Volume of water increases when it boils at 100° C. 1 cm3 of water at 100 ° C becomes 1760 cm3 of steam at 100 ° C.

Hope it helps!!!!!!!!!!!!!!!!!!!!! ~~~~~~~~~~~~~~~~~~~

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3 0

3 years ago

At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/

Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0

3 years ago

(1 point) A pilot flies in a straight path for 1 hour and 30 min. She then makes a course correction, heading 10 degrees to the

kati45 [8]

Answer:

The pilot is 2214.22 miles from her starting position

Explanation:

Since the pilot is traveling at a constant speed of 635 mph, the total distance traveled can be easily found as follows:

$1stleg: d_{1} = 1.5*635 = 952.5 mi\\2ndleg: d_{2} = 2*635 = 1270 mi$

There was a 10 degrees deviation, so the angle between the trajectory of both legs is 170 degrees.

The distance we need to find is that from the start of the first leg to the end of the second leg, those three distances form a triangle and since the side we're interested in is opposite to the 170 degrees angle, we can determine its length by the law of cosines:

$d_{3}^{2} =d_{1}^{2} + d_{2}^{2} -2*d_{1}*d_{2}*cos(170)\\d_{3}^{2} =952.5^{2} + 1270^{2} -2*952.5*1270*cos(170)\\d_{3}=2214.22mi$

The pilot is 2214.22 miles from her starting position

8 0

3 years ago

Phosphorous reacts with chlorine gas to produce phosphorous pentachloride. Calculate the mass of product produced when 25.0 g of

MariettaO [177]

Answer : The mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

Explanation : Given,

Mass of P = 25 g

Mass of $Cl_2$ = 25 g

Molar mass of P = 30.97 g/mole

Molar mass of $Cl_2$ = 71 g/mole

Molar mass of $PCl_5$ = 208.24 g/mole

First we have to calculate the moles of $P$ and $Cl_2$ .

$\text{Moles of }P=\frac{\text{Mass of }P}{\text{Molar mass of }P}=\frac{25g}{30.97g/mole}=0.807moles$

$\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{25g}{71g/mole}=0.352moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2P+5Cl_2\rightarrow 2PCl_5$

From the balanced reaction we conclude that

As, 5 moles of $Cl_2$ react with 2 moles of $P$

So, 0.352 moles of $Cl_2$ react with $\frac{2}{5}\times 0.352=0.1408$ moles of $P$

That means, in the given balanced reaction, $Cl_2$ is a limiting reagent and it limits the formation of products and $P$ is an excess reagent because the given moles are more than the required moles.

Now we have to calculate the moles of $PCl_5$ .

As, 5 moles of $Cl_2$ react with 2 moles of $PCl_5$

So, 0.352 moles of $Cl_2$ react with $\frac{2}{5}\times 0.352=0.1408$ moles of $PCl_5$

Now we have to calculate the mass of $PCl_5$ .

$\text{Mass of }PCl_5=\text{Moles of }PCl_5\times \text{Molar mass of }PCl_5$