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lesya692 [45]
2 years ago
10

What is the net work done on the object over the distance shown?

Physics
1 answer:
enyata [817]2 years ago
3 0

Answer:

0

Explanation:

the positive cancels out the negative area

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A 0.50-kg bomb is sliding along an icy pond (frictionless surface) with a velocity of 2.0 m/s to the west. The bomb explodes int
nevsk [136]

Answer:

2.667m/s to the north and 3.333 m/s to the west

Explanation:

According to law of momentum conservation, the total momentum should be conserved before and after the explosion.

Before the explosion, the momentum was

0.5*2 = 1 kg m/s to the west

Therefore the total momentum after the explosion should be the same horizontally and vertically.

Vertically speaking, it was 0 before the explosion. After the explosion:

0.2*4 + 0.3v = 0

0.3v = -0.8

v = -0.8/0.3 = -2.667 m/s

So the vertical component of the 0.3kg piece is 2.667m/s to the north

Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:

0.2*0 + 0.3V = 1

0.3V = 1

V = 1/0.3 = 3.333 m/s

So the horizontal component of the 0.3kg piece is 3.333 m/s to the west

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3 years ago
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Two application of heat energy​
densk [106]

Answer:

Heat is very important in our daily life in warming the house, cooking, heating the water, and drying the washed clothes. The heat has many usages in the industry as making and processing the food and manufacture of the glass, the paper, the textile, and etc.

Explanation:

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When have you experienced an increase in kinetic<br> energy within a system?
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If a man starts running on a boat with an acceleration a with respect to the boat, there is no external force that acts on the Boat+Man system

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Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage
agasfer [191]

v = 2.45×10^3\:\text{m/s}

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

F = \dfrac{d{p}}{d{t}} (1)

Assuming that the velocity remains constant then

F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}

Solving for v, we get

v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}

The exhaust rate dm/dt is

\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}

\;\;\;\;\;= 1.36×10^4\:\text{kg/s}

Therefore, the velocity at which the exhaust gases exit the engines is

v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}

\;\;\;= 2.45×10^3\:\text{m/s}

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Number 4 is c ,  number 5 is a  ,  number 6 is d and 7 is a 
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