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3 years ago

Help What are energy and work

Physics

1 answer:

mylen [45]3 years ago

8 0

Answer:

<em>In physics, energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat, the object. Energy is a conserved quantity; the law of conservation of energy states that energy can be converted in form, but not created or destroyed.</em>

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Explanation:

<h2><u><em>HOPE THIS HELPS</em></u></h2>

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A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto

sesenic [268]

<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float= $l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water= $10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$ cm

Hence, the depth of barge float=3 cm

<h2 />

4 0

4 years ago

Read 2 more answers

If you were told an atom was an ion, you would know the atom must have a ______?

andreyandreev [35.5K]

Answer:

charge

Explanation:

7r0I and its etc. ,"!×_/;

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3 years ago

7. An 8 kg ball is travelling to the east at 10 ms', collides with a 2 kg ball travelling to the

Ymorist [56]

Answer:

The final velocity of the ball is 7m/s

Explanation:

M1=8kg, V1 =10m/s , M2=2kg , V2=-5m/s

initial momentum before collison

m1v1+m2v2

=8×10 +2×(-5) =80-10 = 70kg m/s

final momentum after collison

=(m1+m2)×v

=(8+2)×v

=10v

According to the law of conversion of momentum

initial momentum =final momentum

70=10v

10v=70

v=70/10

v=7m/s

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3 years ago

Which part(s) of a hair can be analyzed for nuclear DNA?

kiruha [24]

<span>hair follicle
should be your answer</span>

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3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago