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olga2289 [7]
2 years ago
14

The velocity of a 10.0 kg object that has 720 J of kinetic energy is m/s. (Report the answer to two significant figures.)

Physics
2 answers:
RideAnS [48]2 years ago
7 0

the answer is 12 m/s

nikdorinn [45]2 years ago
4 0
Kinetic energy<span> is the </span>energy<span> of motion. An object that has motion - whether it is vertical or horizontal motion - has </span>kinetic energy<span>. It is expressed as:

KE = mv^2 /2

720 = 10.0v^2 /2
v = 12 m/s

Hope this answers the question. Have a nice day.</span>
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the acceleration due to gravity on the moon is 1.6 m/s^2 what is the gravitational potential energy of a 1200 kg lander resting
Mumz [18]

<u>Answer</u>

672,000 Joules

<u>Explanation</u>

Gravitational potential energy (P.E) is the energy possessed by a body that is at a potential height from the ground.

IT is calculated by the formula;

P.E = mgh

Where m ⇒ mass

            g ⇒ acceleration due to gravity

            h ⇒ height from trhe ground.

P.E = 1200 × 1.6 × 350

      = 672,000 Joules.

5 0
2 years ago
The earth's magnetic field, with a magnetic dipole moment of 8.0×1022Am2, is generated by currents within the molten iron of the
Fed [463]

To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.

Part A) By definition we know that magnetic dipole moment is

m = IS

Where,

I = Current

S = Area

m = IA \rightarrow m= I( \pi r^2)

Replacing with our values we have that,

8*10^{22} = I \pi(\frac{3000*10^3}{2})^2

Re-arrange to find I,

I = 1.1317*10^{10} A

Part B) To find the Current density we need to find the cross sectional area of the Wire:

A = \pi r^2 \\A = \pi (\frac{1000*10^3}{2})^2\\A = 7.854*10^{11}m^2

Finally the current density is simply J

J = \frac{I}{A}\\J = \frac{1.1317*10^{10}}{7.854*10^{11}}\\J = 0.0144A/m^2

PART C) Finally to make the comparison with the given values we have to cross-sectional area would be

A = \pi (10-3)^2 \\A = 49\pi

Therefore the current density would be

J = \frac{I}{A}\\J = \frac{30A}{49\pi}\\J = 9.549*10^5 A/m^2

Comparing the two values we can see that the 2mm wire has a higher current density.

4 0
2 years ago
Write difference between upstroke and downstroke of water pump
lakkis [162]

Explanation:

Upstroke is a mechanism which helps to raise the plunger and downstroke helps to help lower the plunger. On the up-stroke of the plunger, the lower valve opens and the upper valve is closed. ... Whereas, on the downstroke, the lower valve closes and the upper one opens.

7 0
2 years ago
Calculate the density of mercury if 500 cm3
notka56 [123]

Answer:

The density of the mercury is 13.2 g/cm³

Explanation:

Density is a measurement that compares the amount of matter an object

has to its volume

Density is equal to mass divided by volume

We need to find the density of mercury if 500 cm³ has a mass of

6.60 kg in g/cm

We must to change The kilogram to grams

The mass of mercury is 6.60 kilograms

1 kilogram = 1000 grams

6.60 kilograms = 6.60 × 1000 = 6600 grams

Density = mass ÷ volume

The volume of the mercury is 500 cm³

The density = 6600 ÷ 500

The density = 13.2 g/cm³

<em>The density of the mercury is 13.2 g/cm³ </em>

5 0
2 years ago
What is the total energy q released in a single fusion reaction event for the equation given in the problem introduction? use c2
kkurt [141]

Answer:

4.3\cdot 10^{-12} J

Explanation:

The fusion reaction in this problem is

4^1_1H \rightarrow ^4_2He +2e^+

The total energy released in the fusion reaction is given by

\Delta E = c^2 \Delta m

where

c=3.0\cdot 10^8 m/s is the speed of light

\Delta m is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

m(^1_1H)=1.007825u is the mass of one nucleus of hydrogen

m(^4_2 He)=4.002603u is the mass of one nucleus of helium

So the mass defect is:

\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u

The conversion factor between atomic mass units and kilograms is

1u=1.66054\cdot 10^{-27}kg

So the mass defect is

\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg

And so, the energy released is:

\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J

6 0
2 years ago
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