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3 years ago

If a weightlifter lifts a 286 kg mass 0.25 meters above his head, how much PEg does the mass have?

Physics

2 answers:

KonstantinChe [14]3 years ago

5 0

Answer:

Answer on edge is 700

Explanation:

vazorg [7]3 years ago

3 0

Answer:

Change in potential energy is 700.7 J.

Explanation:

Given:

Mass of the object is, $m=286$ kg

Height to which the object is raised above his head, $h=0.25$ m

Acceleration due to gravity is, $g=9.8$ m/s²

We know that for a mass $m$ raised to a height $h$ , the change in potential energy is given as:

$\Delta PE=mgh$ , where, $\Delta PE\rightarrow \textrm{Change in Potential Energy}$

Now, plug in the given values and solve.

$\Delta PE=286\times 9.8\times 0.25=700.7\textrm{ J}$

Therefore, the change in the potential energy is 700.7 J.

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625 /5=125/5=25/5=5
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is 5 Over 8

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3 years ago

If a sample of a radioactive isotope has a half-life of 1 year, how much of the original sample will be left at the end of the s

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1/4 of the original

Explanation:

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11 months ago

Which is not expecting acceleration?

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A car on the freeway experiencing a net force of -120 N.

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Here net force on it is due to two given forces

$F_1 = 4320 N$ towards right

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so here net force is given as

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3 years ago

The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

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3 years ago

two children having a disagreement pull on a 50N sled in opposite directions. one pulls with a force of 300N east, the other wit

prohojiy [21]

<span>Since forces are vector quantities, we must indicate direction using positive and negative values. East will be assigned positive and west will be negative. Friction will act as a negative force since it impedes action. To calculate the net force we sum the vector quantities, as follows. Net force equals 50n which is derived by the following calculation: 300n-220n-30n.</span>

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3 years ago

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