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2 years ago

What is the half-life of the imaginary element Lokium? Show your work

Physics

1 answer:

antoniya [11.8K]2 years ago

3 0

Answer:What is the half life of the element Lokium?

The half-live of the element Lokium is 4.

Explanation:

sorry if im wrong, have good day

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A car rounds a flat curve and experiences a centripetal force directed toward the center of the curve and perpendicular to the d

Usimov [2.4K]

<h2>Answer:</h2>

If a car is rounding a flat curve, it experiences a centripetal force that pulls it towards the center of the circle it is rotating in.

Now,

The centripetal force can be balanced by the centrifugal force caused due to the acceleration of the body at the high speed which counters the centripetal force and in turn prevents the car from slipping down the curve.

So,

If the car doesn't hit the gas then the car will fall down from the curve as the Centripetal force will exceed the Centrifugal force of the car.

However, if the car doesn't hit the brake then the car will maintain it's position on the flat curve track as the centrifugal force will counter the effect of centripetal force directed towards the center.

3 0

3 years ago

Read 2 more answers

Jay bought a toy car. To make the car move, he must turn a key attached to a spring. Turning the key winds the spring tight. The

lapo4ka [179]

The chemical energy in Jay's body, to kinetic energy in the car

5 0

3 years ago

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How do you convert seconds to minutes, seconds to hours , minutes to seconds and hours to seconds

Bess [88]

Answer:

You could memorize conversions, or use conversion charts, or do a quick internet search for help.

Explanation:

I'll do a few of the conversions for you

Seconds to minutes:

there are 60 seconds in one minute. So if you are wondering how many seconds are in 3 and 1/2 minutes, you would do this conversion:

60 sec/min times x sec/3.5 min

which can be written as 60 x 3.5 = 210. so "x" seconds would be 210 seconds in 3 1/2 minutes.

you could do the same thing in the opposite direction for minutes to seconds:

1 min has 60 seconds. So in 7.25 minutes, how many seconds are there?

1 min/60 sec times 7.25 min/x sec

which can be written as 7.25 x 60 = 435. so "x" seconds would be 435 seconds in 7.25 minutes.

hours to seconds:

this one is slightly more complicated

In one hour there are 60 minutes, and in one minute there are 60 seconds.

so to convert from hours to seconds you would do this conversion:

1 hr/60 min times 1 min/60 sec. then the "min" would cancel out, and you would be left with the label "hr/sec". to do the math, it would be 1 hr / 60 x 60.

60x60 = 3600. so you would have 1 hr/3600 sec. So in one hour there are 3600 seconds.

so if you want to know how many seconds are in 6.75 hours:

6.75 hr/x sec times 3600 sec/1 hr

6.75 x 3600 = 24,300 so there are 24,300 seconds in 6.75 hours.

I hope this helps :)

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3 years ago

Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109

r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

$E=\Delta mc^2\\\Delta m=\frac{E}{c^2}$ (1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

$E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J$

Hence, by replacing in the equation (1) you have (c=3*10^{8}m/s)

$\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg$

HOPE THIS HELPS!!

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3 years ago

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A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have

denpristay [2]

1) The total mechanical energy of the rock is:
$E=U+K$
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
$E_i=U=mgh$
where $m=4 kg$ is the mass, $g=9.81 m/s^2$ is the gravitational acceleration and $h=5 m$ is the height.
Putting the numbers in, we find the potential energy
$U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J$

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
$E_f=K= \frac{1}{2}mv^2$
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
$K=U=196.2 J$

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
$W=196.2 J-0 J=196.2 J$

6 0

3 years ago