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likoan [24]
3 years ago
15

Explain with method pls..

Physics
1 answer:
seraphim [82]3 years ago
5 0

Answer:

See explanation below.

Explanation:

Given two blocks of masses M and m, And  a force F acting on the block of mass M pulling it outside.

The two masses system can be considered as a whole, With F  being the only external force acting on them.

Here the spring force has been ignored as while considering the two system as a whole, it will be accounted as an internal force for the system

Therefore, The acceleration of the system will be:

a=\frac{Total\:external\:force}{Total\:mass}=\frac{F}{m+M}

Now both the masses are moving with this same acceleration towards the direction of force .

Acceleration of mass m = a (as the system is moving as a whole)

Force on the mass m= mass times acceleration.

Force=ma=\frac{mF}{m+M}

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What is the circle of least confusion?
Artist 52 [7]

Answer and Explanation:

In optics, a CoC(Circle of Confusion) is defined the minimum cross section of a paraxial bundle of rays made by a lens which is sphero-cylindrical type and can be viewed as an optical spot, which do not converge perfectly at the focus  while a point source is being imaged due to spherical aberration.

The Circle of Confusion is also referred to as circle of indistinctness or a blur spot

5 0
4 years ago
What traction of the radioisotope<br>remains in the body after one day?​
r-ruslan [8.4K]

The fraction of radioisotope left after 1 day is (\frac{1}{2})^{\frac{1}{\tau}}, with the half-life expressed in days

Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}

where:

m_0 is the mass of the radioactive sample at t = 0

\tau is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}

And the term on the left represents the fraction of the radioisotope left after a certain time t.

Therefore, after t = 1 days, the fraction of radioisotope left in the body is

\frac{m(1)}{m_0}=(\frac{1}{2})^{\frac{1}{\tau}}

where the half-life \tau must be expressed in days in order to match the units.

Learn more about radioactive decay:

brainly.com/question/4207569

brainly.com/question/1695370

#LearnwithBrainly

5 0
3 years ago
A runner generates 1260 W of thermal energy. If this heat has to be removed only by evaporation, how much water does this runner
sergey [27]

Answer:0.502kg

Explanation:

F4om the relation

Power x time = mass x latent heat of vapourization

P.t=ML

1260 * 15 *60 = M * 22.6 * 10^5

M= 1134000/(22.6 *10^5)

M=0.502kg=502g

3 0
3 years ago
As gravity continues to crush the inner core, the fusion of carbon and oxygen begins. What temperature is needed for the fusion
Olin [163]

Answer:

600,000,000 degree C

Explanation:

This stage is the last stage and is refereed to as supernova. In the beginning of this stage, gravity pulls the inner core and crush it, due to which fusion of atoms starts. Carbon and Oxygen fuse together and the temperature is about of 600,000,000 degree C.

The most heavier atom that can be formed out of this fusion is the iron. The moment all the atoms becomes of iron, no further fusion is possible hence that body emits radiation of high intensity and collapse causing a big supernova.

8 0
3 years ago
A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

KE = \frac{7}{10}*150*(0.200)^2(50)^2

KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0
3 years ago
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