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4 years ago

Ikg.

Physics

1 answer:

Tcecarenko [31]4 years ago

6 0

Answer:

The mass of G1 at all times during this trial was 0.5 kg.

The velocity of G1 + G2 after the collision was -1.24 m/s.

The momentum of G1 after the collision was -2.10 kg · m/s.

Explanation:

i got it right

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What will happen for Potential energy of a body if the mass is dobled keeping height constant

4vir4ik [10]

<h3>Given :-</h3>

The mass of the body is doubled
The height of the body is constant

<h3>Solution :- </h3>

We know that ,

Potential energy = mgh

Therefore,

We can say that,

PE is directly proportional to Mass of the body

According to the question,

PE of the body = 2m * g * h. ...eq( I)

From (I) , we can conclude that, If mass of the body get doubled then its PE will also be doubled .

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3 years ago

Chegg ) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m(t)

stepladder [879]

Answer:

$m(t)=20e^{-0.5t}$

Explanation:

Given:

Initial mass of isotope (m₀) = 20 g

Half life of the isotope $(t_{1/2})$ = (ln 4) years

The general form for the radioactive decay of a radioactive isotope is given as:

$m(t)=m_0e^{-kt}$

Where,

$m(t)\to mass\ after\ 't'\ years\\t\to years\ passed\\k\to rate\ of\ decay\ per\ year$

So, the equation is: $m(t)=20e^{-kt}$

At half-life, the mass is reduced to half of the initial value.

So, at $t=t_{1/2},m(t)=\frac{m_0}{2}$ . Plug in these values and solve for 'k'. This gives,

$\frac{m_0}{2}=m_0e^{-k\times\ln 4}\\\\0.5=e^{-k\times\ln 4}\\\\Taking\ natural\ log\ on\ both\ sides,we\ get:\\\\\ln(0.5)=-k\times \ln 4\\\\k=\frac{\ln 0.5}{-\ln 4}=0.5$

Hence, the equation for the mass remaining is given as:

$m(t)=20e^{-0.5t}$

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4 years ago

In order to determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an in

Strike441 [17]

Answer:

μs = 0.75

μk = 0.58

Explanation:

From a force diagram:

$m*g*sin \theta - Ff=m*a$ (1)

$N-m*g*cos \theta = 0$ (2)

When it starts slipping, friction force is the maximum and acceleration is 0. Replacing these conditions on (1):

$m*g*sin \theta - \mu*m*g*cos \theta=0$ Solving for μs:

$\mu=tan \theta$

μs = tan 36.7° = 0.75

When it moves at constant speed, friction force is kinetic friction and acceleration is 0. With these conditions the coefficient is:

μk = tan 30.1° = 0.58

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4 years ago

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Y_Kistochka [10]

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3 years ago

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My name is Ann [436]

Answer:

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3 years ago

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