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Sedbober [7]
3 years ago
10

Andrew pokes a marble, and the marble rolls down a ramp. The marble moves with speed. Which forces are acting on the marble in t

his situation?
When Andrew pokes the marble, he gives it [A tension/A friction/An applied/] force. As the marble rolls down the ramp, [Tension/Normal/Gravitational}, force pulls it down. The movement of the marble is slowed down by [Gravitational force/Friction/Elastic force] from the ramp
Physics
2 answers:
kvasek [131]3 years ago
5 0

Answer:

When Andrew pokes the marble, he gives it An applied force. As the marble rolls down the ramp, Gravitational force pulls it down. The movement of the marble is slowed down by Friction from the ramp.

Explanation:

At the beginning, Andrew pushes the marble, so is applying a force, which is called applied. force. Then, this force is removed (as Andrew no longer touches the marble), but now the weigth of the marble (the gravitational force) pushes the marble down along the ramp, because the gravitational force acts downward, so it has a donward component along the ramp. At the same time, the surface of the ramp exerts a force of friction against the motion of the marble, and the frictional force is always directed against the motion of the object.

slega [8]3 years ago
4 0
An applied force
gravitational force
friction
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A pure musical torte causes a thin wooden
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39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
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<u>Answer:</u> The final temperature of the solution is 80.14^oC

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The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

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7 0
3 years ago
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From the activity values and the decay constant, the mass of of Strontium in the sample is:

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<h3>What is the decay constant of an element?</h3>

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{λ = ln 2 / t1/2 

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t1/2 is the half-life of the isotope.

The half-life is converted to seconds since the decay constant is asked in per seconds.

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The activity of the element, A, the decay constant, λ and the number of nuclei, N are related as follows:

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1 mole of Strontium-90 contains 6.02×10^23 nuclei.

The mass, m of Strontium in the sample is calculated:

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