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IRISSAK [1]
3 years ago
6

Anna Litical is practicing a centripetal force demonstration at home. She fills a bucket with water, ties it to a strong rope, a

nd spins it in a circle. Anna spins the bucket when it is half-full of water and when it is quarter-full of water. In which case is more force required to spin the bucket in a circle? Explain using an equation as a "guide to thinking.
Physics
1 answer:
dlinn [17]3 years ago
6 0

Answer:

Explanation:

Suppose when bucket is half full it has a mass of 2 m rotating in a circle of  radius r

When Bucket is quarter full then it has a mass of m rotating in a circle of radius r.

When an object moves in a circular path then it experiences an inward force which is given by

F_1=\frac{2mv^2}{r}

where v=velocity of bucket

Force in case 2 is given by

F_2=\frac{mv^2}{r}

Thus F_1=2F_2

therefore force required in half bucket is more than force required in quarter bucket full.

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A spring with a spring constant value of 2500 StartFraction N over m EndFraction is compressed 32 cm. A 1.5-kg rock is placed on
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it's 9m

Explanation:

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If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he
forsale [732]
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q= mC_s \Delta T
where m is the mass of the substance, Cs is its specific heat capacity and \Delta T is the increase of temperature.

If we re-arrange the formula, we get
C_s =  \frac{Q}{m \Delta T}
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
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3 years ago
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S Problem Set<br> 2.) 6.4 x 109 nm to cm
anyanavicka [17]

Answer:

6.4\cdot 10^2 cm

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

1 nm = 10^{-9} m

So we have:

6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m

Now we can convert from metres to centimetres, keeping in mind that

1 m = 10^2 cm

So, we find:

6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm

8 0
3 years ago
The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of
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Answer:

Δh_air=714m

Explanation:

Given data

P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\  g=9.8m/s^{2}

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

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Δh_air=(pHg+ΔhHg)÷p_air

=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m

Δh_air=714m

7 0
3 years ago
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