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1 year ago

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What is the force of gravity between two 50.0kg masses that are separated by 0.300m?3.71x10-8N5.59x10-7N2.78x104N1.85x10-6N

1 answer:

Varvara68 [4.7K]1 year ago

7 0

We will have the following:

$\begin{gathered} F=G\frac{m_1m_2}{r^2}\Rightarrow F=\frac{(6.67\ast10^{-11}m^3\ast kg^{-1}\ast s^{-2})(50kg)(50kg)}{(0.3m)^2} \\ \\ \Rightarrow F=1.852777778...\ast10^{-6}N\Rightarrow F\approx1.85\ast10^{-6}N \end{gathered}$

So, the force is approximately 1.85*10^-6 N.

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The lons entering the mass spectrometer have the same charges. After being accelerated through a potential difference of 8.20 kV

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The calculated magnitude is 6.73 x 10³ V/m.

AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.

Potential difference = 8.20 kV= 8.20 x 10³ V

radius= 19.4/100=0.194 m

total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194

= 1.218 m

therefore Magnitude= 8.20 x 10³ / 1.218

=6.73 x 10³ V/m

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It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon

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Answer: $8.76\times 10^{-3} min^{-1}$

Explanation:

Given

n=5

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According to Avrami equation

$y=1-e^{-kt^n}$

where y=fraction Transformed

k=constant

t=time

$0.3=1-e^{-k(100)^5}$

$e^{-k(100)^5} =0.7$

Taking log both sides

$-k\cdot (10^{10}=\ln 0.7$

$k=3.566\times 10^{-11}$

At this Point we want to compute $t_{0.5}\ i.e.\ y=0.5$

$0.5=1-e^{-kt^n}$

$0.5=e^{-kt^n}$

$0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}$

taking log both sides

$\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5$

$t^5=1.943\times 10^{10}$

$t=114.2 min$

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3 years ago

In order for an object to have kinetic energy it must have a mass and a ?

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Answer:

Velocity

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E = P.E + K.E

The potential energy of a body is due to the height from the surface of the earth.

P.E = mgh

The kinetic energy of the is possessed by the body due to the virtue of its motion,

K.E = ½ mv²

If there is no velocity associated with the body, there is no K.E in the body.

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Larger animals use proportionately less energy than smaller animals; that is, it takes less energy per kg to power an elephant t

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Answer:

Part a)

$Q = 952 cal/day$

Part b)

$P = 46 Watt$

Part c)

$\Delta P = 54 W$

Explanation:

As we know that 5000 kg African elephant requires 70,000 Cal for basic needs per day

so we will have

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$Q = 70,000 Cal$

so we have energy required per kg

$E = \frac{70,000}{5000}$

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Part a)

now we know that per kg the energy required will be same

so we have mass of the human is 68 kg

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Part b)

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Part c)

resting power given in the book is

$P' = 100 W$

so this is less than the power given

$\Delta P = 100 - 46$

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Answer:

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