Question

ipc unit 2: disasters unit assessment

1. on the low end of the scale, meteorites have a mass of at least 100 kg. the mass of earth is approximately 5.972×10²⁴ kg. calculate the force of gravity between earth and a meteorite at the edge of earth’s atmosphere given the distance from outer space to the center of the earth is 6.478×10⁶ m. remember that the gravitational constant, g, is 6.6743×10⁻¹¹ m³/kg·s²

a. 949.8 n
b. 1.423×10⁹ n
c. 6.153×10⁹ n
d. 2.58×10¹⁷ n

1. the perseid meteor shower occurs roughly the same time every year, usually in august. this is because at that time in earth’s orbit our planet passes through a debris trail of particles left by the comet swift - tuttle. as the earth approaches the rocky debris left by the comet, the distance between earth and the debris (increases / remains the same / decreases) and the force of gravity between earth and the debris (increases / remains the same / decreases).
2. in populated areas, avalanche control teams will intentionally trigger smaller (10,000 kg) avalanches over time to prevent larger ones from occurring and potentially becoming a disaster. given that the average acceleration for an avalanche is 6.23 m/s², what is the force of that avalanche?

a. 1605 n
b. 62,300 n
c. 1,605,000 n
d. 62,300,000 n

1. prior to an avalanche being triggered, the mass of snow at the top of a mountain has a large amount of (kinetic / potential) energy. after an avalanche has started, this converts to (kinetic / potential) energy as the snow moves down the mountain slope.
2. as the magnitude of the charges increases, the electrical force (increases / remains the same / decreases). as the distance between two charges increases, electrical force (increases / remains the same / decreases).
3. calculate the potential energy of baseball - sized hail (0.200 kg in mass) prior to falling as precipitation given that the average height at which hail forms is 8352 m above earth’s surface.

a. 16,387,000 j
b. 1,670,000 j
c. 16,387 j
d. 1670 j

Explanation:

Step1: Use Newton's gravity law

$F = G\frac{m_1m_2}{r^2}$
Substitute values: $G=6.6743\times10^{-11}$, $m_1=5.972\times10^{24}\ \text{kg}$, $m_2=100\ \text{kg}$, $r=6.478\times10^6\ \text{m}$

Step2: Compute denominator

$r^2 = (6.478\times10^6)^2 = 4.1964484\times10^{13}$

Step3: Compute numerator

$Gm_1m_2 = 6.6743\times10^{-11} \times 5.972\times10^{24} \times100 = 6.6743\times5.972\times10^{15} \approx 39.86\times10^{15}$

Step4: Final division

$F = \frac{39.86\times10^{15}}{4.1964484\times10^{13}} \approx 949.8\ \text{N}$

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Step1: Analyze distance change

Earth approaches comet debris, so distance decreases.

Step2: Relate to gravity force

Gravity is inversely proportional to $r^2$, so force increases.

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Step1: Use Newton's second law

$F=ma$
Substitute $m=10,000\ \text{kg}$, $a=6.23\ \text{m/s}^2$

Step2: Calculate force

$F =10000\times6.23 = 62300\ \text{N}$

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Step1: Identify initial energy

Snow at rest at height has potential energy.

Step2: Identify final energy

Moving snow has kinetic energy.

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Step1: Relate charge magnitude to force

Electrical force is proportional to charge product, so force increases with charge magnitude.

Step2: Relate distance to force

Electrical force is inversely proportional to $r^2$, so force decreases with distance.

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Step1: Use gravitational potential energy

$PE = mgh$
Substitute $m=0.200\ \text{kg}$, $g=9.81\ \text{m/s}^2$, $h=8352\ \text{m}$

Step2: Compute energy

$PE =0.200\times9.81\times8352 = 0.200\times81933.12 = 16386.624 \approx16387\ \text{J}$

Answer:

1. a. 949.8 N
2. decreases; increases
3. b. 62,300 N
4. potential; kinetic
5. increases; decreases
6. c. 16,387 J