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3 years ago

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A mosquito of mass 0.15 mg is found to be flying at a speed of 50 cm/s with an uncertainty of 0.5 mm/s. (a) How precisely may it

s position be known? (b) Does this inherent uncertainty present any hindrance to the application of classical mechanics?

1 answer:

amid [387]3 years ago

5 0

Answer:

a) Δx = t 0.05 + 0.5

, Δx = 0.5 cm, b) Do not present any problem

Explanation:

The kinematic equation for constant speed is

v = x / t

x = v t

a) the uncertainty can be calculated with

Δx = dx /dv Δv + dx /dt Δt

Δx = t Δv + v Δt

Speed is

v = (50.00 ± 0.05) cm / s

The most common uncertainty for the time of Δt = 0.01 s

We replace

Δx = t 0.05 + 50 0.01

Δx = t 0.05 + 0.5

We must know the time to have an explicit value, if we assume that the measure was t = 1s

Δx = 0.5 cm

b)

Do not present any problem since its value is not very small, we must take as soon as the quantum effects and the velocity are not so high that we must take into account the relativistic effects

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In 8,450 seconds, the number of radioactive nuclei decreases to 1/16 of the number present initially. What is the half-life (in

almond37 [142]

Answer:

<h2>2113 seconds</h2>

Explanation:

The general decay equation is given as $N = N_0e^{-\lambda t} \\\\$ , then;

$\dfrac{N}{N_0} = e^{-\lambda t} \\$ where;

$N/N_0$ is the fraction of the radioactive substance present = 1/16

$\lambda$ is the decay constant

t is the time taken for decay to occur = 8,450s

Before we can find the half life of the material, we need to get the decay constant first.

Substituting the given values into the formula above, we will have;

$\frac{1}{16} = e^{-\lambda(8450)} \\\\Taking\ ln\ of \both \ sides\\\\ln(\frac{1}{16} ) = ln(e^{-\lambda(8450)}) \\\\\\ln (\frac{1}{16} ) = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328$

Half life f the material is expressed as $t_{1/2} = \frac{0.693}{\lambda}$

$t_{1/2} = \frac{0.693}{0.000328}$

$t_{1/2} = 2,112.8 secs$

Hence, the half life of the material is approximately 2113 seconds

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3 years ago

The empire state building is 1,450 feet tall. King Kong weighs 20 tons and he climbs to the very top. If he jumps off the top, w

lozanna [386]

Answer:velocity= $93.12m/s$

Explanation:

We shall use conservation of energy to solve this problem

we have

Potential energy of King Kong at top of building = His kinetic energy at the bottom

Potential energy of an object = $m\times g\times h$ = 20000 kg

Where m is mass of an object(20000 kg)

g is accleration due to gravity = 9.81 $m/s^{2}$

h is the height above the surface of earth = 1450 feet = 441.96 meters

Applying values we get

$(P.E)_{TOP}=(20000\times 9.81\times 441.96) Joules\\\\(P.E)_{TOP}= 86712.552KiloJoules$ ......................(i)

Now Kinetic energy is given by

$(K.E)_{Bottom}=\frac{1}{2}mv^{2}$ .......................(ii)

Equating i and ii we get

86712.552 = $\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2E}{m}}$

Applying values we get

$v=\sqrt{\frac{2\times 86712.552\times 10^{3}}{20\times 10^{3}}}\\\\v= 93.12m/s$

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3 years ago

An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10

Akimi4 [234]

Answer:

a) $F=475.7Hz$

b) $F'=410.899Hz$

Explanation:

From the question we are told that:

Velocity of eagle $V_1=35m/s$

Frequency of eagle $F_1=440Hz$

Velocity of Black bird $V_2=10m/s$

Speed of sound $s=343m/s$

a)

Generally the equation for Frequency is mathematically given by

$F=f_0(\frac{v-v_2}{v-v_1})$

$F=440(\frac{343-10}{343-35})$

$F=475.7Hz$

b)

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$F'=440(\frac{343+10}{343+35})$

$F'=410.899Hz$

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3 years ago

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Answer:

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