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1 month ago

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A hiker walks 9.4 miles at an angle of 60° south of west. Find the west and south components of the walk. Round your answers to

the nearest tenth. Ax = miles Ay = miles

2 answers:

Sav [3.1K]1 month ago

3 0

For this calculation, we apply our understanding of trigonometric functions related to a right triangle in order to determine the x and y components of the walk. We perform the following steps:

sin 60 = ax / 9.4
ax = 8.14

cos 60 = ay / 9.4
ay = 4.7

Ostrovityanka [3.2K]1 month ago

3 0

Solution:

Ax= -4.7

Ay= -8.1

Clarification:

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The first experiment measures inertial mass, while the second experiment measures gravitational mass.

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A student conducts two different experiments to observe resistance to changes in motion, both when at rest and in motion.

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Response:

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Clarification:

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1 month ago

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

Sav [3153]

Response:

Magnitude of the electrostatic force acting on the +32 µC charge, $F_{net} = 12 N$

Clarification:

Let q₁ = +32 µC, located at x₁ = 0

q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m

q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m

Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).

$F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }$

$F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N$

Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).

$F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }$

$F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N$

The resultant electrostatic force on the 32 µC charge is $F_{net} = |F_{2} + F_{3}|$

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2 months ago

"if a stream flow measures 12 meters in 60 seconds, what is the stream's average rate of flow?"

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A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Maru [3345]

Answer:

The final size is nearly the same as the initial size because the increase in size $1.055\times 10^{- 7}$ is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, $m_{p} = 1.67\times 10^{- 27} kg$

Initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This proton energy is $\simeq 250 GeV$

Thus, the speed of the proton, v $\simeq c$

The time to cover 1 km = 1000 m of distance is calculated as:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

According to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width

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1 month ago