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9 days ago

7

Justine is ice-skating at the Lloyd Center what is her final velocity if she accelerates at a rate of 2.0 meters per second for

3.5 seconds

1 answer:

Ostrovityanka [2.2K]9 days ago

7 0

2*3.5 = 7m/s

You need to multiply the acceleration by the time (which must both be in seconds; if not, convert them to the same units).

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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 [2355]

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

3 0

10 days ago

Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet

ValentinkaMS [2425]

Answer:

The rate at which the root beer level is decreasing is 0.08603 cm/s.

Explanation:

The formula for the volume of the cone is:

$V=\frac {1}{3}\times \pi\times r^2\times h$

Where V denotes the cone's volume

r indicates the radius

h signifies the height

The ratio of radius to height remains consistent throughout the cone.

Thus, we have r = d / 2 = 10 / 2 cm = 5 cm

h is 13 cm

Consequently, r / h = 5 / 13

r = {5 / 13} h

$V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h$

$V=\frac {550}{3549}\times h^3$

Additionally, we differentiate the volume expression in relation to time:

$\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}$

Given that $\frac {dV}{dt}$ = -4 cm³/sec (the negative sign indicates outflow)

h equals 10 cm

Hence,

$-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}$

$\frac{55000}{1183}\times \frac {dh}{dt}=-4$

$\frac {dh}{dt}=-0.08603\ cm/s$

The rate at which the root beer level is decreasing is 0.08603 cm/s.

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1 month ago

The flight of a kicked football follows the quadratic function f(x)=−0.02x2+2.2x+2, where f(x) is the vertical distance in feet

Keith_Richards [2256]

The ball covers a horizontal distance of 0.902 meters. The trajectory of a kicked football adheres to a quadratic equation expressed as: f(x), where f(x) indicates the vertical distance in feet, and x signifies how far the ball travels horizontally. To compute the distance the ball will advance before striking the ground, we set the condition f(x) = 0. Upon solving this quadratic equation, we find that the horizontal distance traveled by the ball is: x = -0.902 meters, leading us to conclude that it travels 0.902 meters across the field.

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Which statement is always false for athletes participating in team sports?

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Which statement can never be true for athletes in team sports? The statement that is always false among the listed options for team sports athletes is choice C) Conflict resolution indicates a lack of sportsmanship. The other statements are valid in the context of team sports.

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Answer

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A) Remaining planetesimals formed within the frost line are referred to as ASTEROIDS.

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E) Meteor showers are linked to debris from COMETS.

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21 day ago