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3 years ago

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At a certain factory, 340 kg crates are dropped vertically from a packing machine onto a conveyor belt moving at 1.00 m/s. (A mo

tor maintains the belt's constant speed.) The coefficient of kinetic friction between belt and crate is 0.400. After a short time, slipping between the belt and the crate ceases and the crate then moves along with the belt. Consider the period of time during which the crate is being brought to rest relative to the belt.

1 answer:

Masteriza [31]3 years ago

6 0

Answer:

+b±√b² - 4ac /2a

0.6t ± √36-36/2a

Explanation:

Work done = 1/2 mv² where v = (1.2)²

Therefore, 1/2m(1.2)ω mgh

1/2m (1.2)² = 0.4 × m ×10 5

s = 1.44 / 2.4 = 1.44 / 8

S = ut - 1/2gt²

Where u = 1.2

g = 0.9 × 10

Therefore,

1.8 = 1.2v-2t²

2t²c-1.2t+1.8 = 0

t² - 0.6t + 0.9 = 0

0.6t ± √36-36/2a

Solving this further, we make use of the formula

+b±√b² - 4ac /2a

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3 years ago

Acceleration is defined as the rate of change for which characteristic?

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4 years ago

Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with

IgorLugansk [536]

(a) $2.5\cdot 10^{-6}eV$

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

For the microwave photon,

$\lambda=50.00 cm = 0.50 m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV$

(b) $2.5 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the visible light photon,

$\lambda=500 nm = 5 \cdot 10^{-7}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV$

(c) $2500 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the x-ray photon,

$\lambda=0.5 nm = 5 \cdot 10^{-10}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV$

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3 years ago

You plan to throw stones by using a sling that you whirl over your head. You will whirl the stone in a horizontal circle with a

MariettaO [177]

Answer:

$330.635$ meter per second square

Explanation:

Given

Height ,h $= 2.50$ m

Radius, r $= 1.50$ m

Distance travelled, $d = 15.9$ m

Since, the stone is released and it falls towards the sky, its initial velocity is zero

As per equation

$h = ut + \frac{1}{2} gt^2\\$ , we will evaluate "t"

Substituting the given values in above equation, we get

$2.5 = 0 *t + 0.5 * 9.81 * t^2\\t^2 = \frac{2.5}{0.5*9.81}\\ t = \sqrt{0.5096} \\t = 0.714$

Velocity

$= \frac{d}{t} \\= \frac{15.9}{0.714}\\= 22.27$

centripetal acceleration

$= \frac{v^2}{r} \\= \frac{22.27^2}{1.5}\\ = 330.635\\$

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4 years ago