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Natali5045456 [20]

3 years ago

5

A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti

on force of 12 N. How far does it slide on the snow before coming to rest?

1 answer:

uysha [10]3 years ago

3 0

Answer:The sled slides 16.875m before rest.

Explanation:

$a=\frac{F}{m} =\frac{12N}{20kg}$

a=0.6 m/s²

$Vf=0=Vi-a.t$

$t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg$

$d= Vi.t - \frac{a.t^{2}}{2}$

$d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m$

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Answer:

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Explanation:

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3 years ago

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

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substituting this into the equation

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Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

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$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

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3 years ago

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3 years ago

An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and

svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

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2.

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the pressure at exit is -37.5kPa

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Genrish500 [490]

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3 years ago

Read 2 more answers