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3 years ago

6

A particle, of mass 6 kg, is in equilibrium on a rough horizontal plane under a force o-f magnitude T N, which acts at an angle

15° above the horizontal. Given the coefficient of friction between the particle and the rough horizontal plane is 0.35, what values could T take?

1 answer:

Helga [31]3 years ago

3 0

Answer:

T is less than or equal to 19 N

Explanation:

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Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

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$v_{f}$ -v₀ = v_{e} $ln ( \frac{M_{o} }{M_{f}} )$

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The final mass is the mass of the engines + the mass of the rocket

M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

thrust = 5.26 N

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Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

thrust = v_{e} $\frac{M_{f} - M_{o} }{t_{f} - t_{o} }$

v_{e} = thrust $\frac{\Delta t}{\Delta M}$

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v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

v_{f}-0 = v_{e} $ln ( \frac{M_{o} }{M_{f} } )$

we calculate

v_{f} = 786.93 ln (0.0927 / 0.080)

v_{f} = 115.95 m / s

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3 years ago

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kvv77 [185]

Answer:

$W = 78.48N\\W =0.0784kN\\W = 8kgf\\$

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Explanation:

if

$m=8kg=3.6lb\\$

and g=9.81 m/s2=32.16 ft/s2

and

W=m*g

we can just replace de mass and gravity and we have

$W = 8kg * 9.81 \frac{m}{s^{2} } =78.48N\\W = \frac{78.48N}{1000} =0.0784kN\\W = 8kgf\\$

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