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3 years ago

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If the astronaut throws the tool with a force of 16.0 n , what is the magnitude of the acceleration a of the astronaut during th

e throw? assume that the total mass of the astronaut after she throws the tool is 80.0 kg . express your answer in meters per second squared.

2 answers:

Drupady [299]3 years ago

7 0

The magnitude of the acceleration of the astronaut : <u>0.2 m/s²</u>

<h3>Further explanation </h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object

∑F = m. a

$\large {\boxed {\bold {a = \frac {\sum F} {m}}}$

F = force, N

m = mass = kg

a = acceleration due to gravity, m / s²

the astronaut throws the tool with a force of 16.0 N ⇒ F = 16 N

the total mass of the astronaut after she throws the tool is 80.0 kg

⇒ m = 80 kg

magnitude of the acceleration of the astronaut during the throw

$\rm a=\dfrac{F}{m}\\\\a=\dfrac{16\:N}{80\:kg}\\\\a=\boxed{\bold{0.2\:\dfrac{m}{s^2}}}$

<h3>Learn more </h3>

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Maksim231197 [3]3 years ago

4 0

From Newton’s second law of motion, the value of the force is calculated by multiplying the mass and the acceleration as,

<span> F = mass x acceleration</span>

If we substitute the known values into the equation,

<span> 16 N = 80 kg x acceleration</span>

<span> Acceleration = 0.2 m/s<span>2</span></span>

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<u>Solution</u>:

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$a_{R = \frac{(32.8)^2}{516}$

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=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

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The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

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$\mu =\frac{ma}{mg}$

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Answer:

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Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

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From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

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At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

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$t_{A}$ is the time spent by train A

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From,

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$Time = \frac{Distance}{Velocity}$

Since the time spent by the two trains is equal,

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$\frac{Distance_{A} }{Velocity_{A} } = \frac{Distance_{B} }{Velocity_{B} }$

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This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

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Since it is train B that leaves a station,

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