Answer:
applying 1st eq of motion vf=vi+at here we have to find a=vf-vi/t , a= 1-1/5 , a=0/5 then we got a=0 here(vf value 3.6km/h is converted in standard unit 3.6×1000/3600 so we get vf=1m/s²
Yes. An object of large mass is pulled down onto a surface with a greater force than an object of low mass and, as a consequence, there is greater friction between the surface of the heavy object than between the surface of the light object.
<span>F = ma
</span>Ff = μ*Fn
<span>Fn = Fw
</span>Fw = mg
<span>So we have: </span>
<span>Ff = μmg </span>
<span>And </span>
<span>Ff = ma </span>
<span>So... </span>
<span>μmg = ma </span><span> </span>
<span>μg = a </span>
<span>And we can solve for the acceleration: </span>
<span>(0.15)(9.81 m/s²) = a </span>
<span>a = 1.47 m/s² </span>
The answer to this question is True.
The person must stand at a radius of 0.99 m
Explanation:
In order for the person to stand on the merry go round, the force of friction acting on the person must provide the centripetal force necessary to keep the person in uniform circular motion.
Therefore, we can write:
where:
- the term on the left is the force of friction, and the term on the right is the centripetal force
is the coefficient of friction
m is the mass of the person
is the acceleration of gravity
is the angular velocity
r is the radius of the circular path
Solving the equation for r, we find the radius at which the person must be standing:
Learn more about circular motion:
brainly.com/question/2562955
brainly.com/question/6372960
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