The magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²
<h3>Circular motion</h3>
From the question, we are to determine the magnitude of the centripetal acceleration.
Centripetal acceleration can be calculated by using the formula
Where is the centripetal acceleration
is the velocity
and is the radius
From the given information
and
Therefore,
Hence, the magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²
Learn more on circular motion here: brainly.com/question/20905151
Answer:
An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
Explanation:
When an athlete tries to stop his career, it takes several meters to stop completely, due to the inertia produced.
When trying to push a car, at first it is very difficult, because, due to inertia, the car tends to remain still. But once it is put into motion, the effort is much less to be done, since then inertia causes it to keep moving.
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>D is the correct answer. A Bourdon gage is a popular and commonly used kind of gauge for measuring pressure and vacuum. One use for a Bourdon gage is to indicate steam pressure.</span>