Physics, 22.03.2021 19:50 lovebrain123

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, she travels in a vertical circle with a radius of 15.0 m. She has a mass of 80.0 kg and her motorcycle has mass 30.0 kg. What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere

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Answer from: justinrojo621

v = 12.1 m/s

Explanation:

When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

$F_{c} = N + m*g (1)$

Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

$F_{c} = m*\frac{v^{2}}{r} (2)$

Since the normal force takes any value as needed to make (1) equal to (2), if the speed diminishes, it will be needed less force to keep the equality valid.In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.In this condition, from (1) and (2), we can find the minimum possible value of the speed that still keeps the motorcycle touching the surface, as follows: $v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)$

Answer from: Quest

it may be wrong but i think its a

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

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