Physics, 18.02.2020 01:52 rebeccathecatt

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5.30 kN, and the radius of the circle is 12.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v = 3.60 m/s? (b) What is FB if v = 14.0 m/s? Use g=9.80 m/s2.

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Answer from: wycentliver1859

Explanation:

As the force is given as 5.30 kN or $5.30 \times 1000 N$ . Hence, mass will be calculated as follows.

$F_{w}$ = mg

$5.30 \times 10^{3} = m \times 9.8 m/s^{2}$

m = 540.816 kg

(a) At the top, centripetal force $F_{c}$ is acting upwards and the weight of the riders and car, $F_{w}$ will be acting downwards.

Therefore, force on the car by the boom will be calculated as follows.

$F_{B} = F_{w} - F_{c}$

or, $F_{B} = mg - \frac{mv^{2}}{r}$

= $5.30 \times 1000 N - \frac{540.816 kg \times (3.60)^{2}}{12}$

= 4715.919 N

Hence, the force $F_{B}$ on the car from the boom is 4715.919 N. This means that the car will be hanging on the boom and the boom will exert an upward force.

(b) Now at the top, centripetal force $F_{c}$ will be acting upwards and the weight of cars and car riders will be acting in the downwards direction.

Hence, we will calculate the force on car by the boom as follows.

$F_{B} = F_{w} - F_{c}$

or, $F_{B} = mg - \frac{mv^{2}}{r}$

= $5.30 \times 1000 N - \frac{540.816 kg \times (14.0)^{2}}{12}$

= -3533.33 N

Therefore, car will be pushing on the boom and the boom will exert a downward force.

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.

Answer from: Quest

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me

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