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mariarad [96]
2 years ago
14

What is the centripetal force acting on a 1.5 kg mass moving in a circular path with a centripetal

Physics
1 answer:
zloy xaker [14]2 years ago
8 0

The centripetal force acting on a 1.5 kg mass moving in a circular path is 27N

A centripetal force is a net force acting on an object in order to maintain the object's movement in a circular motion.

According  to Newton's first law, it states an object will continue to proceed its movement in a straight line unless acted upon by an external force.

The centripetal force is the external force at work here and  It's the net force that propels the object in a circular motion.

Using the formula for calculating centripetal force:

\mathbf{F_c = mass (m) \times acceleration (a)}

\mathbf{F_c = 1.5 \ kg \times18 \ m/s^2}

\mathbf{F_c = 27 \ N}

Learn more about centripetal force here:

brainly.com/question/11324711?referrer=searchResults

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Explanation:

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3 years ago
One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet
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0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

               \frac{G M m}{r^{2}}=m \omega^{2} r

The orbit’s period is given by,

               T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}

Where,

T_{e} = Earth’s period

T_{p} = planet’s period

M_{s} = sun’s mass

r_{e} = earth’s radius

Now,

             T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}

As, planet mass is equal to 0.7 times the sun mass, so

            T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}

Taking the ratios of both equation, we get,

             \frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}

            \frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}

           \frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}

Given T_{p}=9.5 \text { days } and T_{e}=365 \text { days }

          \frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}

         r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}

7 0
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natulia [17]

Answer:

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Explanation:

Given data

Mass of merry go round M_m = 120 kg

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Mass of boy M_{boy} = 25 kg

We know that the final velocity is given by

\omega_f = \frac{\frac{1}{2}M_m \omega_i }{M_{boy} + \frac{1}{2} M_m }

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Kinetic energy = 1/2 (mass)*(velocity)^2

ke = 1/2 (1500kg) (17m/s)^2
ke = 216750 kg*m^2/s^2 = 216750 Joules

(sorry, I don't know significant figures. but this is otherwise the correct answer with the right units. hope this helps!! :))
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