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goblinko [34]
3 years ago
6

a particle moves a distance of 190 ft along a straight line. As it moves, it is acted upon by a constant force of magnitude 15 l

b in a direction opposite to that of the motion. What is the work done by the force
Physics
1 answer:
AleksAgata [21]3 years ago
6 0

Answer:

3868.1625 Nm

Explanation:

Given that:

Distance moved = 190 ft

Magnitude of force = 15 lb

Converting units :

1 lb = 4.45 Newton

1 ft = 0.305 meter

Hence,

Distance moved = (190 * 0.305) = 57.95 meter

Magnitude of force = (15 * 4.45) = 66.75 Newton

Workdone = Force * distance

Workdone = 66.75 N * 57.95 m

Workdone = 3868.1625 Nm

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A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If
vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

  • The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
  • The traslational part can be written as follows:

       K_{trans} = \frac{1}{2}* M* v_{cm} ^{2}  (1)

  • The rotational part can be expressed as follows:

       K_{rot} = \frac{1}{2}* I* \omega ^{2}  (2)

  • where I = moment of Inertia regarding the axis of rotation.
  • ω = angular speed of the rotating object.
  • If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

       v = \omega * R (3)

  • For a solid cylinder, I = M*R²/2 (4)
  • Replacing (3) and (4)  in (2), we get:

       K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2}  (5)

  • Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

       K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2}  +\frac{1}{4}* M* v_{cmc}^{2}  =  \frac{3}{4}* M* v_{cmc} ^{2} (6)

  • Repeating the same steps for the spherical shell:

        I_{sph} = \frac{2}{3} * M* R^{2} (7)  

       K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2}  (8)

      K_{sph} = \frac{1}{2}* M* v_{cms} ^{2}  +\frac{1}{3}* M* v_{cms}^{2}  =  \frac{5}{6}* M* v_{cms} ^{2} (9)

  • Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
  • And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
  • Rearranging, and taking square roots on both sides, we get:

       \frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)

  • This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.
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Greg is in a bike race. At mile marker four (out of ten), his speed was measured at 13.5 mph. Which best describes the measured
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Answer:

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Explanation:

we know that P=W÷t

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F=ma

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        =15240 J

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P=W÷t          t=5 mints = 5×60=300 sec

P=15240÷300

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