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Ludmilka [50]
2 years ago
7

On a playground, there is a merry‑go‑round. In order to get it moving, Bonnie applies a force of 39 N . The merry‑go‑round measu

res 5.1 m from the axis of rotation to the edge where Bonnie applies her force. Assuming she applies her force perpendicularly to a line drawn from the axis of rotation, what is the magnitude of the torque Bonnie imparts to the merry‑go‑round?
Physics
1 answer:
hjlf2 years ago
6 0

Answer:

198.9 Nm

Explanation:

Torque = Force * Rperpendicular, since the force applied is 39, and the radius is 5.1, we can build an equation that will lead to 198.9=39*5.1.

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Two friends grab different sides of a textbook and pull with forces of 6.7 N to the east and 4.4 N to the west, respectively. Wh
elena-s [515]

Explanation:

This equation for acceleration can be used to calculate the acceleration of an object that is acted on by a net force. For example, Xander and his scooter have a total mass of 50 kilograms. Assume that the net force acting on Xander and the scooter is 25 Newtons. What is his acceleration? Substitute the relevant values into the equation for acceleration:

<h2>Answer:</h2>

a=Fm=25 N50 kg=0.5 Nkg

The Newton is the SI unit for force. It is defined as the force needed to cause a 1-kilogram mass to accelerate at 1 m/s2. Therefore, force can also be expressed in the unit kg • m/s2. This way of expressing force can be substituted for Newtons in Xander’s acceleration so the answer is expressed in the SI unit for acceleration, which is m/s2:

a=0.5 Nkg=0.5 kg⋅m/s2kg=0.5 m/s2

8 0
3 years ago
after one hour, 1/16 of the initial amount of certain radioactive isotope remains undecayed. the half life of the isotope is:
Anarel [89]
(1/2)^x=1/16
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7 0
4 years ago
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A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx3 , where c = 8.0 N/m3 . The pa
Pie

Answer:4.58 m/s

Explanation:

Given

mass of Particle m=4 kg

F=-cx^3

a=\frac{F}{m}

a=-\frac{cx^3}{m}

a=-\frac{8x^3}{4}

a=-2x^3

v\frac{\mathrm{d} v}{\mathrm{d} x}=-2x^3

vdv=-2x^3dx

integrating

\int_{6}^{v_b}vdv=\int_{1}^{-2}-2x^3dx

\frac{v_b^2-6^2}{2}=-\frac{1}{2}\left [ \left ( -2\right )^4-\left ( 1\right )^4\right ]

\frac{v_b^2-36}{2}=-0.5\times 15

v_b^2=36-15

v_b=\sqrt{21}

v_b=4.58 m/s

6 0
3 years ago
uppose the bottom block has a mass of 0.4 kg and the top block has a mass of 0.1 kg. What force do you need to exert (in newtons
notsponge [240]

Answer:

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Explanation:

Newton's second law is

         F = ma

As in this case the two blocks move with constant speed, it implies that the acceleration is zero, therefore the force applied to the system is zero

       F = 0

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3 years ago
In the absence of air, a penny and a feather dropped from the same height will
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When they are dropped at the same height, taking their mass into account, the feather will touch the ground last.
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