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In-s [12.5K]
3 years ago
13

The law of inertia applies to objects

Physics
2 answers:
Anestetic [448]3 years ago
6 0

Answer:

<em>at</em><em> </em><em>rest</em><em> </em><em>and</em><em> </em><em>in</em><em> </em><em>motion</em>

Explanation:

<em>The</em><em> </em><em>law</em><em> </em><em>of</em><em> </em><em>inertia</em><em> </em><em>applies</em><em> </em><em>to</em><em> </em><em>objects</em><em> </em><em>at</em><em> </em><em>rest</em><em> </em><em>and</em><em> </em><em>in</em><em> </em><em>motion</em>

abruzzese [7]3 years ago
6 0

Answer:

Explanation: At rest and in motion because an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

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At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th
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Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

 ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r  

So (86kg* 101^2)/(167) =

Fc=5253 N

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3 years ago
How much force is needed to accelerate a 15kg bowling ball at 2 m/s^2
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The work done to compress a spring with a force constant of 290.0 N/m a total of 12.3 mm is: a) 3.57 J b) 1.78 J c) 0.0219 J d)
iren2701 [21]

Answer:

Work done, W = 0.0219 J

Explanation:

Given that,

Force constant of the spring, k = 290 N/m

Compression in the spring, x = 12.3 mm = 0.0123 m

We need to find the work done to compress a spring. The work done in this way is given by :

W=\dfrac{1}{2}kx^2

W=\dfrac{1}{2}\times 290\times (0.0123)^2

W = 0.0219 J

So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.

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How large is the area of water found on Mars?
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More than five million cubic kilometers of ice have been identified.


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A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

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