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Elenna [48]
2 years ago
14

Consider a runner in lane 1 with a speed v1 and a runner in lane 4 with a speed v4 , where the ratio of the speeds v4/v1 is 2 .

The radius of lane 4 is b times the radius of lane 1, and both runners have the same mass. Which of the following claims is correct about the net forces F1 and F4 exerted on the runners in lanes 1 and 4, respectively?
a. F4 is equal to 2/b F1.
b. F4 is equal to 4/b F1.
c. F4 is equal to 2/b^2 F1.
d. F4 is equal to 4/b^2 A F1.
Physics
1 answer:
kondor19780726 [428]2 years ago
7 0

Answer:

F_4=\frac{4}{b}F_1

Explanation:

Centripetal force is the net force acting on a body which makes it move along a curved path. This force is always towards the center of curvature.

The centripetal force is given by:

F = mv² / r

where m is the mass of the body, v is the velocity of the body, r is the radius, F is the centripetal force and v²/r is the centripetal acceleration.

Given that:

\frac{v_4}{v_1}=2\\\\v_4=2v_1\\\\ Also, radius\ of\ lane\ 4(r_4)=2*radius\ of\ lane\ 1\\\\Let\ r_1\ be\ radius\ of\ lane\ 1,m=mass\ of\ runner\ in\ lane\ 1=mass\ of\ runner\ in\ lane\ 4\ \\and\ v_4=velocity\ of\ runner\ in\ lane\ 4.\  Hence:\\\\r_4=b*r_1=br_1\\\\The\ centripetal\ force\ for\ lane\ 4(F_4)\ is:\\\\F_4=\frac{mv_4^2}{r_4}\\\\ F_4=\frac{m(2v_1)^2}{br_1}\\\\F_4= \frac{4mv_1^2}{br_1} \\\\But\ F_1=\frac{mv_1^2}{r_1}\\\\Hence\ F_4=\frac{4}{b}F_1

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<h2>The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>

Explanation:

One period means time taken to complete one revolution.

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Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

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