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3 years ago

URGENT PLEASE HELP

Physics

2 answers:

nalin [4]3 years ago

7 0

Answer:

This depends on what angle they are approaching each other before they collided.The two simple cases are if they are running in the same direction or opposite direction from each other. For either case, use the conservation of momentum equation to solve: M_total*V_result = M1*V1 + M2*V2

Explanation:

Here are two possible solutions.

Head-on collision: M1=78, V1=8.5, M2=72, V2=-7.5 (that's negative because he's running the other way), M_total = 78+72 = 150, so V_result = (78*8.5 - 72*7.5)/150 = 0.82 m/s. Sanity check, they weigh about the same and so most of their velocity should cancel out.

Running the same way: change the sign of V2 to positive so V_result = (78*8.5 + 72*7.5)/150 = 8.02 m/s. Sanity check, they weigh about the same and the resultant speed is between the two starting velocities.

<em>hope it helps:)</em>

aivan3 [116]3 years ago

7 0

Answer:

‍♂️

Explanation:

‍♂️

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We know the electric field ( $\vec{E}$ ) is given by $\vec{E} = - \nabla V$ , 'V' being the potential.

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How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

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Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

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This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

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2 years ago

**URGENT PLEASE HELP**

URGENT PLEASE HELP