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Pie
3 years ago
5

On a roller coaster, riders can experience a force of up to 4 g. What is the maximum acceleration of the roller coaster?

Physics
1 answer:
yKpoI14uk [10]3 years ago
7 0

m/s^2 is 39.2266

is the answer If thats what you needed


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The change in the wind patterns on the earth that causes an increase in ocean temperatures is
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An ice boat is coasting along a frozen lake. Friction between the ice and the boat is negligible, and so is air resistance. Noth
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(A) No

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(A) since there is nothing propelling the boat and the friction between the ice and the boat and also air resistance is negligible the net force of the system in the horizontal direction is zero and hence there is no change in the horizontal momentum of the boat.

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Question: How is the energy in the food we eat an example of chemical potential energy?
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3 years ago
Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an
ddd [48]

Answer:

Al's mass is 102.92  kg  

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

F_{net} = 0

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

F_{net} = \frac{dp_{horizontal}}{dt} = 0

p_{horizontal_i }= p_{horizontal_f}

where the suffix i and f means initial and final respectively.

The initial momentum will be:

p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}

But, as they are at rest, initially

p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0

p_{horizontal_i} = 0

So, this means:

p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0

We know that the have an combined mass of 195 kg:

m_{total} = m_{Al} + m_{Jo} = 195 \ kg.

so:

m_{Jo} = 195 \ kg - m_{Al}.

m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0

m_{Al} \  v_{Al_f} - m_{Al} \  v_{Jo_f}= - 195 \ kg \  v_{Jo_f}

m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}

m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } {  v_{Al_f} - v_{Jo_f} }

m_{Al} = \frac{195 \ kg  \ v_{Jo_f} } {    v_{Jo_f} - v_{Al_f} }

Now, we can use the values:

v_{Al_f}= 10.2 \frac{m}{s}

v_{Jo_f}= - 11.4 \frac{m}{s}

where the minus sign appears as they are moving at opposite directions

m_{Al} = \frac{195 \ kg  ( - 11.4 \frac{m}{s} ) } {   (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }

m_{Al} = 102.92 \ kg

and this is the Al's mass.

5 0
2 years ago
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