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Artemon [7]
3 years ago
9

A roller coaster has a vertical loop with radius 25.7 m. With what minimum speed should the roller-coaster car be moving at the

top of
the loop so that the passengers do not lose contact with the seats?
m/s
Physics
1 answer:
gogolik [260]3 years ago
7 0

Answer:

15.88m/s

Explanation:

At the top of the roller coaster you will have three forces acting on the roller-coaster. See the image below. Fc is the centripetal force (for an object in circular motion), Fg is the gravitational force, and Fn is the normal force. To achieve the minimum speed we assume the roller-coaster is barely touching the vertical loop and so the normal force is zero. This leaves two acting forces.

F_g = F_c\\mg = \frac{m\times v^2}{r}\\v = \sqrt{gr} = \sqrt{9.81 \times 25.7}  = 15.88 m/s

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200 Hz = 200 cycles per sec 

<span>1 cycle, the period = 1/200 = 0.005 seconds, or 5 milli seconds.</span>
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Pre-questioning identifies a purpose for reading.
nadya68 [22]

Answer:

True

Explanation:

Pre-questioning may help a reader focus on information s/he hopes to find in the reading selection.

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A particle has a charge of -4.25 nC.
SpyIntel [72]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m

The distance from the electric field is 1.71436 m

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Answer rain gauge measures rain shadow units millimetres

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2 years ago
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7. An 8 kg ball is travelling to the east at 10 ms', collides with a 2 kg ball travelling to the
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Answer:

The final velocity of the ball is 7m/s

Explanation:

M1=8kg,  V1 =10m/s , M2=2kg , V2=-5m/s

initial momentum before collison

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final momentum after collison

=(m1+m2)×v

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=10v

According to the law of conversion of momentum

initial momentum =final momentum

70=10v

10v=70

v=70/10

v=7m/s

3 0
3 years ago
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