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Hatshy [7]
3 years ago
12

A fairgrounds ride spins its occupants inside a flying saucer-shaped container. if the horizontal circular path the riders follo

w has a 5.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?
Physics
1 answer:
Valentin [98]3 years ago
3 0

We know that the acceleration due to gravity g is: g = 9.81 m/s^2

So the centripetal acceleration (w) is:

w^2 = 1.5 g / r

w^2 = 1.5 * (9.81 m/s^2) / 5 m

w = 1.716 rad / s

To convert to rad to rev:

w = (1.716 rad / s) * (1 rev / 2π rad) * (60 s/min)

<span>w = 16.4 rev/min </span>

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Data given:

m=12840kg

a=0.490m/s²

Formula:

F=ma

Solution:

F= 12840kg×0.490m/s²

F=6291.6N

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Modern observations have shown that the geometry of the universe is flat and the universe mist be infinite.

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8 0
2 years ago
Knowing the constant g what will the gravitational force between two masses be if the gravitational force between them is 36n an
charle [14.2K]
The gravitational force between two masses is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the two masses
r is the separation between the two masses

We see that the force is proportional to the inverse of the square of the distance: F \sim  \frac{1}{r^2}
therefore, if the distance is tripled:
r'=3r
The force decreases by a factor 1/9:
F \sim  \frac{1}{(3r)^2}= \frac{1}{9}  \frac{1}{r^2}

Since the original force was 36 N, the new force will be
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6 0
3 years ago
A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

\mathrm{m}_{1}=600 \mathrm{kg}

\mathrm{m}_{2}=900 \mathrm{kg}

\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

600 \times 5=900 \times v_{2}

3000=900 \times v_{2}

\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

5 0
4 years ago
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