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OverLord2011 [107]
3 years ago
12

The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be

greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop (in m/s) if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?
Physics
1 answer:
kap26 [50]3 years ago
3 0

Answer:

v = 14.86 m/s

Explanation:

As we know that the force equation at the top is given as

\frac{mv^2}{R} = ma

now we know that

a_c = 1.5 g

so we have

\frac{v^2}{R} = 1.5 g

v = \sqrt{1.5 Rg}

so we will have

v = \sqrt{1.5(15)(9.81)}

v = 14.86 m/s

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CO2 increases in the winter

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Air at the poles tends to flow close to the surface toward the equator. What can you conclude about the characteristics of this
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15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remai
alexandr1967 [171]

The alpha particle is emitted at 4235 m/s

Explanation:

We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:

p_i = p_f\\ Mu=m_1 v_1 + m_2 v_2 =  

where:  

M =222u is the mass of the original nucleus

v=420 m/s is the initial velocity of the nucleus

m_1 = 4 u is the mass of the alpha particle

v_1 is the final velocity of the alpha particle

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v_2 = 350 m/s is the final velocity of the nucleus

Solving for v_1, we  find the final velocity of the alpha particle:

v_1 = \frac{Mu-m_2 v_2}{m_1}=\frac{(222)(420)-(218)(350)}{4}=4235 m/s

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4 0
3 years ago
A ball is launched from ground level at 20 m/s at an angle of 40° above the
DedPeter [7]

(a) The ball's height <em>y</em> at time <em>t</em> is given by

<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :

0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )

<em>t</em> = 0   or   (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 <em>g t</em>

<em>t</em> = (40 m/s) sin(40º) / <em>g</em>

<em>t</em> ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So

0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>

where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :

<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)

<em>y</em> ≈ 8.4 m

8 0
2 years ago
Someone please help me with these questions! (The ones in the picture) Please I am super confused!
lozanna [386]

Answer:

c-d

Explanation:

3 0
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