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Dmitrij [34]
3 years ago
5

What happens to gravitational potential energy as a roller coaster moves down a hill?

Physics
1 answer:
Kaylis [27]3 years ago
5 0
It is converted to kinetic energy.
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A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of grav
erastovalidia [21]

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes 90-(\theta+5) with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

\sum F_x = 0

W+1.4-Fm cos(70)

We know that W is equal to

W= 0.06*100N = 6N

Substituting,

Fm cos (70) = W+1.4N

Fm cos (70) = 6N + 1.4N

Fm = \frac{7.4}{cos(70)}

Fm = 21.636N

<em>For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that</em> \beta  = \theta

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3 years ago
Question for the day!
Alex17521 [72]

Answer:

1. <--> A.

2. <--> C.

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explanation: i know my science!

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Air bubbles being released when breathing from the mouth
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2 years ago
Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v
wlad13 [49]
1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we callv_f the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
where
m_A=7 kg is the mass of ball A
m_B=2 kg is the mass of ball B
v_{Ai}=6 m/s is the initial velocity of ball A
v_{Bi}=-12 m/s is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find v_f, we find that the final velocity of the balls is
v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s
and the positive sign means the two balls are going to the right.


2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
where
v_{fA} is the final velocity of ball A
v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s
and the total momentum after the collision is:
p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

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3 years ago
A hummingbird flies in circles around a bush. if the velocity of the hummingbird is 720 cm/s and the radius of the circular path
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57,600 cm/s ^2 ...........
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