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igor_vitrenko [27]
3 years ago
13

A ball is thrown upward. As it passes 5.0 m height it is traveling at 4.0 m/s up. What was its initial upward velocity? (a) 7.0

m/s ( b) 9.1 m/s (c) 10.7 m/s (d) 12.1 m/s (e) 14.6 m/s
Physics
1 answer:
sveticcg [70]3 years ago
5 0

Answer:

c) 10.7m/s

Explanation:

From the exercise we know that at 5m the ball  is traveling at 4m/s

To calculate its initial velocity we need to solve the following equation:

v_{y}^{2}=v_{oy}^{2}+2g(y-y_{o})

Since the initial height is 0

Solving for v_{o}

v_{oy}=\sqrt{v_{y}^{2}-2gy}=\sqrt{(4m/s)^2-2(-9.8m/s^2)(5m)}=10.7m/s

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Men and women exhibit different preferences in mate selection because both have different attraction cues and priorities for their partner. These cues and priorities leads to difference in preference of life partner.

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0.015m^3

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2 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
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