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Taya2010 [7]
2 years ago
5

Wile E. Coyote wants to launch Roadrunner into the air using a long lever asshown below. The lever starts at rest before the Coy

ote steps onto one end and pivots around its center of mass. Roadrunner is standing on the other end, which is held up by a small piece of ground underneath. The lever has mass 1.2 kg and length 16 m, the Roadrunner has mass 3.2 kg, and there is no friction at the pivot point of the lever.
a) Draw free body diagrams for the lever, Coyote, and Roadrunner. Make sure to label where all of the forces act on the lever’s free body diagram.
b) What is the minimal mass that Coyote must have to cause the lever to start rotating?
c) The coyote actually has a mass of 18 kg. If the roadrunner goes flying when the lever gets to a 45∘ angle from the horizontal, what is the roadrunner’s speed at that time, assuming the pivot is frictionless?
d) (Optional/challenge) What is the angular acceleration just after Coyote steps onto the lever?

Physics
1 answer:
algol132 years ago
4 0

Answer:

Explanation:

Find attached the solution

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The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s2. Neptune has mass 1.02×1026 kg and radiu
larisa [96]

Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N

(b) The apparent weight of the object at Neptune's equator is 50.4N

Explanation: Please see the attachments below

5 0
2 years ago
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec
bixtya [17]

Answer:

1340.2MW

Explanation:

Hi!

To solve this problem follow the steps below!

1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid

W=αhQ

α=specific weight for water =9.81KN/m^3

h=height=220m

Q=flow=690m^3/s

W=(690)(220)(9.81)=1489158Kw=1489.16MW

2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator

Wr=(0.9)(1489.16MW)=1340.2MW

the maximum possible electric power output is 1340.2MW

3 0
2 years ago
A diver in the pike position (legs straight, hands on ankles) usually makes only one or one-and-a-half rotations. To make two or
Korolek [52]

Answer:

from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

Explanation:

This can be explained on the basis of conservation of angular momentum.

This means the initial and the final angular velocity is conserved. Consider initial position (1)in the pike and final position in the be truck position. So there inertia's will also be different.

⇒I_1\omega_1 = I_2\omega_2

\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}

also,

I_1= mr_1^2

I_2= mr_2^2

since, r_2^2

I_2^2

therefore,

\omega_1^2

So, from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

6 0
2 years ago
A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle. The pulley is initially not
stiks02 [169]

Answer:

2

Explanation:

2

5 0
2 years ago
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