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3 years ago

To move the Center of Gravity (CG) forward, you could do which of the following?

Physics

2 answers:

Sergio039 [100]3 years ago

7 0

Answer:

I think it's no. 3

Decrease the volume of the horizontal and vertical stabilizer material.

Explanation:

Lemme know if it's correct

nadezda [96]3 years ago

5 0

Answer:

all of the above

Explanation:

You might be interested in

What is the effective resistance of a car’s starter motor when 150 A flows through it as the car battery applies 12.0 V to the m

QveST [7]

Answer:

From ohms law,

V=IR

R=V/I =12.0/150 =0.08 ohm.

8 0

3 years ago

A bicycle takes 8.0 seconds to accelerate at a constant rate from rest to a speed of 4.0 m/s. If the mass of the bicycle and rid

Inessa05 [86]

Acceleration = (change in speed) / (time for the change)
Acceleration = (4 m/s) / (8 seconds)
Acceleration = 0.5 m/s²

Force = (mass) x (acceleration)
Force = (85 kg) x (0.5 m/s²)
Force = 42.5 Newtons

4 0

3 years ago

Read 2 more answers

Carter found the rock shown in the picture below near a river

noname [10]

Answer:

there's no picture

Explanation:

and what would the question be anyways?

4 0

3 years ago

Read 2 more answers

You leave the doctor's office after your annual checkup and recall that you weighed 689 N in her office. You then get into an el

son4ous [18]

Answer: 0.455 m/s^2

Explanation:

f = ma = m x 9.8=689 (the mass is constant)

m = 78.3

now on the elevator

f = 721 = m x a

a= 10.26

the elevator is moving down because there is a increase in weight. the acceleration of elevator = 10.26-9.8 = 0.455

8 0

3 years ago

A person's center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass mm at the top of a

muminat

Answer:

$\mathbf{v_{max} = \sqrt{gL}}$

Explanation:

Considering an object that moving about in a circular path, the equation for such centripetal force can be computed as:

$\mathbf {F = \dfrac{mv^2}{2}}$

The model for the person can be seen in the diagram attached below.

So, along the horizontal axis, the net force that is exerted on the person is:

$mg cos \theta = \dfrac{mv^2}{L}$

Dividing both sides by "m"; we have :

$g cos \theta = \dfrac{v^2}{L}$

Making "v" the subject of the formula: we have:

$v^2 = g Lcos \theta$

$v=\sqrt{ gL cos \theta$

So, when $\theta$ = 0; the velocity is maximum

∴

$v_{max} = \sqrt{gL \ cos \theta}$

$v_{max} = \sqrt{gL \ cos (0)}$

$v_{max} = \sqrt{gL \times 1}$

$\mathbf{v_{max} = \sqrt{gL}}$

Hence; the maximum walking speed for the person is $\mathbf{v_{max} = \sqrt{gL}}$

3 0

3 years ago