All categories

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

How is force proportional to mass?

sukhopar [10]

Answer:

a = F / m

Explanation:

force same -> mass variable

more mass -> less force

8 0

3 years ago

The force applied when using a simple machine ??

beks73 [17]

Answer : I hope this helps !

The Effort Force is the force applied to a machine. Work input is the work done on a machine. The work input of a machine is equal to the effort force times the distance over which the effort force is exerted.

8 0

2 years ago

A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wave

Ann [662]

Answer:

$\lambda$ = 40 cm

Explanation:

given data

string length = 30 cm

solution

we take here equation of length that is

L = $n \times \frac{1}{4} \lambda$ ...............1

total length will be here

$L = \frac{\lambda}{2} + \frac{\lambda}{4}\\$

$L = \frac{3 \lambda }{4}$

so $\lambda$ will be

$\lambda = \frac{4L}{3}\\\lambda = \frac{4\times 30}{3}$

$\lambda$ = 40 cm

5 0

3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago

A certain mass of a gas occupies 2 litres at 27 °C and 100

Zielflug [23.3K]

Answer:

75k

Explanation:

You can see solution in the picture

3 0

2 years ago