6. The rate at which velocity changes is called

Help i am having a mental breakdown help! need this done...

guajiro [1.7K]

1. he traveled a total of 24 miles
2. peter is not moving between 10 and 30 minutes

4 0

3 years ago

About 3% of the water on Earth is freshwater. Only about 40% of that freshwater is available for human use. Why is so much fresh

Pie

A) it is frozen i think

7 0

3 years ago

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In this model, the velocity of the spacecraft at position 2 is the velocity of the craft at position 4. at position 1, the direc

Free_Kalibri [48]

1. The velocity of the spacecraft at position 2 is greater than the velocity of the craft at position 4.

This is due the gravity field of the Earth is used to accelerate the craft. This is true when in a specific point the direction of the movement of the craft is the same direction of the movement of the planet.

In this case the craft will be “catched” by the Earth’s gravitational field, making the craft to enter a circular orbit.

2. At point 1, the direction of the spacecraft changes because of the gravitational force between earth and the spacecraft.

As explained in the first answer, this is the exact point where the trajectory of the spacecraft enters into a circular orbit because of the attraction due gravity of the Earth and therefore changes its direction.

3. Position 3 represents the orbital path of Earth

Being this the orbital path of the Earth and considering the trajectory of the craft, the condition of accelerating the craft is accomplished. If the orbital path of the Earth were the opposite, the effect on the craft would be braking.

Note all of these is related to the gravitational assistance, this consists in a maneuver in which the energy of the gravitational field of a planet or satellite is used to obtain an acceleration or braking of the probe or craft, changing its trajectory.

To learn more about velocity of the spacecraft : brainly.com/question/11900446

#SPJ4

7 0

1 year ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago

How many like does it take to get to the center of a tosipop

dmitriy555 [2]

252-364 licks

I'm not sure if it's correct or not

4 0

3 years ago

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