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3 years ago

Describe one long-term goal that you have and at least two things you will need to do in order to accomplish that goal.

1 answer:

4 0

Graduate from college ! I’m a freshman at SDSU :) two things would be to stay dedicated to my studies and not give into distractions!!!

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A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed

oksian1 [2.3K]

Explanation:

a. The velocity of the wind as a vector in component form will be represented as v vector:

$v=30j$

b.The velocity of the jet relative to the air as a vector in component form will be represented as u vector

$u=475i$

c. The true velocity of the jet as a vector will be represented as w:

$w=u+v$

$w=475i+30j$

d. The true speed of the jet will be calculated as:

$IwI=\sqrt{(475)^2+(30)^2}$

$IwI=\sqrt{225625+900}$

$IwI=\sqrt{226525}$

$IwI=476 mi/h$

e. The direction of the jet will be:

$tita=tan^{-1}\frac{30}{475}$

$tita=tan^{-1}(0.0632)$

$tita=3.62degrees,or,N86.38degreesS$

7 0

3 years ago

What swine breed is white in color with pointed ears

Annette [7]

Answer:

landrace

Explanation:

5 0

3 years ago

Read 2 more answers

Do the data for the first part of the experiment support or refute the first hypothesis? Be sure to explain your answer and incl

Sphinxa [80]

The data for the first part of the experiment support the first hypothesis. As the force applied to the cart increased, the acceleration of the cart increased. Since the increase in the applied force caused the increase in the cart's acceleration, force and acceleration are directly proportional to each other, which is in accordance with Newton's second law.

5 0

3 years ago

Read 2 more answers

A box of negligible mass rests at the left end of a 2.00-m, 25.0-kg plank (Fig. P11.43). The width of the box is 75.0 cm, and sa

Sati [7]

Answer:

Required mass of sand is 20 kg

Explanation:

Given:

Mass of the plank = 25 kg

Distance of the Center of gravity of the Plank from the fulcrum = $\frac{2}{2}-0.50 = 0.5m$

Distance of the Center of gravity of the sand box from the fulcrum = $\frac{2}{2}-\frac{0.75}{2}= 0.625m$

Balancing the torque due to the plank and the sand box with respect to the fulcrum

Torque = Force × perpendicular distance

thus, we get

(25 × g) × 0.5 = weight of sand × 0.625

where, g is the acceleration due to gravity

(25 × g) × 0.5 = (mass of sand × g) × 0.625

mass of sand = 20 kg

<u>Hence, the required mass of the sand is </u><u>20 kg</u>

8 0

3 years ago

A uniform disk of mass M and radius R has rotational inertia about its central axis of Idisk = 1/2 MR2. A uniform rod of mass m

expeople1 [14]

Answer:

5/6 MR²

Explanation:

In the composite object, the distance from the disk's axis to the rod's axis is R/2. Using parallel axis theorem, the rod's new moment of inertia is:

I = 1/12 MR² + M (R/2)²

I = 1/12 MR² + 1/4 MR²

I = 1/3 MR²

The total moment of inertia is therefore:

I = 1/2 MR² + 1/3 MR²

I = 5/6 MR²

4 0

4 years ago