I have to submit my homework soon and I am not familiar with these concepts. can anybody do then for me?

Physics

1 answer:

AlladinOne [14]2 years ago

6 0

#4). The concepts are: A). acceleration is always in the direction of the force, and B). friction always acts in the direction opposite to motion. (that's B)

#5). The concepts are: A). the NET force is the sum of all the individual forces acting (on this car, it's 1600N forward). and B). Force = (mass) x (acceleration). So Acceleration = (force) / (mass). For this car, that's (1600N forward) / (800 kg) .

#6). The concept is: As long as you don't exceed the "proportionality limit" of a spring, its extension is proportional to the load on it. That means that the change in extension is always proportional to the change in the load. So now, look at the table: As long as the load is 10N or less, the spring stretched 3cm longer for every 2N more of load. But if the load is somewhere between 10N and 12N, that relationship disappears. Something changes between 10N and 12N of load. The spring's "elastic limit" is somewhere in that slot.

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If an airplane is flying directly north at 300.0 km/h, and a crosswind is hitting the airplane at 50.0 km/h from the east, what

Rashid [163]

Answer:

magnitude = 304.14 km/h

direction: $9.46^o$ West of North

Explanation:

The final plane's vector velocity will be the result of the vector addition of one pointing North of length 300 km/h, another one pointing West of length 50 km/h.

To find the magnitude of the final velocity vector (speed) we need to apply the Pythagorean theorem in a right angle triangle with sides: 300 and 50, and find its hypotenuse:

$|v|=\sqrt{300^2+50^2}=\sqrt{92500} = 304.14$ km/h

The actual direction of the plane is calculated using trigonometry, in particular with the arctan function, since the tangent of the angle can be written as:

$tan(\theta)=\frac{50}{300} = \frac{1}{6} \\\theta = arctan(\frac{1}{6} ) = 9.46^o$