Question

URGENT. please help me with these physics problems.

Answer 1

Answers:

4a) A race car going around a track 200 times in a NASCAR race

For this sentence, the cases that apply are:

• 2-dimension problem, assuming that the movement is done in a circular track in which we have to work with 2 coordinates.

• It implies acceleration (A race car needs to accelerate a lot in order to win the race) and periodic movement, because the car is going around a track an specific amount of time

4b) An apple dropping off a branch through the air

For this sentence, the cases that apply are:

• 1-dimension problem, assuming that the movement is done only in the Y-axis

• Acceleration (Gravity acceleration)

4c) A basketball flying through the air on its way to the basket

For this sentence, the cases that apply are:

• 2-dimension problem, because we are talking about a parabolic-projectile movement, which implies both axis X and Y.

• Acceleration, including gravity acceleration

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Now, for the next problem, it is important to note that Distance and Displacement are different concepts in physics. Displacement is a vector magnitude, while Distance a scalar magnitude.

5a) Distance traveled

To solve this problem, let's begin by the Distance, as shown in the figure attached with the Cartesian coordinate system, the blue lines represent the distance the student walked: 10 m N, 30m E, 60 S. In this case we only have to sum each distance:

10m+30m+60m= 100m

5b) Displacement

The green line represents the displacement, which is a line from the origin vector (0,0) to the end of the track vector (30,-60)

We have to do a vector sum to find the value of the resulting Displacement vector.

Remember that this 2-dimensions vector has X component and Y component, so, we have to sum Xs with Xs and Ys with Ys:

(0,0)+ (30,-60)=(30,-70) >>>>>This is the resulting Displacement vector.

But, we need to obtain the module of this vector to express the displacement in a scalar magnitude in meters. This can be done by using the Pythagorean Theorem:

`|(X,Y)|=\sqrt{X^(2)+ Y^(2) }`

In order to obtain the module of (30,-70), we have to apply the theorem above:

`|(30,-70)|=\sqrt{30^(2)+ (-70)^(2) }`

Obtaining the final result:

Displacement= 76,157 m