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Step2247 [10]
4 years ago
6

Rosa studies the position-time graph of two race cars. A graph titled Position versus Time shows time in hours on the x axis, nu

mbered 0 to 5, position in kilometers on the y axis, numbered 0 to 50. The graph has two straight lines, one from point (0, 20) up to (2.8, 50), the second line starts at the origin and goes to point (5, 20). Rosa concludes that truck X has a greater velocity than truck Y, and that truck Y had a head start. She also concluded that the graph shows both trucks moving at a constant velocity and that neither truck changes direction. Which statement best describes Rosa’s error?
A Both trucks change direction after they start moving.
B Truck X had a head start, not truck Y.
C Truck Y has a greater velocity than truck X.
D The trucks are not traveling at constant velocities.
Physics
2 answers:
Harlamova29_29 [7]4 years ago
5 0

Answer:

B. Truck X had a head start, not truck Y.

Explanation:

Let's see what data can we obtain from this information.

Truck X's position changed from point (0,20) to point (2.8,50). That means that it started on the 20th kilometer and in the next 2.8 hours reached 50 km. It's velocity, then, is v1 = (s2 - s1) / t

v1 = (50 - 20) / 2.8

v1 = 10.7 km/h

Since it started from 20th km that means it had head start and since its line on the graph is straight, that means that its velocity was constant and that it didn't change direction.

Truck Y's position changed from the origin (0,0) to the point (5,20). That means it crossed 20 km in 5 hours, so its velocity is v2 = 20 / 5

v2 = 4 km/h

Again, since its line on the graph is straight, its velocity is constant and it did not change direction.

Now, knowing this, we can see that Rosa's mistake was to claim that truck Y had head start.

fgiga [73]4 years ago
4 0

Answer:

Answer is C

Explanation:

Just took the test and got it right

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1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

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m=1.9\cdot 10^4 kg is the rocket's mass

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- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

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We can therefore rewrite eq.(1) as

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Answer:

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