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olganol
3 months ago
11

A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.

The driver takes the turtle to a town 1.1 km to the north with an average speed of 12 m/s. What is the magnitude of the average velocity of the turtle for its entire journey?
Physics
1 answer:
kicyunya [3.2K]3 months ago
8 0
The average velocity is 3.59 m/s. To determine this, we first calculate the total displacement. The turtle initially walks 18m south and is then carried 1.1 km (or 1100m) north, resulting in a net displacement of 1082m north (1100m - 18m). The time taken includes the duration of walking and the time spent in the truck. The turtle's walking time amounts to 3.5 minutes, equating to 210 seconds. For the truck's travel time, using the formula Time = Distance / Speed, we find the truck time for the 1100m distance at a speed of 12m/s to be approximately 91.67s. Therefore, the total time taken sums to 301.67s. The average velocity, calculated as total displacement divided by total time, thus equals 3.59 m/s to the north.
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Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on
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Answer:

2.87 m

Explanation:

Given parameters:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Force applied on the ball by compressed air (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assumed)

Final velocity of the ball at the tube's exit (v) =?

Acceleration of the ball (a) =?

The ball's weight is derived from multiplying mass and gravity. Therefore,

Weight (W) = mg=0.06\times 9.8=0.588\ N

Thus, the total force acting on the ball equals the net of upward force minus the weight.

Net force = Air force - Weight

F_{net}=F-mg\\F_{net}=3.0-0.588 = 2.412\ N

According to Newton's second law, net force equals the mass multiplied by acceleration.

F_{net}=ma\\\\a=\frac{F_{net}}{m}=\frac{2.412\ N}{0.06\ kg}=40.2\ m/s^2

Acceleration (a) is calculated as 40.2 m/s².

Using the motion equation, we find:

v^2=u^2+2ah\\\\v^2=0+2\times 40.2\times 0.7\\\\v=\sqrt{56.28}=7.5\ m/s

Let’s denote the maximum height achieved as 'H'.

Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.

A decrease in kinetic energy equals an increase in potential energy.

\frac{1}{2}mv^2=mgH\\\\H=\frac{v^2}{2g}

Substituting the values, we solve for 'H', yielding:

H=\frac{56.28}{2\times 9.8}\\\\H=\frac{56.28}{19.6}=2.87\ m

Hence, the ball ascends to a height of 2.87 m above the top of the tube.

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