Question

an experimenter records data for the speed and time of rolling carts traveling along a track. the table shows the data for four trials. use this table for questions 2 - 3.

1. in which trial did the cart experience the greatest average acceleration?

a. trial 2 c. trial 4
b. trial 3 d. trial 1

1. in which trial did the cart experience centripetal acceleration?

a. trial 1 c. trial 3 e. none of these
b. trial 2 d. trial 4 f. all of these

1. montez read that bobcats can reach speeds of up to 30 mi/h. what does the speed of a bobcat indicate?

e. the time it takes to reach its maximum speed
f. the total amount of time it travels in one direction
g. the distance it travels to move between two points
h. the distance it can travel in a certain amount of time

1. mr. krabs wants to determine how fast a krabby patty travels when it is catapulted over the restaurant. he uses a stopwatch to record, in seconds, the time it takes for the patty to land on the other side. which unit would be most appropriate to measure the length of the distance traveled by said patty?

a. inches c. kilometers e. feet
b. millimeters d. centimeters f. meters

1. a moving object collides with a stationary object. which of the following statements is true according to newtons law of motion?

d. each object exerts a force on the other, and the two forces are the same in magnitude and direction.
e. each object exerts a force on the other, and the two forces are equal and in opposite directions.
f. each object exerts a force on the other, and the two forces are proportional to the masses of the ob-

1. which is an example of centripetal acceleration?

e. an object at rest
f. an object moving at 20 m/s backward, then 20 m/s to the right
g. an object moving in a circle at 20m/s
h. an object moving in a straight line at a steady speed of 20m/s forward

1. a glowing pumpkin rolls off the table that is 1 meter high. assuming normal gravity of 9.8 m/s², how long will it take before it crashes to the ground? show your work using the formula.

t = \sqrt{\frac{2h}{g}}

Explanation:

Step1: Substitute the values into the formula

Given \(h = 1\) meter and \(g=9.8\ m/s^{2}\), substitute into \(t = \sqrt{\frac{2h}{g}}\). So we have \(t=\sqrt{\frac{2\times1}{9.8}}\).

Step2: Calculate the value

First, calculate \(\frac{2\times1}{9.8}=\frac{2}{9.8}\approx0.204\). Then \(t=\sqrt{0.204}\approx 0.45\) s.

Answer:

\(t\approx0.45\) s