Test Investigation Review

A marble is placed at the top of a frictionless track. The top of the track has a height of 1.20 meters. The mass of the marble is 0.300 grams. What is the maximum velocity of the marble, assuming the bottom of the track is 0.300 meters above the ground?

• 4.2 m/s

• 3.8 m/s

• 5.0 m/s

• 2.5 m/s

Answer explanation

Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (0.5mv^2) at the bottom. The height difference is 1.20m - 0.30m = 0.90m. Solving gives a maximum velocity of 4.2 m/s.

Tags

NGSS.HS-PS2-1

Using the given equations, calculate the maximum velocity of an object when it falls from a height of 1.20 m to 0.300 m. What is the maximum velocity?

• 3.00 m/s

• 4.20 m/s

• 5.00 m/s

• 6.00 m/s

Answer explanation

To find the maximum velocity, use the equation v = sqrt(2gh), where g = 9.81 m/s² and h = 1.20 m - 0.300 m = 0.90 m. This gives v = sqrt(2 * 9.81 * 0.90) = 4.20 m/s, which is the correct answer.

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-4

Choose the expression that is a valid representation of the elastic potential energy of a spring. Assume F is the force on a spring according to Hooke’s law.

• PE_S = 2F/x

• PE_S = k/2x

• PE_S = Fx/2

Answer explanation

The elastic potential energy of a spring is given by PE_S = Fx/2, where F is the force and x is the displacement. This formula derives from integrating Hooke's law, confirming that this choice correctly represents the energy stored in a spring.

• 1.15 x 10^{-28} J

• 5.9 x 10^-28 J

• 3.45 x 10^{-28} J

• 4.60 x 10^{-28} J

Answer explanation

The change in electrostatic potential energy is calculated using the formula U = k * (q1 * q2) / r, where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them. The correct answer is 5.9 x 10^-28 J.

Tags

NGSS.HS-PS2-4

A worker is tasked with moving to a new office. They pack up all of their belongings in boxes and move to the new room. One of the boxes is particularly heavy. The worker exerts a force in the direction of motion equal to 29.0 N. They manage to push it 12.0 meters before asking for assistance from another worker. How much work has been done so far?

• 348 J

• 3480 J

• 34.8 J

• 3.48 J

Answer explanation

Work is calculated using the formula: Work = Force x Distance. Here, Work = 29.0 N x 12.0 m = 348 J. Therefore, the correct answer is 348 J.

Tags

NGSS.MS-PS2-1

NGSS.MS-PS2-2

If the work done by a gas in a container is equal to 5 J and the change in volume is +15 cm³, is energy flowing into the system, out of the system, or neither?

• Energy is neither flowing in nor out of the system, resulting in an increase in pressure.

• Energy is neither flowing in nor out of the system, resulting in a decrease in pressure.

• Energy is flowing into the system, resulting in an increase in pressure.

• Energy is flowing out of the system, resulting in a decrease of pressure.

Answer explanation

The work done by the gas is positive, indicating it is expanding. Since the energy is not being added to the system, it is flowing out, leading to a decrease in pressure.

What is the highest point on the roller coaster?

Answer explanation

The highest point on a roller coaster is typically the first drop or peak, where the coaster gains maximum height before descending. This point is crucial for the thrill and speed of the ride.

Tags

NGSS.HS-PS3-2

The motion of the roller coaster from point A to point D can be described as:

• A gradual ascent followed by a steep drop

• A series of loops and twists

• A straight path with no elevation change

• A gentle slope with a sudden stop

Answer explanation

The correct choice, 'A gradual ascent followed by a steep drop', accurately describes the typical motion of a roller coaster, where it climbs to a height before experiencing a rapid descent.

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-2

The students are riding the steel roller coaster, as shown. Assume the total mass of the roller coaster carts and the riders is 4200.0 kg. If the tallest part of the roller coaster track is at 91.44 meters and the roller coaster reaches a maximum velocity of 27.0 m/s at the bottom of the hill, what is the height of the lowest part of the track? Assume the track is frictionless.

• 0 meters

• 10 meters

• 20 meters

• 30 meters

Answer explanation

Using conservation of energy, potential energy at the top (mgh) equals kinetic energy at the bottom (0.5mv^2). Setting h=91.44m and solving gives h=0m at the lowest point, confirming the correct answer is 0 meters.

Assume the total mass of the roller coaster carts and the riders is 4200.0 kg. If the tallest part of the roller coaster track is at 91.44 meters and the roller coaster reaches a maximum velocity of 27.0 m/s at the bottom of the hill, what is the height of the lowest part of the track? Assume the track is frictionless.

• 90.1 m

• 54.3 m

• 17.1 m

• 28.5 m

Answer explanation

Using conservation of energy, potential energy at the top equals kinetic energy at the bottom. PE = mgh and KE = 0.5mv^2. Setting them equal gives h = (v^2)/(2g) + h_lowest. Solving yields h_lowest = 54.3 m.

Tags

NGSS.HS-PS3-1