A 5kg bucket hangs from a ceiling on a rope. A student attaches a spring scale to the buckets handle and pulls horizontally on t

Answer: small barrel gun

Explanation:

It is noted that short barrel guns have a higher muzzle velocity for bullets compared to longer barrel guns.

Acceleration is determined by the change in velocity with respect to time.

For short barrel guns, the bullet reaches its muzzle velocity more quickly, leading to greater acceleration than that of bullets from long barrel guns.

Answer:

b. The loop's current consistently flows in a counterclockwise direction.

Explanation:

As a magnet descends through a wire loop, it generates an induced current within that loop. This induced current arises due to the magnet's movement, leading to a variation in magnetic flux. Lenz's law states that the induced current will act to counteract the change that produces it. In this scenario, the only feasible resistance to the magnet’s fall is through inducing a similar pole on the loop to counteract its downward motion. An induced current that circulates counterclockwise in the wire loop mimics the polarity of a northern pole, thereby repelling the magnet's descent. Furthermore, as the magnet passes the wire loop, this induced north pole will seek to attract the magnet's south end in an effort to halt its downward progression.

Case A:

A.75 kg 65 N/m 1.2 m

m = weight of the car = 0.75 kg

k = spring's stiffness = 65 N/m

h = elevation of the hill = 1.2 m

x = spring's compression = 0.25 m

Applying the principle of energy conservation from the Top of the hill to the Bottom of the hill

Energy at the Top of the hill equals Energy at the Bottom of the hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s

Case B:

B.60 kg 35 N/m.9 m

m = weight of the car = 0.60 kg

k = spring's stiffness = 35 N/m

h = height of the hill = 0.9 m

x = spring's compression = 0.25 m

Using energy conservation from the Top of the hill to the Bottom of the hill

Top hill energy = Bottom hill energy

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s

Case C:

C.55 kg 40 N/m 1.1 m

m = weight of the car = 0.55 kg

k = spring's stiffness = 40 N/m

h = height of the hill = 1.1 m

x = spring's compression = 0.25 m

Using conservation of energy from the Top of the hill to the Bottom of the hill

Energy at the Top of the hill = Energy at the Bottom of the hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s

Case D:

D.84 kg 32 N/m.95 m

m = weight of the car = 0.84 kg

k = spring's stiffness = 32 N/m

h = height of the hill = 0.95 m

x = spring's compression = 0.25 m

Using energy conservation from the Top of the hill to the Bottom of the hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s

thus, the closest result is from case C at 5.1 m/s

Answer:

-utilize precisely the same apparatus

-maintain identical measures (release height)

Explanation: