What would be the initial velocity of the projectile in above problem A) 257m/s

Question

What would be the initial velocity of the projectile in above problem A) 257m/s B) 4.41 m/s C) 10.3 m/s D) 14 m/s A 75 gram ball is fired into a 200 gram pendulum and the center of gravity of the pendulum ball to a height of 2.5 cm. The length from the pivot point to the center of gravity of the pendulum-ball is 30cm. Determine the initial of the ball. A) 257m/s B) 4.41 m/s C) 10.3 m/s D) 14 m/s A projectile is fired horizontally off a table at a height of 95 cm. It lands 2.10 meters away. What is the initial velocity of the projectile? A) 4.77 m/s B) 4.41 m/s C) 10.3 m/s D) 14 m/s A 550 gram mass is suspended from a spring. The mass is stretched 1 cm and released. A period of 0.50s in measured. What is the spring constant for the spring? A) 13.8N/m B) 43.4 N/m C)27.6 N/m D)869 N/m In the Hooke's Law experiment the slope of the plotted data for a graph of the Force to stretch the spring vs. the elongation of the spring represents. A) the initial tension B) the spring constant C) the work to stretch the spring D) the equilibrant A 200 gram mass is placed upon a meter stick at the 40 cm mark with the pivot point of the meter stick. How much torque will it produce? A) 17640000 dyne cm B) 45645641 dyne cm C) 12345654 dyne cm From the above problem, a 250 gram mass has to be placed on the meter stick to balance it. Where would the mass be placed? A) 32 cm B) 20 cm C) 15 cm D) 30 cm On a hot summer day, you want to cool your warm soda, and you place ice cubes in your cup. What kind of energy is being exchanged between the soda and ice? A) Kinetic B) potential C) Heat D) Mechanical In the simple pendulum experiment, you determined the period is directly related to the _____ of the pendulum bob. A) Length B) Gravity C) Mass D) Buoyancy In the simple pendulum experiment, you determined the period is not related to the _____ of the pendulum bob. A) Length B) Gravity C) Mass D) Buoyancy When does swimming occur most? A) Summer B) Winter It was pleasure being your instructor, the answer is A

Explanation / Answer

10)

2.5cm = 0.025m.

200g = 0.2kg.

75g = 0.075kg.

Final pendulum/ ball mass = (0.2 + 0.075) = 0.275kg.

GPE of pendulum when at height = (mgh) = (0.275 x 9.8 x 0.025) = 0.067375 Joules.

This is the result of its initial KE, which is the same.

Initial V of pendulum = sqrt(2KE/m) = sqrt(2 x 0.067375)/0.275 = 0.7m/s

(0.7 x 0.275) / 0.075 = 2.57 m/s

12)

T = 2pi*sqrt(m/k)

T/2pi = sqrt(m/k)

T^2/4pi^2 = m/k

k = (4pi^2m) / T^2

k = (4pi^2*0.55)/(0.5)^2

k = 86.85 N/m

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