Science_requirments.docx
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PHYSICS Aemeryn A. Montoya 9-Makiling
PROJECTILES: 1. An object is projected the ground with an initial velocity of 34.6 m/s at 40 ° horizontal. Find; a. Horizontal and vertical components of its initial velocity. Vox= 34.6 m/s cos 40° = 26.5 m/s Voy=34.6 m/s sin 40 ° = 22.2 m/s b. Time to reach the maximum height. T= 22.2 m/s 9.8 m/s = 2.3 s c. Time of flight 2(2.3 s) = 4.6 s d. Range R= (26.5 m/s)(4.6 s) = 121.9m e. Velocity upon striking the ground - 34.6 m/s and 40° 2. Ma’am Josie throw the whiteboard marker with an initial velocity of 25.8 m/s at 35° horizontal. Find; a. Horizontal and vertical components of its initial velocity. Vox= 25.8 m/s cos 35°= 21.1 m/s Voy= 25.8 m/s sin 35°= 14.8 m/s b. Time to reach the maximum height. 14.8 m/s 9.8 m/s = 1.5 s c. Time of flight 2(1.5 s) = 3s d. Range
R= (21.1 m/s)(3s)= 63.3m e. Velocity upon striking the ground. - 25.8 m/s and 35° 3. Xhary is driving in the forest and she accidentally fell on the cliff. Her car has an initial velocity of 32.6 m/s at 23° horizontal. Find; a. Horizontal and vertical components of initial velocity. Vox= 32.6 m/s cos 23°= 30.0 m/s Voy= 32. 6 m/s sin 23°= 12.7 m/s b. Time to reach the maximum height. 12.7 m/s 9.8 m/s =1.3 s c. Time of flight 2(1.3 s)= 2.6 s d. Range R= (30.0)(2.6 s)= 78m e. Velocity upon striking the ground. - 32.6 m/s and 23° MOMENTUM: 1. What is the momentum of a 23.6 kg truck travelling 29m/s to the left? p= mv p= (23.6kg)(29m) = 684.4 kg m/s
2. Calculate the momentum of 550 g sniper bullet travelling at 700 m/s. P= mv 550m/s 1000 = 0.55 kg P= (0.55 kg)(700 m/s) = 385 kg m/s 3. What is the momentum of 70 kg person sprinting at 8 m/s?
P= mv P= (70 kg)(8 m/s) =560 kg m/s
IMPULSE: 1. A force of 68.9 N acts on a 45.0 s GIVEN: F= 68.9 N T= 45.0 s FORMULA: J= ft SUBSTITUTION: J= (68.9 N)(45.0s) ANSWER: J= 3,100.5 Ns 2. An object is sitting on a frictionless surface. An unknown constant force of 69.2 N pushes the mass for 2 seconds. GIVEN: F= 69.2 N T= 2 s FORMULA: J=Ft SUBSTITUTION: J=(69.2 N)(2 s) ANSWER: 138.4 Ns 3. A force of 30000 N is exerted for 4.00 s. GIVEN: F= 30000 N T= 4.00 s FORMULA: J=Ft SUBSTITUTION: J=(30000 N)(4.00 s) ANSWER: 120000 Ns WORK: 1. Luis and his mother is playing in the garden. Luis rides in the wagon while his mother is pulling the handle with the force of 65N and in a 50 ° at a distance of 25m. Describe how much work is done by Luis and his mother. GIVEN: d= 25M F= 65N FORMULA: W=Fd x cos ° SUBSTITUTION: W= (65N)(25M)cos50 ° ANSWER: 1044.5 J
2. A 20 kg object experiences a horizontal with the force of 5 N, moving it in a 15 ° at a distance of 30 m, horizontally. How much work is done? GIVEN: d=30 m F= 5N FORMULA: W=Fd x cos ° SUBSTITUTION: W= (5N)(30m)cos15 ° ANSWER: 144.9 J 3. A crate is moved across a frictionless floor by a rope THAT is inclined 30 degrees above horizontal. The tension in the rope is 50 N. How much work is done in moving the crate 10 meters? GIVEN: d=10 m F= 50 N FORMULA: W=Fd x cos ° SUBSTITUTION: W= (50N)(10 m)cos 30 ° ANSWER:433.0 J
POWER: 1. An electric motor lifts an elevator that weighs 120.6N a distance of 800 m in 200s . What is the power of the motor in watts? GIVEN: F= 120.6 N D= 800 m T= 200 s FORMULA: W=Fd P=W/t SUBSTITUTION: W= (120.6 N)(800 m) =96480J P= 96480 J/ 200 s ANSWER: 482.4 Watts 2. A chair that weighs 575 N is lifted a distance of 200 m straight up by a rope. The job is done in 100s. What power is developed in watts? GIVEN: F= 575 N D= 200 m T= 100 S
FORMULA: W=Fd P=W/t SUBSTITUTION: W= (575 N)(200 m) =115000 J P= 115000J/100S ANSWER: 1150 Watts 3. The tired dolphin does 0.50 J of work in 2.0 seconds. The power rating of this dolphin is found by: GIVEN: W= 0.50J T= 2.0 s FORMULA: P=W/t SUBSTITUTION: P= O.50 J/2.0 s ANSWER: 0.25 Watts
ENERGY: 1. An object has a kinetic energy of 25 J and a mass of 34 kg. How fast is the object? GIVEN: KE= 25 J M= 34 kg FORMULA: 𝑉 2 = √2KE/m SUBSTITUTION: 𝑉 2 = 2(25 J)/34 Kg 𝑉 2 = √1.5 ANSWER: 1.2 J 2. An object moving with a speed 0f 35 m/s and has a kinetic energy of 1,500 J. What is the mass of the object? GIVEN: 𝑉 2 = 35 m/s KE=1500 J FORMULA: m= 2KE/𝑉 2 SUBSTITUTION: m= 2(1500 J)/352 m/s ANSWER: 2.4 Kg 3. An object has a kinetic energy of 25J and a mass of 34 kg. How fast is the object?
GIVEN: KE= 25 J M= 34 Kg FORMULA: 𝑉 2 = √2KE/m SUBSTITUTION: 𝑉 2 = 25 J/ 34Kg =√0.7 ANSWER: 0.8 J
PROJECTILES: 1. An object is projected the ground with an initial velocity of 34.6 m/s at 40 ° horizontal. Find; a. Horizontal and vertical components of its initial velocity. Vox= 34.6 m/s cos 40° = 26.5 m/s Voy=34.6 m/s sin 40 ° = 22.2 m/s b. Time to reach the maximum height. T= 22.2 m/s 9.8 m/s = 2.3 s c. Time of flight 2(2.3 s) = 4.6 s d. Range R= (26.5 m/s)(4.6 s) = 121.9m e. Velocity upon striking the ground - 34.6 m/s and 40° 2. Ma’am Josie throw the whiteboard marker with an initial velocity of 25.8 m/s at 35° horizontal. Find; a. Horizontal and vertical components of its initial velocity. Vox= 25.8 m/s cos 35°= 21.1 m/s Voy= 25.8 m/s sin 35°= 14.8 m/s b. Time to reach the maximum height. 14.8 m/s 9.8 m/s = 1.5 s c. Time of flight 2(1.5 s) = 3s d. Range
R= (21.1 m/s)(3s)= 63.3m e. Velocity upon striking the ground. - 25.8 m/s and 35° 3. Xhary is driving in the forest and she accidentally fell on the cliff. Her car has an initial velocity of 32.6 m/s at 23° horizontal. Find; a. Horizontal and vertical components of initial velocity. Vox= 32.6 m/s cos 23°= 30.0 m/s Voy= 32. 6 m/s sin 23°= 12.7 m/s b. Time to reach the maximum height. 12.7 m/s 9.8 m/s =1.3 s c. Time of flight 2(1.3 s)= 2.6 s d. Range R= (30.0)(2.6 s)= 78m e. Velocity upon striking the ground. - 32.6 m/s and 23° MOMENTUM: 1. What is the momentum of a 23.6 kg truck travelling 29m/s to the left? p= mv p= (23.6kg)(29m) = 684.4 kg m/s
2. Calculate the momentum of 550 g sniper bullet travelling at 700 m/s. P= mv 550m/s 1000 = 0.55 kg P= (0.55 kg)(700 m/s) = 385 kg m/s 3. What is the momentum of 70 kg person sprinting at 8 m/s?
P= mv P= (70 kg)(8 m/s) =560 kg m/s
IMPULSE: 1. A force of 68.9 N acts on a 45.0 s GIVEN: F= 68.9 N T= 45.0 s FORMULA: J= ft SUBSTITUTION: J= (68.9 N)(45.0s) ANSWER: J= 3,100.5 Ns 2. An object is sitting on a frictionless surface. An unknown constant force of 69.2 N pushes the mass for 2 seconds. GIVEN: F= 69.2 N T= 2 s FORMULA: J=Ft SUBSTITUTION: J=(69.2 N)(2 s) ANSWER: 138.4 Ns 3. A force of 30000 N is exerted for 4.00 s. GIVEN: F= 30000 N T= 4.00 s FORMULA: J=Ft SUBSTITUTION: J=(30000 N)(4.00 s) ANSWER: 120000 Ns WORK: 1. Luis and his mother is playing in the garden. Luis rides in the wagon while his mother is pulling the handle with the force of 65N and in a 50 ° at a distance of 25m. Describe how much work is done by Luis and his mother. GIVEN: d= 25M F= 65N FORMULA: W=Fd x cos ° SUBSTITUTION: W= (65N)(25M)cos50 ° ANSWER: 1044.5 J
2. A 20 kg object experiences a horizontal with the force of 5 N, moving it in a 15 ° at a distance of 30 m, horizontally. How much work is done? GIVEN: d=30 m F= 5N FORMULA: W=Fd x cos ° SUBSTITUTION: W= (5N)(30m)cos15 ° ANSWER: 144.9 J 3. A crate is moved across a frictionless floor by a rope THAT is inclined 30 degrees above horizontal. The tension in the rope is 50 N. How much work is done in moving the crate 10 meters? GIVEN: d=10 m F= 50 N FORMULA: W=Fd x cos ° SUBSTITUTION: W= (50N)(10 m)cos 30 ° ANSWER:433.0 J
POWER: 1. An electric motor lifts an elevator that weighs 120.6N a distance of 800 m in 200s . What is the power of the motor in watts? GIVEN: F= 120.6 N D= 800 m T= 200 s FORMULA: W=Fd P=W/t SUBSTITUTION: W= (120.6 N)(800 m) =96480J P= 96480 J/ 200 s ANSWER: 482.4 Watts 2. A chair that weighs 575 N is lifted a distance of 200 m straight up by a rope. The job is done in 100s. What power is developed in watts? GIVEN: F= 575 N D= 200 m T= 100 S
FORMULA: W=Fd P=W/t SUBSTITUTION: W= (575 N)(200 m) =115000 J P= 115000J/100S ANSWER: 1150 Watts 3. The tired dolphin does 0.50 J of work in 2.0 seconds. The power rating of this dolphin is found by: GIVEN: W= 0.50J T= 2.0 s FORMULA: P=W/t SUBSTITUTION: P= O.50 J/2.0 s ANSWER: 0.25 Watts
ENERGY: 1. An object has a kinetic energy of 25 J and a mass of 34 kg. How fast is the object? GIVEN: KE= 25 J M= 34 kg FORMULA: 𝑉 2 = √2KE/m SUBSTITUTION: 𝑉 2 = 2(25 J)/34 Kg 𝑉 2 = √1.5 ANSWER: 1.2 J 2. An object moving with a speed 0f 35 m/s and has a kinetic energy of 1,500 J. What is the mass of the object? GIVEN: 𝑉 2 = 35 m/s KE=1500 J FORMULA: m= 2KE/𝑉 2 SUBSTITUTION: m= 2(1500 J)/352 m/s ANSWER: 2.4 Kg 3. An object has a kinetic energy of 25J and a mass of 34 kg. How fast is the object?
GIVEN: KE= 25 J M= 34 Kg FORMULA: 𝑉 2 = √2KE/m SUBSTITUTION: 𝑉 2 = 25 J/ 34Kg =√0.7 ANSWER: 0.8 J
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