1. While traveling along a highway a driver slows from 24 m/s to 15 m/s in 12 seconds. What is the automobile’s acceleration? (Remember that a negative value indicates a slowing down or deceleration)
Known: v1 = 24 m/s
v2 = 15 m/s
t = 12 seconds
Asked : acceleration (a)?
Answer: a = (v2 - v1) / t
a = (15 -24) / 12
a = - 9 / 12
a = - 3/4 or -0.75 m/s2
So, The acceleration is -0.75 m/s2 (sign (-); indicate that the automobile is slowing down).
2. A parachute on a racing dragster opens and changes the speed of the car from 85 m/s to 45 m/s in a period of 4.5 seconds. What is the acceleration of the dragster?
Known: v1 = 85 m/s
v2 = 45 m/s
t = 4.5 seconds
Asked: acceleration (a)?
Answer: a = (v2 - v1) / t
a = (45 - 85) / 4.5
a = - 40 / 4.5
a = -8.89 m/s2
So, The acceleration is -8.89 m/s2 (sign (-); indicate that the automobile is slowing down).
3. A car traveling at a speed of 30.0 m/s encounters an emergency and comes to a complete stop. How much time will it take for the car to stop if it decelerates at -4.0 m/s2?
Known: v1 = 30 m/s
v2 = 0 m/s (comes to a complete stop)
a = -4 m/s2
Asked: time (t)?
Answer: a = (v2 - v1) / t
-4 = (0 - 30) / t
-4 = - 30 / t
t = -30/-4
t = 7.5 seconds
So, the time is 7.5 s.
4. If a car can go from 0 to 60 km/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50 km/hr?
Known:
first case; v1 = 0 km/hr; v2 = 60 km/hr
t = 8 s;
Second case; v1 = 50 km/hr;
t = 5 s;
Asked: v2?
Answer: First of all, you have to find the acceleration at first case;
before it, be carefull with the units. You have to make it in SI units.
first case; v1 = 0 km/hr
= 0 m/s;
v2 = 60 km/hr
= (60 x 1000)/3600 m/s
= 16.67 m/s
Second case; v1 = 50 km/hr
= (50 x 1000)/3600 m/s
= 13.89;
Find the acceleration at first case;
a = (v2 - v1) / t
= (16.67 - 0) / 8
= - 16.67 / 8
= 2.08375 m/s2
Then, calculate v2(final velocity) on second case;
a = (v2 - v1) / t
2.08375 = (v2 - 13.89) / 5
2.08375 x 5= (v2 - 13.89)
v2 = 24.30875 m/s
So, the final velocity v2 = 24.30875 m/s on second case.
5. A cart rolling down an incline for 5.0 seconds has an acceleration of 4.0 m/s2. If the cart has a beginning speed of 2.0 m/s, what is its final speed?
Known: t = 5 seconds
a = 4.0 m/s2
v1 = 2 m/s
Asked: final velocity (v2)?
Answer: v2 = v1 + a . t
= 2 + 4 . 5
= 2 + 20
= 22 m/s
So, The final velocity is 22 m/s.
6. A car traveling at 72 km/h in 2 s is accelerating 5 m/s2. Determine the speed and the distance!
Known: v1 = 72 km/h =20 m/s
t = 2 s
a = 5.0 m/s2
Asked: a). final velocity (v2)?
b). distance (d)?
Answer: a). v2 = v1 + a . t
= 20 + 5 . 2
= 20 + 10
= 30 m/s
b). d = v1 . t + 1/2 a . t2
= 20 . 2 + 1/2 . 5 . (2)2
= 40 + 10
= 50 m
So, The final velocity is 30 m/s and the distance is 50 m.
7. A car moving at a speed of 72 km/h do braking so the speed regularly reduced to 18 km/h in 5 seconds. What is the distance at t = 5 s?
Known: v1 = 72 km/h = 20 m/s
v2 = 18 km/h = 5 m/s
t = 5 s
Asked: distance (d)?
Answer: We have to calculate the acceleration first before calculate the distance.
v2 = v1 + a . t
5 = 20 + a . 5
5a= 5 - 20
5a= -15
a = -3 m/s2
Then,
d = v1 . t + 1/2 . a . t2
= 20 . 5 + 1/2 . -3 . (5)2
= 100 - 37.5
= 62.5 m
So, The distance is 62.5 m.
8. How long does it take for a car to change its velocity from 10 m/s to 25 m/s if the acceleration is 5 m/s2?
Known: v1 = 10 m/s
v2 = 25 m/s (comes to a complete stop)
a = 5 m/s2
Asked: time (t)?
Answer: a = (v2 - v1) / t
5 = (25 - 10) / t
5 = 15 / t
t = 15/5
t = 3 seconds
So, the time is 3 s.
9. How long will it take for an apple falling from a 29.4m-tall tree to hit the ground?
Known: d = 29.4 m
v1 = 0 m/s (in the beginning is at rest)
a = g = 10 m/s2 (gravitational acceleration)
Asked: time (t)?
Answer:
d = v1 . t + 1/2 . a . t2
29.4 = 0 . t + 1/2 . 10 . (t)2
29.4 = 0 + 5t2
t2 = 29.4 / 5
t = 5.881/2
t = 2.42 s
So, the time is 2.42 s.
10. If a car has a constant acceleration of 4 m/s2, starting from rest, how fast is it traveling after 5 seconds?
Known: t = 5 seconds
a = 4.0 m/s2
v1 = 0 m/s (starting from rest)
Asked: final velocity (v2)?
Answer: v2 = v1 + a . t
= 0 + 4 . 5
= 0 + 20
= 20 m/s
So, The final velocity is 20 m/s.
11. If a car has a constant acceleration of 4 m/s2, starting from rest, how far has it traveled by the time it reaches the speed of 40 m/s?
Known: a = 4.0 m/s2
v1 = 0 m/s (starting from rest)
v2 = 40 m/s
Asked: d?
Answer: We have to find the time in the beginning, later we can calculate the distance.
v2 = v1 + a . t
40 = 0 + 4 . t
40 = 4t
t = 10 s
we get t = 10 s, then
d = v1 . t + 1/2 . a . t2
= 0 . 10 + 1/2 . 4 . (10)2
= 200 m
So, The distance is 200 m.
12. A car is at velocity of 20 km/h. If the car traveled 120 km in 3 hours at constant acceleration, what is its final velocity?
Known: v1 = 20 km/h (starting from rest)
d = 120 km
t = 3 hours
Asked: v2?
Answer: We have to find the acceleration in the beginning, later we can calculate the final velocity.
d = v1 . t + 1/2 . a . t2
120 = 20 . 3 + 1/2 . a . (3)2
120 = 60 + 4.5a
4.5a = 120 - 60
4.5a = 60
a = 13.33 m/s2
Then,
v2 = v1 + 13.33 . 3
= 20 + 40
= 60 km/h
So, the final velocity is 60 km/h.
13. How long will it take for a falling object to reach 108 m/s if its initial velocity is 10 m/s?
Known: v1 = 10 m/s
v2 = 108 m/s
a = g = 10 m/s2 (gravitational acceleration)
Asked: time (t)?
Answer:
v2 = v1 + a . t
108 = 10 + 10 . t
10t = 108 - 10
10t = 98
t = 9.8 s
So, the time is 9.8 s.
14. What is the final velocity of an apple if it falls from a 100m-tree?
Known: v1 = 0 m/s (Hanging on the tree)
d = 100 m
a = g = 10 m/s2 (gravitational acceleration)
Asked: v2?
Answer:
v22 = v12 + 2. a . d
v22 = 02 + 2. 10 . d
v22 = 20 . 100
v2 = (2000)1/2
v2 = 20.(5)1/2
So, the final velocity is 20.(5)1/2 m/s
15. What is the displacement of a car whose initial velocity is 5 m/s and then accelerated 2 m/s2 for 10 seconds?
Known: v1 = 5 m/s
a = 2 m/s2
t = 10 s
Asked: d?
Answer:
d = v1 . t + 1/2 . a . t2
= 5 . 10 + 1/2 . 2 . (10)2
= 50 + 100
= 150 m
So, the displacement of a car is 150 m
16. What is the final velocity of a car that accelerated 10 m/s2 from rest and traveled 180m?
Known: v1 = 0 m/s (from rest)
d = 180 m
a = 10 m/s2
Asked: v2?
Answer:
v22 = v12 + 2. a . d
v22 = 02 + 2. 10 . 180
v22 = 3600
v2 = (3600)1/2
v2 = 60 m/s
So, the final velocity is 60 m/s
17. If a car accelerated from 5 m/s to 25 m/s in 10 seconds, how far will it travel?
Known: v1 = 5 m/s
v2 = 10 m/s
t = 10 s
Asked: d?
Answer:
We have to find the acceleration in the beginning, later we can calculate the distance.
v2 = v1 + a . t
10 = 5 + a . 10
10a = 10 - 5
10a = 5
a = 0.5 m/s2
Then,
d = v1 . t + 1/2 . a . t2
= 5 . 10 + 1/2 . 0.5 . (10)2
= 50 + 25
= 75 m
So, the distance of car is 75 m.
18. Renata Gas is driving through town at 25.0 m/s and begins to accelerate at a constant rate of -1.0 m/s2. Eventually Rennata comes to a complete stop. Use kinematic equations to calculate the distance that Renata travels while decelerating.
Known: v1 = 25 m/s
v2 = 0 m/s (complete to stop)
a = -1.0 m/s2
Asked: d?
Answer:
We have to find the time in the beginning, later we can calculate the distance.
v2 = v1 + a . t
0 = 25 + -1 . t
t = 25 s
Then,
d = v1 . t + 1/2 . a . t2
= 25 . 25 + 1/2 . -1 . (25)2
= 625 - 312.5
= 312.5 m
So, the distance of car is 312.5 m.
19. An object that is not moving initially given force F so that the speed is 10 m / s and the distance of 15 meters. Calculate the length of time the object is given force F!
Known: v1 = 0 m/s (from rest)
v2 = 10 m/s
d = 15 m
Asked: t?
Answer: We calculate the acceleration at first,
v22 = v12 + 2. a . d
102 = 02 + 2. a . 15
100 = 30a
a = 100/30
a = 3.33 m/s2
Then,
v2 = v1 + a . t
10 = 0 + 3.33 . t
t = 3 s
So, the time is 3 s.
Thursday, November 24, 2016
Monday, September 26, 2016
Soal dan Pembahasan Vektor Fisika kelas 10 SMA
1. Two vectors of 6 units in magnitude and 8 units in magnitude make an angle of 90o. The resultant force of these vectors is . . .
a. 10 units
b. 14 units
c. 16 units
d. 20 units
e. 24 units
Result: FR = (Fx2 + Fy2)1/2
= (62 + 82)1/2
= (36 + 64)1/2
= (100)1/2
= 10 N
2. The magnitude of vector A is 3 units and vector B is 4 units. If the addition of vectors A and B (A + B) is 5 units, the angle made by these two vectors is . . .
a. 30o
b. 45o
c. 60o
d. 73o
e. 90o
Result: R = (A2 + B2 + 2 A.B cos α)1/2
5 = (32 + 42 + 2.3.4 cos α)1/2
25 = (32 + 42 + 2.3.4 cos α)
25 = (9 + 16 + 24 cos α)
25 = 25 + 24 cos α
25 - 25 = 24 cos α
0 = 24 cos α
cos α = 0
α = arc cos-1 0
α = 90o
3. One vector of 10 units in magnitude directed to the right and another of 5 units in magnitude is directed to the left act on an object. To maintain the object at rest, we need an additional vector of . . .
a. 10 units in magnitude that is directed to the right
b. 15 units in magnitude that is directed to the left
c. 5 units in magnitude that is directed to the right
d. 5 units in magnitude that is directed to the left
e. zero length
Result: ∑F = 0
Fx+ + Fx- + Fr = 0 10 + (-5) + Fr = 0
Fr = 5 - 10
Fr = -5 (minus sign shows that the direction is to the left).
4. Look at the following scheme. The magnitude of the component of resultant vector to X and Y directions are . . .
a. Fx = (3√3 - 2√2), Fy = (3 + 2√2)
b. Fx = (3√3 - 2√2), Fy = (3√2 + 2√2)
c. Fx = (3√3 + 2√2), Fy = (3 + 2√2)
d. Fx = (3√3 + 2√2), Fy = (3√2 + 2√2)
e. Fx = (3√3 - 2√2), Fy = (3√3 + 2√2)
Result: Fx = 6 units cos 30 + 4 units cos 135
Fx = 6 units 0.5√3 - 4 units 0.5√2
Fx = 3√3 - 2√2;
Fy = 6 units sin 30 + 4 units sin 135
Fy = 6 units 0.5 + 4 units 0.5√2
Fy = 3 + 2√2.
The answer is A.
5. Vector A of 20 units in magnitude makes an angle of 45o. with respect to the +X direction. The components of this vector in the X and Y directions are . . .
a. Ax = 20 units, Ay = 20 units
b. Ax = 20√2 units, Ay = 20√2 units
c. Ax = 10 units, Ay = 10 units
d. Ax = 10√2 units, Ay = 10√2 units
e. Ax = 10√3 units, Ay = 10√2 units
Result:
Ax = 20 units cos 45
Ax = 20 units 0.5√2
Ax = 10√2
Ay = 20 units sin 45
Ay = 20 units 0.5√2
Ay = 10√2
The answer is D.
6. Look at the following figure. The resultant force of the three vectors is . . .
a. 10 N
b. 20 N
c. 30 N
d. 40 N
e. 70 N
Result:
Fx = 30 N cos 30 + 10 N cos 90 + 30 N cos 150
Fx = 30 N (0.5√3) + 10 N (0)+ 30 N (-0.5√3)
Fx = 15√3 - 15√3
Fx = 0;
Fy = 30 N sin 30 + 10 N sin 90 + 30 N sin 150
Fy = 30 N (0.5) + 10 N (1) + 30 N (0.5)
Fy = 15 + 10 + 15
Fy = 40
R = √(Fx2+Fy2)
= √(02+402)
= 40
The answer is D.
7. The resultant of the following three vectors is . . .
a. 3 units
b. 5 units
c. 5√5 units
d. 10 units
e. 10√3 units
Result:
Fx = 10units cos 60 + 5units cos 180 + 5units cos 300
Fx = 10units (0.5) + 5units (-1) + 5units (0.5)
Fx = 5 - 5 + 2.5
Fx = 2.5;
Fy = 10units sin 60 + 5units sin 180 + 5units sin 300
Fy = 10units (0.5√3) + 5units (0) + 5units (-0.5√3)
Fy = 5√3 + 0 - 2.5√3
Fy =2.5√3
R = √(Fx2+Fy2)
= √(2.52+2.5√32)
= 5
The answer is B.
8. Two vectors, F1 and F2, make an angle of 105o. The resultant of these two vectors makes an angle of 60o with respect to vector F2. If the magnitude of vector F2 is 8 units, the magnitude of vector F1 is . .
a. 10 units
b. 6 units
c. 4√6 units
d. 2√2 units
e.√6 units
Result:
F1 /sin α = F2 /sin β
F1 / sin (60) = 8 / sin (105 - 60)
F1 / sin (60) = 8 / sin (45)
F1 = 8 . sin (60) / sin (45)
F1 = 8 . 1/2√3 / 1/2√2
F1 = 4√6 units
The answer is C.
9. Look at the following figure. The resultant of the two vectors in the figure is . . .
a. 3 units
b. 6 units
c. 8 units
d. 10 units
e. 11 units
Result:
10. An object is placed on a smooth surface, It is pulled with a force of 20 N making an angle of 60 with respeect to the surface. If the mass of the object is negative, the norm of the vertical force needed to maintain the object at rest is . . . N
a. 20√3
b. 20√2
c. 20
d. 10√3
e. 10√2
Result:
11. Three forces A, B, and C, act on a massless body simultaneously, as shown in the following figure. The component of resultant force in the X and Y directions is . . .
a. Rx = 5√3 units, Ry = 5√2 units
b. Rx = 0 units, Ry = 5 units
c. Rx = 5√3 units, Ry = -5√2 units
d. Rx = 5 units, Ry = 10 units
e. Rx = -5√3 units, Ry = 15 units
Result:
Fx = A cos 30 + B cos 210 + C cos 150
Fx = 10 cos 30 + 5 cos 210 + 5 cos 150
Fx = 10 (0.5√3) + 5 (-0.5√3) + 5 (-0.5√3)
Fx = 5√3 - 2.5√3 - 2.5√3
Fx = 0;
Fy = A sin 30 + B sin 210 + C sin 150
Fy = 10 (0.5) + 5 (-0.5) + 5 (0.5)
Fy = 5 - 2.5 + 2.5
Fy =5
R = √(Fx2+Fy2)
= √(02+52)
= 5
The answer is B.
12. Vector A has 6 units in magnitude and vector C has 5 units in magnitude. If A + B = 2C and the angle made by vectors A and B is 90o. The magnitude of vector B is . . .
a. 4 units
b. 6 units
c. 8 units
d. 10 units
e. 25 units
Result:
13. Vector B is 10 N in magnitude and vector C is 20 N in length. If A + B = C and the angle between A and C is 30o. The angle between vectors B and C is . . .
a. 30o
b. 45o
c. 60o
d. 90o
e. 120o
Result:
14. An object is pulled to the right with a force of 30 N that makes an angle of 30o with respect to the surface of the floor. The magnitude and direction of an additional vector that must be given to the object to maintain the object at rest is . . .
a. 20 N to the right
b. 24 N to the left
c. 10 N to the right
d. 10 N to the left
e. 18 N to the up
Result:
15. Two vectors have the same magnitude. If the fraction of addition and subtraction of these vectors is √3, the angle made by these two vectors is . . .
a. 30o
b. 37o
c. 45o
d. 60o
e. 120o
Result:
II. STRUCTURED QUESTIONS
Answer all the questions on the answer sheet provided. Calculators are allowed.
1. The figure below shows two vectors, A and B. Determine the resultant.
2. Compute the magnitude and the direction of the resultant of the four vectors.
3. Vector A has a magnitude of 3 units and vector C has 5 units. If 2A + B = 2C and the angle made by vectors A and B is 60°. What is the magnitude of vector B ?
4. Vector B has a magnitude of 20 N and vector C has 25 N in length. If A + B = C and the angle between vectors A and C is 30°, determine the angle between vectors B and C.
a. 10 units
b. 14 units
c. 16 units
d. 20 units
e. 24 units
Result: FR = (Fx2 + Fy2)1/2
= (62 + 82)1/2
= (36 + 64)1/2
= (100)1/2
= 10 N
2. The magnitude of vector A is 3 units and vector B is 4 units. If the addition of vectors A and B (A + B) is 5 units, the angle made by these two vectors is . . .
a. 30o
b. 45o
c. 60o
d. 73o
e. 90o
Result: R = (A2 + B2 + 2 A.B cos α)1/2
5 = (32 + 42 + 2.3.4 cos α)1/2
25 = (32 + 42 + 2.3.4 cos α)
25 = (9 + 16 + 24 cos α)
25 = 25 + 24 cos α
25 - 25 = 24 cos α
0 = 24 cos α
cos α = 0
α = arc cos-1 0
α = 90o
3. One vector of 10 units in magnitude directed to the right and another of 5 units in magnitude is directed to the left act on an object. To maintain the object at rest, we need an additional vector of . . .
a. 10 units in magnitude that is directed to the right
b. 15 units in magnitude that is directed to the left
c. 5 units in magnitude that is directed to the right
d. 5 units in magnitude that is directed to the left
e. zero length
Result: ∑F = 0
Fx+ + Fx- + Fr = 0 10 + (-5) + Fr = 0
Fr = 5 - 10
Fr = -5 (minus sign shows that the direction is to the left).
4. Look at the following scheme. The magnitude of the component of resultant vector to X and Y directions are . . .
b. Fx = (3√3 - 2√2), Fy = (3√2 + 2√2)
c. Fx = (3√3 + 2√2), Fy = (3 + 2√2)
d. Fx = (3√3 + 2√2), Fy = (3√2 + 2√2)
e. Fx = (3√3 - 2√2), Fy = (3√3 + 2√2)
Result: Fx = 6 units cos 30 + 4 units cos 135
Fx = 6 units 0.5√3 - 4 units 0.5√2
Fx = 3√3 - 2√2;
Fy = 6 units sin 30 + 4 units sin 135
Fy = 6 units 0.5 + 4 units 0.5√2
Fy = 3 + 2√2.
The answer is A.
5. Vector A of 20 units in magnitude makes an angle of 45o. with respect to the +X direction. The components of this vector in the X and Y directions are . . .
a. Ax = 20 units, Ay = 20 units
b. Ax = 20√2 units, Ay = 20√2 units
c. Ax = 10 units, Ay = 10 units
d. Ax = 10√2 units, Ay = 10√2 units
e. Ax = 10√3 units, Ay = 10√2 units
Result:
Ax = 20 units cos 45
Ax = 20 units 0.5√2
Ax = 10√2
Ay = 20 units sin 45
Ay = 20 units 0.5√2
Ay = 10√2
The answer is D.
6. Look at the following figure. The resultant force of the three vectors is . . .
b. 20 N
c. 30 N
d. 40 N
e. 70 N
Result:
Fx = 30 N cos 30 + 10 N cos 90 + 30 N cos 150
Fx = 30 N (0.5√3) + 10 N (0)
Fx = 15√3 - 15√3
Fx = 0;
Fy = 30 N sin 30 + 10 N sin 90 + 30 N sin 150
Fy = 30 N (0.5) + 10 N (1) + 30 N (0.5)
Fy = 15 + 10 + 15
Fy = 40
R = √(Fx2+Fy2)
= √(02+402)
= 40
The answer is D.
7. The resultant of the following three vectors is . . .
a. 3 units
b. 5 units
c. 5√5 units
d. 10 units
e. 10√3 units
Result:
Fx = 10units cos 60 + 5units cos 180 + 5units cos 300
Fx = 10units (0.5) + 5units (-1) + 5units (0.5)
Fx = 5 - 5 + 2.5
Fx = 2.5;
Fy = 10units sin 60 + 5units sin 180 + 5units sin 300
Fy = 10units (0.5√3) + 5units (0) + 5units (-0.5√3)
Fy = 5√3 + 0 - 2.5√3
Fy =2.5√3
R = √(Fx2+Fy2)
= √(2.52+2.5√32)
= 5
The answer is B.
8. Two vectors, F1 and F2, make an angle of 105o. The resultant of these two vectors makes an angle of 60o with respect to vector F2. If the magnitude of vector F2 is 8 units, the magnitude of vector F1 is . .
a. 10 units
b. 6 units
c. 4√6 units
d. 2√2 units
e.√6 units
Result:
F1 /sin α = F2 /sin β
F1 / sin (60) = 8 / sin (105 - 60)
F1 / sin (60) = 8 / sin (45)
F1 = 8 . sin (60) / sin (45)
F1 = 8 . 1/2√3 / 1/2√2
F1 = 4√6 units
The answer is C.
9. Look at the following figure. The resultant of the two vectors in the figure is . . .
a. 3 units
b. 6 units
c. 8 units
d. 10 units
e. 11 units
Result:
10. An object is placed on a smooth surface, It is pulled with a force of 20 N making an angle of 60 with respeect to the surface. If the mass of the object is negative, the norm of the vertical force needed to maintain the object at rest is . . . N
a. 20√3
b. 20√2
c. 20
d. 10√3
e. 10√2
Result:
11. Three forces A, B, and C, act on a massless body simultaneously, as shown in the following figure. The component of resultant force in the X and Y directions is . . .
a. Rx = 5√3 units, Ry = 5√2 units
b. Rx = 0 units, Ry = 5 units
c. Rx = 5√3 units, Ry = -5√2 units
d. Rx = 5 units, Ry = 10 units
e. Rx = -5√3 units, Ry = 15 units
Result:
Fx = A cos 30 + B cos 210 + C cos 150
Fx = 10 cos 30 + 5 cos 210 + 5 cos 150
Fx = 10 (0.5√3) + 5 (-0.5√3) + 5 (-0.5√3)
Fx = 5√3 - 2.5√3 - 2.5√3
Fx = 0;
Fy = A sin 30 + B sin 210 + C sin 150
Fy = 10 (0.5) + 5 (-0.5) + 5 (0.5)
Fy = 5 - 2.5 + 2.5
Fy =5
R = √(Fx2+Fy2)
= √(02+52)
= 5
The answer is B.
12. Vector A has 6 units in magnitude and vector C has 5 units in magnitude. If A + B = 2C and the angle made by vectors A and B is 90o. The magnitude of vector B is . . .
a. 4 units
b. 6 units
c. 8 units
d. 10 units
e. 25 units
Result:
13. Vector B is 10 N in magnitude and vector C is 20 N in length. If A + B = C and the angle between A and C is 30o. The angle between vectors B and C is . . .
a. 30o
b. 45o
c. 60o
d. 90o
e. 120o
Result:
14. An object is pulled to the right with a force of 30 N that makes an angle of 30o with respect to the surface of the floor. The magnitude and direction of an additional vector that must be given to the object to maintain the object at rest is . . .
a. 20 N to the right
b. 24 N to the left
c. 10 N to the right
d. 10 N to the left
e. 18 N to the up
Result:
15. Two vectors have the same magnitude. If the fraction of addition and subtraction of these vectors is √3, the angle made by these two vectors is . . .
a. 30o
b. 37o
c. 45o
d. 60o
e. 120o
Result:
II. STRUCTURED QUESTIONS
Answer all the questions on the answer sheet provided. Calculators are allowed.
1. The figure below shows two vectors, A and B. Determine the resultant.
4. Vector B has a magnitude of 20 N and vector C has 25 N in length. If A + B = C and the angle between vectors A and C is 30°, determine the angle between vectors B and C.
Monday, May 30, 2016
Tuesday, May 24, 2016
Monday, May 23, 2016
Sunday, May 22, 2016
Wednesday, May 18, 2016
7th Physics Heat Daily test 2015/2016 Code A
1. What is the heat required to increase the temperature of 7,500 g of aluminum from temperature of 473 K to 753 K? If the specific heat of aluminum is 900 J/kg °C (5 points)
2. A kettle aluminum made of mass 0.35 kg of the initial temperature 15°C, then it is given heat of 25,000 J. What is the temperature of the kettle now? (Specific heat of aluminum is 900 J/kg°C) (5 points)
3. A cup of hot tea is let first until warm before it is drunk. The temperature of the hot tea decreases from 80 °C to be 45 °C by releasing heat of 60 kJ. What is the mass of tea in the cup? (Suppose the specific heat of tea is equal to the specific heat of water) (5 points)
4. Heat of 32,500 J can increase the temperature of the body from 15°C until 45°C. What is the heat capacity of the body?(5 points)
5. Look at the graph and calculate the heat required 100 g of ice to reach point E (heat specific heat of ice is 2,100 J/kg°C and melting heat is 336,000 J/kg and evaporating heat is 2,260,000 J/kg.
(15 points)
6. How much power does it take to lift 30 N, 10 m high in 5 s? (5 points)
7. You move a 25 N object 5 meters. How much work did you do? (5 points)
8. Give 3 ways of heat transfer. (3 points)
9. Give 3 applications of heat. (3 points)
Untuk mendapatkan soal dalam bentuk .docx kalian dapat download disini
2. A kettle aluminum made of mass 0.35 kg of the initial temperature 15°C, then it is given heat of 25,000 J. What is the temperature of the kettle now? (Specific heat of aluminum is 900 J/kg°C) (5 points)
3. A cup of hot tea is let first until warm before it is drunk. The temperature of the hot tea decreases from 80 °C to be 45 °C by releasing heat of 60 kJ. What is the mass of tea in the cup? (Suppose the specific heat of tea is equal to the specific heat of water) (5 points)
4. Heat of 32,500 J can increase the temperature of the body from 15°C until 45°C. What is the heat capacity of the body?(5 points)
5. Look at the graph and calculate the heat required 100 g of ice to reach point E (heat specific heat of ice is 2,100 J/kg°C and melting heat is 336,000 J/kg and evaporating heat is 2,260,000 J/kg.
(15 points)
6. How much power does it take to lift 30 N, 10 m high in 5 s? (5 points)
7. You move a 25 N object 5 meters. How much work did you do? (5 points)
8. Give 3 ways of heat transfer. (3 points)
9. Give 3 applications of heat. (3 points)
Untuk mendapatkan soal dalam bentuk .docx kalian dapat download disini
Saturday, May 14, 2016
Thursday, May 12, 2016
Wednesday, May 11, 2016
Monday, May 2, 2016
Thursday, January 7, 2016
Kunci Jawaban Uji Kompetensi Bab 6 Dinamika Rotasi dan Keseimbangan Benda Tegar
1. Perhatikan empat persegi panjang di bawah. Tentukan torsi dari gaya F1, F2, F3, F4 dan F5, yang masing-masing 10 N terhadap poros
a) Poros melalui O,
b) Poros melalui A.
Jawaban:
2. Pada sebuah roda dengan jari-jari 40 cm bekerja gaya-gaya, seperti ditunjukkan pada gambar. Tentukan torsi total terhadap poros yang melalui O.
Jawaban:
3. Empat buah partikel seperti ditunjukkan pada gambar dihubungkan oleh sebuah batang kaku ringan yang massanya dapat diabaikan. Tentukan momen inersia sistem partikel terhadap proses:
a. sumbu AA1,
b. sumbu BB1
4. Jari-jari (ruji) sepanjang 0,5 m, seperti tampak pada gambar, memiliki massa yang dapat diabaikan terhadap delapan partikel bermassa 3,0 kg. tentukan momen inersia sistem terhadap :
a. poros melalui pusat jari-jari
b. poros AA1
Jawaban:
a) Poros melalui O,
b) Poros melalui A.
Jawaban:
2. Pada sebuah roda dengan jari-jari 40 cm bekerja gaya-gaya, seperti ditunjukkan pada gambar. Tentukan torsi total terhadap poros yang melalui O.
3. Empat buah partikel seperti ditunjukkan pada gambar dihubungkan oleh sebuah batang kaku ringan yang massanya dapat diabaikan. Tentukan momen inersia sistem partikel terhadap proses:
a. sumbu AA1,
b. sumbu BB1
4. Jari-jari (ruji) sepanjang 0,5 m, seperti tampak pada gambar, memiliki massa yang dapat diabaikan terhadap delapan partikel bermassa 3,0 kg. tentukan momen inersia sistem terhadap :
a. poros melalui pusat jari-jari
b. poros AA1
Jawaban:
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