Homework 03 – Khaleefoh, Fahad – Due: Feb 11, 11:00 PM
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 1 Question 1, chap 3, sect 4. part 1 of 2 10 points
A vector representing 120 N is oriented at 41° with the horizontal. What is the magnitude of its horizontal component? Answer in units of N.
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The problem involves finding the components of a vector based on its magnitude and direction. Specifically, the vector has a magnitude of 120 N and is inclined at an angle of 41° from the horizontal. To determine the horizontal component, we use trigonometry, namely the cosine function, which relates the adjacent side of a right triangle (the horizontal component) to the hypotenuse (the vector magnitude).
The horizontal component (F_x) is calculated as:
F_x = F × cos(θ)
Where F is the magnitude of the vector (120 N) and θ is the angle with the horizontal (41°). Substituting the values:
F_x = 120 N × cos(41°)
Calculating cos(41°):
cos(41°) ≈ 0.7547
Thus, the horizontal component is:
F_x ≈ 120 N × 0.7547 ≈ 90.56 N
Therefore, the magnitude of the horizontal component is approximately 90.56 N.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 2 Question 2, chap 3, sect 4. part 2 of 2 10 points
What is the magnitude of the vector’s vertical component? Answer in units of N.
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Continuing from the previous problem, to find the vertical component (F_y) of the vector, we use the sine function, since it relates the opposite side of the right triangle to the hypotenuse (the vector magnitude).
The vertical component (F_y) is given by:
F_y = F × sin(θ)
Plugging in the known values:
F_y = 120 N × sin(41°)
Calculating sin(41°):
sin(41°) ≈ 0.6561
Therefore:
F_y ≈ 120 N × 0.6561 ≈ 78.73 N
Hence, the magnitude of the vertical component of the vector is approximately 78.73 N.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 3 Question 3, chap 3, sect 4. part 1 of 2 10 points
A truck travels 1170 m uphill along a road that makes a constant angle of 5.05° with the horizontal. Find the magnitude of the truck’s horizontal component of displacement. Answer in units of m.
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The problem involves decomposing the displacement of the truck into horizontal and vertical components. Given that the truck travels along an inclined path (length of 1170 m) at an angle of 5.05° to the horizontal, the horizontal component can be calculated using the cosine of the incline angle.
The horizontal component (d_x) is:
d_x = d × cos(θ)
Where d is 1170 m and θ is 5.05°.
Calculating cos(5.05°):
cos(5.05°) ≈ 0.9958
Thus:
d_x ≈ 1170 m × 0.9958 ≈ 1165.51 m
Therefore, the horizontal component of the displacement is approximately 1165.51 m.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 4 Question 4, chap 3, sect 4. part 2 of 2 10 points
Find the magnitude of the truck’s vertical component of displacement. Answer in units of m.
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Using the same setup as the previous problem, the vertical component of the displacement (d_y) is found via the sine function, which relates the opposite side of the triangle (vertical displacement) to the hypotenuse (the path length). The calculation is:
d_y = d × sin(θ)
Substituting the known values:
d_y = 1170 m × sin(5.05°)
Calculating sin(5.05°):
sin(5.05°) ≈ 0.0880
Thus:
d_y ≈ 1170 m × 0.0880 ≈ 102.96 m
Hence, the vertical component of displacement is approximately 102.96 m.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 5 Question 5, chap 3, sect 6. part 1 of 3 10 points
A particle starts from the origin at t = 0 with an initial velocity having an x component of 15.7 m/s and a y component of −23.9 m/s. The particle moves in the xy plane with an x component of acceleration only, given by 2.88 m/s². Determine the x component of velocity after 2.14 s. Answer in units of m/s.
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This problem involves calculating the x component of the velocity of a particle under constant acceleration, with initial velocity components. Since the y component involves no acceleration, only the x component changes over time.
The x component of velocity (v_x) at time t is given by:
v_x = v_{x0} + a_x × t
Where:
- v_{x0} = initial x velocity = 15.7 m/s
- a_x = acceleration in x direction = 2.88 m/s²
- t = 2.14 s
Calculating v_x:
v_x = 15.7 m/s + (2.88 m/s² × 2.14 s)
First, compute the acceleration term:
2.88 × 2.14 ≈ 6.1632
Adding to initial velocity:
v_x ≈ 15.7 + 6.1632 ≈ 21.86 m/s
The x component of velocity after 2.14 seconds is approximately 21.86 m/s.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 6 Question 6, chap 3, sect 6. part 2 of 3 10 points
Find the speed of the particle after 2.14 s.
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To find the speed of the particle after 2.14 seconds, we consider both the x and y components of its velocity. The x component has been calculated in the previous step, and the y component remains constant because there is no acceleration in the y direction.
The y component of velocity (v_{y}) is initially −23.9 m/s and does not change:
v_y = v_{y0} = -23.9 m/s
Using the value from earlier:
v_x ≈ 21.86 m/s
The magnitude of the velocity vector — the speed — is given by:
v = √(v_x² + v_y²)
Calculating:
v = √(21.86² + (−23.9)²) ≈ √(477.57 + 571.21) ≈ √1048.78 ≈ 32.4 m/s
The speed of the particle after 2.14 seconds is approximately 32.4 m/s.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 7 Question 7, chap 3, sect 6. part 3 of 3 10 points
Find the magnitude of the displacement vector of the particle after t = 2.14 s. Answer in units of m.
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The displacement involves both the x and y components of the particle's position after 2.14 seconds. For the x component, since initial x velocity and acceleration are known, we use the kinematic equation:
d_x = v_{x0} × t + 0.5 × a_x × t²
Substituting values:
d_x = 15.7 × 2.14 + 0.5 × 2.88 × (2.14)²
Calculations:
15.7 × 2.14 ≈ 33.638
0.5 × 2.88 × 4.58 (since 2.14² ≈ 4.58)
= 1.44 × 4.58 ≈ 6.605
Adding:
d_x ≈ 33.638 + 6.605 ≈ 40.24 m
The y component of displacement is:
d_y = v_{y0} × t = (−23.9) × 2.14 ≈ -51.15 m
The magnitude of the displacement vector is:
√(d_x² + d_y²) ≈ √(40.24² + (−51.15)²) ≈ √(1619 + 2616) ≈ √4235 ≈ 65.07 m
Thus, the magnitude of the displacement vector after 2.14 seconds is approximately 65.07 m.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 8 Question 8, chap 3, sect 7. part 1 of 2 10 points
Vector ~A is 2.06 units long and points in the positive y direction. Vector ~B has a negative x component 6.83 units long, a positive y component 2.1 units long, and no z component. Find ~A · ~B. Answer in units of units2.
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Given vectors:
- ~A has a magnitude of 2.06 units and points purely in the positive y direction. Therefore, its components are:
- ~A_x = 0
- ~A_y = 2.06
- ~A_z = 0
- ~B has components:
- ~B_x = -6.83
- ~B_y = 2.1
- ~B_z = 0
The dot product of vectors ~A and ~B is calculated as:
~A · ~B = A_x × B_x + A_y × B_y + A_z × B_z
Substituting the components:
~A · ~B = 0 × (−6.83) + 2.06 × 2.1 + 0 × 0 = 0 + 4.326 + 0 = 4.326
Therefore, ~A · ~B ≈ 4.33 units2.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 9 Question 9, chap 3, sect 7. part 2 of 2 10 points
What is the angle between ~A and ~B? Answer in units of °.
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The angle θ between vectors ~A and ~B can be found using the dot product formula:
cosθ = (~A · ~B) / (|A| × |B|)
Calculating the magnitudes:
|A| = 2.06 units (since it points only in y direction)
|B| = √(B_x² + B_y²) = √((-6.83)² + (2.1)²) ≈ √(46.64 + 4.41) ≈ √51.05 ≈ 7.14 units
Using the dot product and magnitudes:
cosθ = 4.33 / (2.06 × 7.14) ≈ 4.33 / 14.72 ≈ 0.294
Then, θ = arccos(0.294) ≈ 73.02°
Thus, the angle between ~A and ~B is approximately 73.02°.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 10 Question 10, chap 3, sect 8. part 1 of 2 10 points
Given: Two vectors ~A = Ax î + Ay ĵ and ~B = Bx î + By ĵ , where Ax = −5, Ay = 1, Bx = 2, and By = 1. Find the z component of ~A × ~B.
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The cross product of two vectors in three dimensions is given by the determinant:
~A × ~B = |î ĵ k̂|
|A_x A_y 0|
|B_x B_y 0|
Calculating the z component of the cross product (since î and ĵ components will be zero):
(A_x)(B_y) - (A_y)(B_x) = (−5)(1) - (1)(2) = -5 - 2 = -7
This is the z component of the cross product vector. The x and y components are zero.
Therefore, the z component of ~A × ~B is −7.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 11 Question 11, chap 3, sect 8. part 2 of 2 10 points
Find the angle between ~A and ~B. Answer in units of °.
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Using the previous calculation of the dot product (~A · ~B = 4.33) and the magnitudes of vectors ~A and ~B (~A = 2.06 units, ~B ≈ 7.14 units), the angle θ is:
cosθ = (~A · ~B) / (|A| × |B|)
Plugging in the values:
cosθ = 4.33 / (2.06 × 7.14) ≈ 4.33 / 14.72 ≈ 0.294
θ = arccos(0.294) ≈ 73.02°
Thus, the angle between vectors ~A and ~B is approximately 73.02°.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 12 Question 12, chap 3, sect 8. part 1 of 1 10 points
Given: Two vectors ~A = Ax î + Ay ĵ , where Ax = −5, Ay = 6.3, Bx = 2, and By = 6. Find the z component of ~A × ~B.
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The cross product's z component is:
(A_x)(B_y) - (A_y)(B_x) = (−5)(6) - (6.3)(2) = -30 - 12.6 = -42.6
Hence, the z component of ~A × ~B is −42.6.
homework 03 – KHALEEFOH, FAHAD – Due: Feb , 11:00 pm (Central time) 13 Question 13, chap 3, sect 4. part 1 of 3 10 points
The weight of a ball rolling down an inclined plane can be broken into two components: one acting parallel to the plane and the other acting perpendicular to the plane. At what inclination angle of the plane are these two components equal?