Question

multiple choice
identify the choice that best completes the statement or answers the question.
1 which of the following is an example of a vector quantity?
a. velocity b. temperature c. volume d. mass

1. in the figure above, which diagram represents the vector addition ( c = a + b )?

a. i b. ii c. iii d. iv

1. in the figure above, which diagram represents the vector subtraction ( c = a - b )?

a. i b. ii c. iii d. iv

1. a stone is thrown with an initial speed of 15 m/s at an angle of ( 53^{circ} ) above the horizontal from the top of a 35 - m building. if ( g = 9.8 mathrm{~m} / mathrm{s}^{2} ) and air resistance is negligible, then what is the magnitude of the horizontal component of velocity as the rock strikes the ground?

a. ( 7.5 mathrm{~m} / mathrm{s} ) b. ( 9.0 mathrm{~m} / mathrm{s} ) c. ( 12 mathrm{~m} / mathrm{s} ) d. ( 29 mathrm{~m} / mathrm{s} )

1. jack pulls a sled across a level field by exerting a force of ( 110 mathrm{~n} ) at an angle of ( 30^{circ} ) with the ground. what are the ( x ) and ( y ) components, respectively, of this force with respect to the ground?

a. ( 64 mathrm{~n}, 190 mathrm{~n} ) b. ( 190 mathrm{~n}, 64 mathrm{~n} ) c. ( 95 mathrm{~n}, 55 mathrm{~n} ) d. ( 55 mathrm{~n}, 95 mathrm{~n} )

1. find the resultant of these two vectors: ( 2.00 \times 10^{2} ) units due east and ( 4.00 \times 10^{2} ) units ( 30.0^{circ} ) north of west.

a. 300 units, ( 29.8^{circ} ) north of west b. 581 units, ( 20.1^{circ} ) north of east c. 546 units, ( 59.3^{circ} ) north of west d. 248 units, ( 53.9^{circ} ) north of west

1. a ball is launched from ground level at ( 30 mathrm{~m} / mathrm{s} ) at an angle of ( 35^{circ} ) above the horizontal. how far before it is at ground level again?

a. ( 14 mathrm{~m} ) b. ( 21 mathrm{~m} ) c. ( 43 mathrm{~m} ) d. ( 86 mathrm{~m} )

Answer:

1. Question 1:

• Brief Explanations:
• A vector quantity has both magnitude and direction. Velocity has direction (e.g., \(5\ m/s\) east), while temperature, volume, and mass are scalar quantities (only magnitude).
• Answer:
• a. velocity

1. Question 2:

• Brief Explanations:
• For vector addition \(\vec{C}=\vec{A}+\vec{B}\), we place the tail of \(\vec{B}\) at the head of \(\vec{A}\), and \(\vec{C}\) is from the tail of \(\vec{A}\) to the head of \(\vec{B}\). Diagram II follows this rule.
• Answer:
• b. II

1. Question 3:

• Brief Explanations:
• For vector subtraction \(\vec{C}=\vec{A}-\vec{B}=\vec{A}+(-\vec{B})\). We reverse the direction of \(\vec{B}\) and then add it to \(\vec{A}\) (place the tail of \(-\vec{B}\) at the head of \(\vec{A}\)). Diagram III follows this rule.
• Answer:
• c. III

1. Question 4:

• Step - by - Step Format:
• Explanation:
• In projectile motion (when air - resistance is negligible), the horizontal component of velocity \(v_x = v_0\cos\theta\). Given \(v_0 = 15\ m/s\) and \(\theta = 53^{\circ}\).
• We know that \(\cos53^{\circ}\approx0.6\).
• Step1: Calculate the horizontal component of velocity
• \(v_x=v_0\cos\theta\)
• Substitute \(v_0 = 15\ m/s\) and \(\cos\theta=\cos53^{\circ}\approx0.6\) into the formula: \(v_x = 15\times0.6\)
• Step2: Get the result
• \(v_x = 9.0\ m/s\)
• Answer:
• b. \(9.0\ m/s\)

1. Question 5:

• Step - by - Step Format:
• Explanation:
• The \(x\) - component of the force \(F_x=F\cos\theta\) and the \(y\) - component \(F_y = F\sin\theta\). Given \(F = 110\ N\) and \(\theta = 30^{\circ}\), \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\approx0.866\), \(\sin30^{\circ}=0.5\)
• Step1: Calculate the \(x\) - component of the force
• \(F_x=F\cos\theta\)
• Substitute \(F = 110\ N\) and \(\cos\theta=\cos30^{\circ}\approx0.866\) into the formula: \(F_x=110\times0.866\approx95\ N\)
• Step2: Calculate the \(y\) - component of the force
• \(F_y=F\sin\theta\)
• Substitute \(F = 110\ N\) and \(\sin\theta = \sin30^{\circ}=0.5\) into the formula: \(F_y=110\times0.5 = 55\ N\)
• Answer:
• c. \(95\ N,55\ N\)

1. Question 6:

• Step - by - Step Format:
• Explanation:
• Let \(\vec{A}=2.00\times10^{2}\) units (east) and \(\vec{B}=4.00\times 10^{2}\) units at \(30^{\circ}\) north of west.
• The \(x\) - component of \(\vec{A}\): \(A_x = 200\) units, the \(x\) - component of \(\vec{B}\): \(B_x=-400\cos30^{\circ}=- 400\times\frac{\sqrt{3}}{2}=-346.4\) units, the \(y\) - component of \(\vec{B}\): \(B_y = 400\sin30^{\circ}=200\) units.
• The \(x\) - component of the resultant \(\vec{R}\): \(R_x=A_x + B_x=200-346.4=-146.4\) units, the \(y\) - component of the resultant \(\vec{R}\): \(R_y = B_y=200\) units.
• Step1: Calculate the magnitude of the resultant
• \(R=\sqrt{R_x^{2}+R_y^{2}}\)
• \(R=\sqrt{(-146.4)^{2}+200^{2}}=\sqrt{21432.96 + 40000}=\sqrt{61432.96}\approx248\) units
• Step2: Calculate the direction of the resultant
• \(\tan\varphi=\frac{R_y}{\vert R_x\vert}\)
• \(\tan\varphi=\frac{200}{146.4}\approx1.366\), \(\varphi=\arctan(1.366)\approx53.9^{\circ}\) north of west
• Answer:
• d. \(248\) units, \(53.9^{\circ}\) north of west

1. Question 7:

• Step - by - Step Format:
• Explanation:
• The range formula for projectile motion \(R=\frac{v_0^{2}\sin2\theta}{g}\). Given \(v_0 = 30\ m/s\), \(\theta = 35^{\circ}\), so \(2\theta = 70^{\circ}\), \(\sin70^{\circ}\approx0.94\), \(g = 9.8\ m/s^{2}\)
• Step1: Calculate the range
• \(R=\frac{v_0^{2}\sin2\theta}{g}\)
• Substitute \(v_0 = 30\ m/s\), \(\sin2\theta=\sin70^{\circ}\approx0.94\), \(g = 9.8\ m/s^{2}\) into the formula: \(R=\frac{30^{2}\times0.94}{9.8}=\frac{900\times0.94}{9.8}=\frac{846}{9.8}\approx86\ m\)
• Answer:
• d. \(86\ m\)