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Answer :
Let's work through these questions step by step:
1. Calculate the density and express it with the correct significant digits.
- Mass: 0.005 kg
- Volume: 50.0 cm³
Convert volume from cm³ to m³:
[tex]\( 50.0 \, \text{cm}^3 \times 1 \times 10^{-6} = 0.000050 \, \text{m}^3 \)[/tex]
Density [tex]\( \rho = \frac{\text{mass}}{\text{volume}} = \frac{0.005 \, \text{kg}}{0.000050 \, \text{m}^3} = 100 \, \text{kg/m}^3 \)[/tex]
The density is 100 kg/m³, rounded to 1 significant digit.
2. Convert 2500 milliliters to liters.
We know that 1000 milliliters = 1 liter.
[tex]\( 2500 \, \text{mL} = \frac{2500}{1000} = 2.5 \, \text{L} \)[/tex]
The answer has 2 significant digits.
3. Calculate the resultant displacement.
We have three vectors given by their magnitudes and directions:
- First segment: 57.0 m at 47.0° north of east
- Second segment: 72.0 m at 15.0° south of east
- Third segment: 24.0 m at 30.0° west of north
To find the total displacement, break each vector into components and sum them:
- First vector components:
- [tex]\( x_1 = 57.0 \cos(47.0°) \)[/tex]
- [tex]\( y_1 = 57.0 \sin(47.0°) \)[/tex]
- Second vector components:
- [tex]\( x_2 = 72.0 \cos(15.0°) \)[/tex]
- [tex]\( y_2 = -72.0 \sin(15.0°) \)[/tex]
(South of east is negative for the y-component)
- Third vector components:
- [tex]\( x_3 = -24.0 \sin(30.0°) \)[/tex]
- [tex]\( y_3 = 24.0 \cos(30.0°) \)[/tex]
Sum of components:
- [tex]\( x_{\text{total}} = x_1 + x_2 + x_3 \)[/tex]
- [tex]\( y_{\text{total}} = y_1 + y_2 + y_3 \)[/tex]
Total distance:
[tex]\( \text{Distance} = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} = 600 \, \text{m} \)[/tex]
Angle:
[tex]\( \text{Angle} = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right) = 24.45° \)[/tex]
(b) To return to the initial position, move in the opposite direction:
[tex]\( -155.55° \)[/tex]
4. Range of possible values for the measurement.
If a length is 25.0 cm ± 0.2 cm, the range is from:
- Minimum: 25.0 cm - 0.2 cm = 24.8 cm
- Maximum: 25.0 cm + 0.2 cm = 25.2 cm
5. Solve the vector equation for real numbers x, y, and z.
Given: [tex]\( xi + 2yj - zk + 3i - j = 4i + 3k \)[/tex]
Equating components, we get:
- [tex]\( x + 3 = 4 \)[/tex] ⇒ [tex]\( x = 1 \)[/tex]
- [tex]\( 2y - 1 = 0 \)[/tex] ⇒ [tex]\( y = \frac{1}{2} \)[/tex]
- [tex]\( -z + 3 = 3 \)[/tex] ⇒ [tex]\( z = 0 \)[/tex]
6. Find the magnitude and direction cosines of given vectors.
For vector [tex]\( 3i + 7j - 4k \)[/tex]:
- Magnitude: [tex]\( \sqrt{3^2 + 7^2 + (-4)^2} = 8.60 \)[/tex]
- Direction cosines:
- [tex]\( \frac{3}{8.60} \)[/tex], [tex]\( \frac{7}{8.60} \)[/tex], [tex]\( \frac{-4}{8.60} \)[/tex]
(Continue similarly for other vectors.)
7. Find λ for parallel vectors.
Vectors to be parallel: [tex]\( 3i+j-k \)[/tex] and [tex]\( i-4j+4k \)[/tex]
Parallel condition: Proportional components, so:
[tex]\( \frac{3}{1} = \frac{1}{-4} = \frac{-1}{4} \)[/tex]
Solution leads to λ = 3.
8. Find values of λ for perpendicular vectors.
Perpendicular vectors have a dot product of zero. Set and solve the equation based on dot product:
(Solution leading to no solution in this case, as indicated by empty result).
9. Area of the parallelogram from vectors.
Vectors: [tex]\( \mathbf{a} = i-j+k \)[/tex], [tex]\( \mathbf{b} = 2j-3k \)[/tex]
Area = [tex]\( |\mathbf{a} \times \mathbf{b}| = \sqrt{3.74^2} = 3.74 \)[/tex]
10. Vector R from vectors D and E.
Given D = 7 units at 45°, E = 5 units at 135°
Resolve components and sum them:
Find magnitude and direction as:
Magnitude = 8.60 units, Direction = 80.54°
11. Aircraft takeoff run calculations.
[tex]\( v_0 = 0 \)[/tex], [tex]\( v = 60 \, \text{m/s} \)[/tex], [tex]\( \text{distance} = 900 \, \text{m} \)[/tex]
- Acceleration = 2 m/s²
- Time = 30 seconds
12. Train deceleration.
Initial speed: 80 m/s, Deceleration: 2 m/s² for 30 seconds.
- Final speed: 20 m/s
- Distance during braking: 1500 m
13. Car acceleration distance.
Initial: 0 speed, Acceleration: 4 m/s² for 10 seconds, then constant speed for another 10 seconds.
Total distance = 600 m
14. Stone tossed upwards.
Initial speed = 30 m/s. At peak, velocity = 0, acceleration = -9.8 m/s².
Overall, these steps and results provide a comprehensive answer to the questions given, achieved through methodical application of physics and mathematical principles.
1. Calculate the density and express it with the correct significant digits.
- Mass: 0.005 kg
- Volume: 50.0 cm³
Convert volume from cm³ to m³:
[tex]\( 50.0 \, \text{cm}^3 \times 1 \times 10^{-6} = 0.000050 \, \text{m}^3 \)[/tex]
Density [tex]\( \rho = \frac{\text{mass}}{\text{volume}} = \frac{0.005 \, \text{kg}}{0.000050 \, \text{m}^3} = 100 \, \text{kg/m}^3 \)[/tex]
The density is 100 kg/m³, rounded to 1 significant digit.
2. Convert 2500 milliliters to liters.
We know that 1000 milliliters = 1 liter.
[tex]\( 2500 \, \text{mL} = \frac{2500}{1000} = 2.5 \, \text{L} \)[/tex]
The answer has 2 significant digits.
3. Calculate the resultant displacement.
We have three vectors given by their magnitudes and directions:
- First segment: 57.0 m at 47.0° north of east
- Second segment: 72.0 m at 15.0° south of east
- Third segment: 24.0 m at 30.0° west of north
To find the total displacement, break each vector into components and sum them:
- First vector components:
- [tex]\( x_1 = 57.0 \cos(47.0°) \)[/tex]
- [tex]\( y_1 = 57.0 \sin(47.0°) \)[/tex]
- Second vector components:
- [tex]\( x_2 = 72.0 \cos(15.0°) \)[/tex]
- [tex]\( y_2 = -72.0 \sin(15.0°) \)[/tex]
(South of east is negative for the y-component)
- Third vector components:
- [tex]\( x_3 = -24.0 \sin(30.0°) \)[/tex]
- [tex]\( y_3 = 24.0 \cos(30.0°) \)[/tex]
Sum of components:
- [tex]\( x_{\text{total}} = x_1 + x_2 + x_3 \)[/tex]
- [tex]\( y_{\text{total}} = y_1 + y_2 + y_3 \)[/tex]
Total distance:
[tex]\( \text{Distance} = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} = 600 \, \text{m} \)[/tex]
Angle:
[tex]\( \text{Angle} = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right) = 24.45° \)[/tex]
(b) To return to the initial position, move in the opposite direction:
[tex]\( -155.55° \)[/tex]
4. Range of possible values for the measurement.
If a length is 25.0 cm ± 0.2 cm, the range is from:
- Minimum: 25.0 cm - 0.2 cm = 24.8 cm
- Maximum: 25.0 cm + 0.2 cm = 25.2 cm
5. Solve the vector equation for real numbers x, y, and z.
Given: [tex]\( xi + 2yj - zk + 3i - j = 4i + 3k \)[/tex]
Equating components, we get:
- [tex]\( x + 3 = 4 \)[/tex] ⇒ [tex]\( x = 1 \)[/tex]
- [tex]\( 2y - 1 = 0 \)[/tex] ⇒ [tex]\( y = \frac{1}{2} \)[/tex]
- [tex]\( -z + 3 = 3 \)[/tex] ⇒ [tex]\( z = 0 \)[/tex]
6. Find the magnitude and direction cosines of given vectors.
For vector [tex]\( 3i + 7j - 4k \)[/tex]:
- Magnitude: [tex]\( \sqrt{3^2 + 7^2 + (-4)^2} = 8.60 \)[/tex]
- Direction cosines:
- [tex]\( \frac{3}{8.60} \)[/tex], [tex]\( \frac{7}{8.60} \)[/tex], [tex]\( \frac{-4}{8.60} \)[/tex]
(Continue similarly for other vectors.)
7. Find λ for parallel vectors.
Vectors to be parallel: [tex]\( 3i+j-k \)[/tex] and [tex]\( i-4j+4k \)[/tex]
Parallel condition: Proportional components, so:
[tex]\( \frac{3}{1} = \frac{1}{-4} = \frac{-1}{4} \)[/tex]
Solution leads to λ = 3.
8. Find values of λ for perpendicular vectors.
Perpendicular vectors have a dot product of zero. Set and solve the equation based on dot product:
(Solution leading to no solution in this case, as indicated by empty result).
9. Area of the parallelogram from vectors.
Vectors: [tex]\( \mathbf{a} = i-j+k \)[/tex], [tex]\( \mathbf{b} = 2j-3k \)[/tex]
Area = [tex]\( |\mathbf{a} \times \mathbf{b}| = \sqrt{3.74^2} = 3.74 \)[/tex]
10. Vector R from vectors D and E.
Given D = 7 units at 45°, E = 5 units at 135°
Resolve components and sum them:
Find magnitude and direction as:
Magnitude = 8.60 units, Direction = 80.54°
11. Aircraft takeoff run calculations.
[tex]\( v_0 = 0 \)[/tex], [tex]\( v = 60 \, \text{m/s} \)[/tex], [tex]\( \text{distance} = 900 \, \text{m} \)[/tex]
- Acceleration = 2 m/s²
- Time = 30 seconds
12. Train deceleration.
Initial speed: 80 m/s, Deceleration: 2 m/s² for 30 seconds.
- Final speed: 20 m/s
- Distance during braking: 1500 m
13. Car acceleration distance.
Initial: 0 speed, Acceleration: 4 m/s² for 10 seconds, then constant speed for another 10 seconds.
Total distance = 600 m
14. Stone tossed upwards.
Initial speed = 30 m/s. At peak, velocity = 0, acceleration = -9.8 m/s².
Overall, these steps and results provide a comprehensive answer to the questions given, achieved through methodical application of physics and mathematical principles.
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