Answer:
see explanation
Explanation:
Given that,
velocity of 1.50 km/s = 1.50 × 10³m/s
acceleration of 2.00 ✕ 1012 m/s2
electric field has a magnitude of strength of 18.0 N/C
![\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]](https://tex.z-dn.net/?f=%5Cbar%20F%3D%20q%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%5Cbar%20B%5D%5C%5C%5C%5C%5Cbar%20F%20%3D%20%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%28%20B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20z%20%29%5D%5C%5C%5C%5C%5C%5Cm%20%5Cbar%20a%20%3D%20%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%28%20B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20z%20%29%5D)
![9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)]](https://tex.z-dn.net/?f=9.1%20%5Ctimes%2010%5E-%5E3%5E1%20%5Ctimes%202%5Ctimes%2010%5E1%5E2%20%5Chat%20k%3D-1.6%5Ctimes10%5E-%5E1%5E9%20%5Chat%20k%20%5B18%5Chat%20k%2B%201.5%5Ctimes%2010%5E3%20%5Chat%20i%20%5Ctimes%20%28B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20k%29%5D)
Answer:
a) The maximum height the ball will achieve above the launch point is 0.2 m.
b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.
Explanation:
a)
For the height reached, we use 3rd equation of motion:
2gh = Vf² - Vo²
Here,
Vo = 3.75 m/s
Vf = 0m/s, since ball stops at the highest point
g = -9.8 m/s² (negative sign for upward motion)
h = maximum height reached by ball
therefore, eqn becomes:
2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²
<u>h = 0.2 m</u>
b)
To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:
2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²
(Vo)² = 19.6 m²/s²
Vo = √19.6 m²/s²
<u>Vo = 4.43 m/s</u>
Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)
<u>Vo = 0.174 in/ms</u>
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Answer:
A & B
Explanation:
A & B Would be the right answer since Morse code cannot be represented through the height of the fire.
PHysical properties are properties that do not involve changes in the molecular level. In this case, malleability and magnetism are physical.
Chemical properties involve changes to the substance at the molecular level or change its identity. In this case, reacting with oxygen and nonflammability is a chemical property.
Answer:
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
Explanation:
Given:
Radius of sphere (r) = 12 cm = 0.12 m
Distance from the electric field R = 24 cm = 0.24 m
Magnitude (E) = 640 N/C
Find:
Charge density on the sphere
Computation:
Charge on the sphere (q) = (1/K)ER² (K = 9 × 10⁹)
Charge on the sphere (q) = [1/(9 × 10⁹)](640)(0.24)²
Charge on the sphere (q) = 4 × 10⁻⁹ C
Charge density on the sphere = q / [4πr²]
Charge density on the sphere = [4 × 10⁻⁹] / [4(3.14)(0.12)²]
Charge density on the sphere = [4 × 10⁻⁹] / [0.18]
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
