What is universal laws of gravitation
1 answer:
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Answer:
a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m
Explanation:
a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.
So y - y₀ = ut - 1/2gt²
y - y₀ = (v₀sinθ)t - 1/2gt²
substituting the values of the variables into the equation, we have
0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²
- 60 = 50t - 4.9t²
So, 4.9t² - 50t - 60 = 0
Using the quadratic formula to find t,
Since t cannot be negative, t = 11.29 s
b. We first need to find the impact vertical velocity component. Using
v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,
v = v₀sinθ - gt
= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s
= 50.01 m/s - 110.64 m/s
= -60.63 m/s
Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.
The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)
= √((-60.63 m/s)² + (72.77 m/s)²)
= √((3676 m²/s² + 5295.48 m²/s²)
= √(8971.48 m²/s²
= 94.72 m/s
The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°
So the impact velocity is 94.72 m/s at -39.8°
c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m
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Reference the diagram below for clarification.
1.
We must begin by knowing the following rules for resistors in series and parallel.
In series:
In parallel:
We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.
Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
2.
We can use Ohm's law to solve for the current in the circuit.
3.
For resistors in series, both resistors receive the SAME current.
Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.
In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.
4.
Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.
