1) Data:
Vo = 20 m/s
α = 37°
Yo = 0
Y = 3m
2) Questions: V at Y = 3m and X at Y = 3 m
3) Calculate components of the initial velocity
Vox = Vo * cos(37°) = 15.97 m/s
Voy = Vo * sin(37°) = 12.04 m/s
4) Formulas
Vx = constant = 15.97 m/s
X = Vx * t
Vy = Voy - g*t
Y = Yo + Voy * t - g (t^2) / 2
5) Calculate t when Y = 3m (first time)
Use g ≈ 9.8 m/s^2
3 = 12.04 * t - 4.9 t^2
=> 4.9 t^2 - 12.04t + 3 = 0
Use the quadratic equation to solve the equation
=> t = 0.28 s and t = 2.18s
First time => t = 0.28 s.
6) Calculate Vy when t = 0.28 s
Vy = 12.04 m/s - 9.8 * 0.28s = 9.3 m/s
7) Calculate V:
V = √ [ (Vx)^2 + (Vy)^2 ] = √[ (15.97m/s)^2 + (9.30 m/s)^2 ] = 18.48 m/s
tan(β) = Vy/Vx = 9.30 / 15.97 ≈ 0.582 => β ≈ arctan(0.582) ≈ 30°
Answer: V ≈ 18.5 m/s, with angle ≈ 30°
8) Calculate X at t = 0.28s
X = Vx * t = 15.97 m/s * 0.28s = 4,47m ≈ 4,5m
Answer: X ≈ 4,5 m
Answer:
15 V
Explanation:
From the question,
For series connection: (i) Both resistor have a common current flowing through the (ii) The combined resistance = R1+R2
Rt = R1+R2.................. Equation 1
Given: R1 = 5 ohms, R2 = 7.5 ohms.
Rt = 5+7.5 = 12.5 ohms.
Applying Ohm's law,
V = IRt................... Equation 2
Where V = Voltage, I = current.
make I The subject of the equation
I = V/Rt.............. Equation 3
Given: V = 25 V, Rt = 12.5 ohms.
Substitute into equation 3
I = 25/12.5
I = 2 A.
Now,
Voltage drop across the 7.5 ohms resistor = R2×I
Voltage drop across the 7.5 ohms resistor = 7.5×2
Voltage drop across the 7.5 ohms resistor = 15 V
Answer: 83.8 m/s
Explanation:
momentum of shell = momentum of cannon
M(s) × V(s) = M(c) × V(c)
M(s) = mass of shell, V(s) = velocity of shell
M(c) = mass of cannon, V(c) = velocity of cannon
100kg × 75m/s = 90kg × V(c)
7500 = 90 × V(c)
7500 ÷ 90 = V(c)
83.3 = V(c)
The answer is unbalanced because the numbers 100n and 300n don’t equal each other. If the answer was balanced the numbers would have to equal each other. Also, the answer net force would only be the answer if we were adding the numbers. I hope this helped you! :)
