Answer:
28.5 m/s
18.22 m/s
Explanation:
h = 20 m, R = 20 m, theta = 53 degree
Let the speed of throwing is u and the speed with which it strikes the ground is v.
Horizontal distance, R = horizontal velocity x time
Let t be the time taken
20 = u Cos 53 x t
u t = 20/0.6 = 33.33 ..... (1)
Now use second equation of motion in vertical direction
h = u Sin 53 t - 1/2 g t^2
20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)
t = 1.17 s
Put in equation (1)
u = 33.33 / 1.17 = 28.5 m/s
Let v be the velocity just before striking the ground
vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s
vy = uSin 53 - 9.8 x 1.17
vy = 28.5 x 0.8 - 16.66
vy = 6.14 m/s
v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2
v = 18.22 m/s
If the object's <em>velocity is constant</em> ... (it's speed isn't changing AND it's moving in a straight line) ... then the net force on the object is zero.<em> (D)</em>
Either there are no forces at all acting on the object, OR there are forces on it but they're 'balanced' ... when you add up all of their sizes and directions, they just exactly cancel each other out, and they have the SAME EFFECT on the object as if there were no forces at all.
Answer:
i know the questin but i got to try and find it
Explanation:
Answer:
Impedance = 93.75 ohms
Current = 1.81 A
Explanation:
Resistance = R = 80 ohms
Inductance = L = 0.2 H
Inductive reactance = XL = 
= ωL = (2πf) L
= 2 (3.14) (60)(0.2) = 75.398 Ohms
Capacitive reactance = 1 / ωC = 1/(2πf)C = 1 / [(2π)(60)(0.1 × 10⁻3)]
= 26.526 Ohms
Impedance = Z = 
= 
= 93.747 ohms
Voltage = 
× 120 = 169.7056 V
Current = I = V ÷ R = (169.7056) ÷ 93,747 = 1.81 A
Answer:
I have no clue what's really going on I'm just here to get answer maybe I will just try to get an answer but I have no clue I'm sorry I am confused and dint really know what to do here.
