The projectile (ball) reaches an instantaneous vertical speed (Vy) of zero at maximum height.
so, V(max height) = ¬Г(Vx)^2+(Vy)^2
in this case V(max height) = Vx, where Vy=0
The maximum height, Yf, can be solved using Vfy^2=Viy^2 + 2gy. At maximum height Vfy=0.
At a definite point, the bridge would begin oscillating to the matching rhythm as that
of the marching footsteps.
This oscillation would touch a determined peak when the bridge can
no longer tolerate its own
power and later collapses. So, soldiers are
systematic to break their steps
while passing a bridge.
Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
Answer:
a = -2.4 m/s²
Explanation:
Given,
The initial speed of the bus, u = 24 m/s
The final speed of bus, v = 12 m/s
Time taken to reach final speed is, t = 5.0 s
The acceleration of the body is given by the change in velocity by time
a = (v - u) / t
= (12 - 24) / 5
= -2.4 m/s²
The negative sign in the acceleration indicates that the bus is decelerating.
Therefore, the acceleration of the bus is, a = -2.4 m/s²
