Answer:
<em>The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.</em>
Explanation:
First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. <em>Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight</em>
Answer:
(a) 693.12 torr
(b) 68.5 kilopascals
(c) 0.862 atmosphere
(d) 1.306 atmospheres
(e) 36.74 psi
Explanation:
(a) 0.912 atm = 0.912 atm × 760 torr/1 atm = 693.12 torr
(b) 0.685 bar = 0.685 bar × 100 kPa/1 bar = 68.5 kPa
(c) 655 mmHg = 655 mmHg × 1 atm/760 mmHg = 0.862 atm
(d) 1.323×10^5 Pa = 1.323×10^5 Pa × 1 atm/1.01325×10^5 Pa = 1.306 atm
(e) 2.50 atm = 2.50 atm × 14.696 psi/1 atm = 36.74 psi
Answer:
118800 seconds
Explanation:
Given :
Voltage, V = 1.2 V
Resistance, R = 22 Ω
Applying Ohm's law, we get
Voltage, V = IR
Current 
I = 0.0545 A
Rate = 1800 mAh
Time taken, 
= 33 hr
= 118800 s
Answer:
Minimum capacitance = 200 μF
Explanation:
From image B attached, we can calculate the current flowing through the capacitors.
Thus;
Since V=IR; I = V/R = 5/500 = 0.01 A
Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V
So minimum capacitance will be determined from;
i(t) = C(dv/dt)
So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]
C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF
