As per the question, the velocity of the airplane [v] = 660 miles per hour.
The total time taken by airplane [t] = 3.5 hours.
We are asked to determine the total distance travelled by the airplane during that period.
The distance covered [ S] by a body is the product of velocity with the time.
Mathematically distance covered = velocity × total time
S = v × t
= 660 miles/hour ×3.5 hours
= 2310 miles.
Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.
The answer would be Exosphere because, there are 3 main regions that circulate oxygen through the Earths system, which are the Biosphere, Atmosphere, and the Lithosphere.
Answer:
725.2 N
Explanation:
Since it is not stated the scale, the person or both accelerated or experience weightlessness, the net force acting on the bathroom scale is the weight of the person acting downward as the person stands on the scale .
Weight = mass of a body × acceleration due to gravity
= 74 kg × 9.8 m/s²
= 725.2 N
The product of (voltage) times (current, in Amperes) is POWER.
<span>1) The differential equation that models the RC circuit is :
(d/dt)V_capacitor </span>+ (V_capacitor/RC) = (V_source/<span>RC)</span>
<span>Where the time constant of the circuit is defined by the product of R*C
Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s
2) C<span>harge of the capacitor 1.57 time constants
1.57*(2.742) = 4.3048 s
The solution of the differential equation is
</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T) + </span>V_capac(∞)
Since the capacitor is initially uncharged V_capac(0) = 0
And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V
This means,
V_capac (t) = (-9V)e ^(-t /T) + 9V
The charge in a capacitor is defined as Q = C*V
Where C is the capacitance and V is the Voltage across
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V
V_capac (4.3048 s) = 7.1275 V
Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C
3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V
Q (∞) = 89.9mF*(9V) = 0.8091 C
