Answer:
Net force exerted on the radio is 27.5 Newton.
Given:
Mass = 5.5 kg
Acceleration = 5
To find:
Force exerted on the radio = ?
Formula used:
F = ma
Where F = net force
m = mass
a = acceleration
Solution:
According to Newton's second law of motion,
F = ma
Where F = net force
m = mass
a = acceleration
F = 5.5 × 5
F = 27.5 Newton
Hence, Net force exerted on the radio is 27.5 Newton.
Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
Answer:
An aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²
Explanation:
Applying Bernoulli's equation, we determine the highest pressure on the aircraft.
where;
P is the highest pressure on the aircraft
is the density of air = 1.204 kg/m³ at sea level temperature.
V is the velocity of the aircraft = 220 m/s
P = 0.5*1.204*(220)² = 29136.8 N/m²
Therefore, an aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²