Answer:
The height of the package when it was released was 1102.5 meters
Explanation:
A package is dropped from a hovering helicopter
Dropped means the initial velocity of the package is zero
The package takes 15 seconds to strike the ground
Ignore the air resistance
We need to know the height of the package when it was released
, where h is the height of the package
from the ground, u is the initial velocity, g is the acceleration of gravity
and t is the time
u = 0 , t = 15 seconds , g = 9.8 m/s²
h = 1102.5 meters
<em>The height of the package when it was released was 1102.5 meters</em>
Answer:
0.6 A
Explanation:
As the motor gears towards full speed, the voltage of the circuit tends to become the difference that exists between the line voltage and the that of the back emf. Remembering Ohm's law, we then apply it to get the final, lower current that is based on the reduced voltage Ef. We use the provided resistance in the question, that is 3 ohms.
Ef = (120 V) - (117 V) = 3 V
If = Ef/R = (3 V) / (5.0 Ω) = 0.6 A
The answer is D.<span>longitudinal</span>
Answer:
M L1 = m L2 torques must be zero around the fulcrum
M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg
Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.
Let V = the volume of the sample.
Then
W = ρVg
or
V = W/(ρg)
= (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
= 4.4972 x 10⁻⁵ m³
Answer: The volume is 44.972 x 10⁻⁶ m³
