Answer:
18.63 N
Explanation:
Assuming that the sum of torques are equal
Στ = Iα
First wheel
Στ = 5 * 0.51 = 3 * (0.51)² * α
On making α subject of formula, we have
α = 2.55 / 0.7803
α = 3.27
If we make the α of each one equal to each other so that
5 / (3 * 0.51) = F2 / (3 * 1.9)
solve for F2 by making F2 the subject of the formula, we have
F2 = (3 * 1.9 * 5) / (3 * 0.51)
F2 = 28.5 / 1.53
F2 = 18.63 N
Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.
Answer:
The answer is "False"
Explanation:
The geologic time scale is the "schedule" for occasions in Earth history. It partitions time into named units of unique time called in descending order of duration "eons, eras, periods, epochs, and ages". The specification of those geologic time units depends on stratigraphy, which is the relationship and order of rock layers. The fossil structures that happen in the stones, nonetheless, give the central methods for setting up a geologic time scale, with the circumstance of the development and vanishing of far and wide species from the fossil record being used to outline the beginnings and endings of ages,, periods, and different stretches.
Geologic time is the broad time period involved by the geologic history of Earth. Formal geologic time starts toward the beginning of the Archean Eon (4.0 billion to 2.5 billion years back) and proceeds to the current day.
Answer:
Use Fc centripetal force as positive and W the weight as negative
N = m v^2 / R + m g
v^2 = (N - m g) R / m
v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2
v = 18.1 m/s
Note: N - m g is the net force producing the centripetal force
Given :
A box weighing 12,000 N is parked on a 36° slope.
To Find :
What will be the component of the weight parallel to the plane that balances friction.
Solution :
The component of that will be parallel to the plane to balance friction is :
Therefore, component of force to balance friction is F sin 36° .
Hence, this is the required solution.
The net force acting on the airplane is 25N.
Forces acting on the paper airplane when it is in the air:
- The forward force generated by the engine, propeller, or rotor is called thrust. It resists or defeats the drag force. It operates generally perpendicular to the longitudinal axis. However, as will be discussed later, this is not always the case.
- Drag is an airflow disruption generated by the wing, rotor, fuselage, and other projecting surfaces that causes a backward, decelerating force. Drag acts backward and perpendicular to the relative wind, opposing thrust.
- Weight is the total load carried by airplane, including the weight of the crew, fuel, and any cargo or baggage. Due to the influence of gravity, weight pulls the airplane downward.
- Lift—acts perpendicular to the flight path through the center of lift and opposes the weight's downward force. It is produced by the air's dynamic influence on the airfoil.
Given.
Weight of the paper airplane, F1 = 16N
The force of air resistance, F2 = 9N
Net force = F1 + F2
Net force = 25N
Thus, the net force acting on the airplane is 25N.
Learn more about the net force here:
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