What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light i

Referencing the diagram below, we can deduce from the geometry that x = 2.5 - 0.55 = 1.95 m, leading to cos θ = 1.95/2.5 = 0.78. Therefore, θ = cos⁻¹ 0.78 = 38.74°. According to the free body diagram, the tension in the chain measures 450 N. Here, F denotes the centripetal force and W signifies Dee's weight. The tension's components are as follows: Horizontal component = 450 sin(38.74°) = 281.6 N, directed to the left, and Vertical component = 450 cos(38.74°) = 351.0 N, directed upward. Answers: Horizontal: 281.6, directed left. Vertical: 351.0 N, directed upward.

Answer:

Explanation:

a )

Each blade resembles a rod with its axis positioned near one end.

The moment of inertia for one blade is:

= 1/3 x m l²

where m stands for the mass of the blade

l represents the length of each blade.

Total moment of inertia for 3 blades is:

= 3 x  x m l²

ml²

2 )

Details provided include:

m = 5500 kg

l = 45 m

Substituting these values produces:

moment of inertia of one blade:

= 1/3 x 5500 x 45 x 45

= 37.125 x 10⁵ kg.m²

Moment of inertia for 3 blades:

= 3 x 37.125 x 10⁵ kg.m²

= 111.375 x 10⁵ kg.m²

c )

Angular momentum

= I x ω

I denotes the moment of inertia of the turbine

ω symbolizes angular velocity

ω = 2π f

f indicates the rotational frequency of the blades

d )

We have I = 111.375 x 10⁵ kg.m² (Calculated)

f = 11 rpm (revolutions per minute)

= 11 / 60 revolutions per second

ω = 2π f

=  2π x  11 / 60 rad / s

Calculating angular momentum yields

= I x ω

111.375 x 10⁵ kg.m² x  2π x  11 / 60 rad / s

= 128.23 x 10⁵  kgm² s⁻¹.

Answer:

The answer I think fits this question is D.

Explanation:

I hope this is helpful and accurate.

Answer:

4.05 m/s

Explanation:

We will express the varying velocities as vectors.

Newton moves northward at 3.90 m/s from Daniel's stationary position.

V_n = 3.9 j

Assuming Pauli runs relative to Daniel at velocity X.

The relative velocity of Newton as seen by Pauli will be

3.9 j - X

Given that

the relative velocity of Newton with respect to moving Pauli = 1.1 i (1.1 towards the east).

Thus,

3.9 j - X = 1.1 i

X = -1.1 i + 3.9 j.

Magnitude of X

X² = 1.1² + 3.9²

X = 4.05 m/s

Therefore, Pauli runs relative to Daniel at 4.05 m/s.

The direction will be west of north at an angle θ,

Tan θ = 1.1 / 3.9

Part A: -,+

The elevator descends, dictating the velocity's direction, which aligns with the movement's path. With the positive direction defined as upward, the velocity is negative. After the emergency button is pressed, the elevator begins to stop, indicating that the velocity is decreasing. Therefore, the acceleration is oriented in the opposite way to the velocity, making its sign +.

Part B: +, -

The ball ascends, which determines the positive direction of the velocity. Hence, the velocity is positive.

Once the ball is thrown, the only force acting on it is gravity, opposing the ball’s ascent and causing it to descend. This means the acceleration is directed downwards, resulting in a negative sign.

Part C: 0, -

Throughout the ball's trajectory from the moment it's thrown until it drops to the ground, gravity constantly exerts downward acceleration (-).

After the throw, the ball's velocity will decline due to gravity. When it reaches a velocity of 0, it achieves its peak height. At this specific moment, the ball begins to descend again under the influence of gravity. However, at the peak height, the ball's velocity is 0.