A 1200-ft equal-tangent crest vertical curve is currently designed for 50 mi/h. A civil engineering student contends that 60 mi/

1. τbiceps = +(Positive) 2. τforearm = -(Negative) 3. τball = -(Negative) Explanation: The attached figure illustrates the following: 1. For the biceps, τbiceps indicates that torque is calculated as Torque = r x F, where r and F are vectors. Here, r corresponds to the vector from the elbow to the biceps. In the figure, the force from the biceps is directed upwards. Applying the right-hand rule from r to F results in counterclockwise torque, which is considered positive (+). 2. The torque related to the weight of the forearm, τforearm, uses the same torque formula, with r being the vector from the elbow to the forearm. The weight acts downward, causing a clockwise torque that is negative (-). 3. Similarly, for the weight of the ball, τball, the downward force from the ball's weight generates a clockwise torque, which also registers as negative (-).

V - wind speed;
53° - 35° = 18°
v² = 55² + 40² - 2 · 55 · 40 · cos 18°
v² = 3025 + 1600 - 2 · 55 · 40 · 0.951
v² = 440.6
v = √440.6
v = 20.99 ≈ 21 m/s
Conclusion: The wind speed calculates to 21 m/s.  

Answer:

The duration, t = 3.53 seconds

Explanation:

The following information is provided:

The equation to calculate the time t is expressed as:

...... (1)

Where

s denotes the distance in feet

We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet

Substituting the value of s into equation (1) yields:

t = 3.53 seconds

Thus, the time taken by the object is 3.53 seconds, which provides the required answer.

Answer:

≈ 3077.34

Explanation:

To calculate the flux of F (vector field) across surface S, where

F(x,y,z) = y i − x j + k

and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We will evaluate the following integral:

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + k

The normal vector N is computed as

Where:

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, >. < 2v sin v, -2v cos v, u >

F(S(u,v)).N = 2uv + u

Thus

≈ 3077.34

Answer:

The period of the pendulum measuring 16 m is double that of the 4 m pendulum.

Explanation:

Recall that the period (T) of a pendulum with length (L) is defined by:

where "g" denotes the local gravitational acceleration.

Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:

Thus, the duration of the 16 m pendulum is two times that of the 4 m one.