The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2

Answer:

(a). The coefficient of permeability is .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

Insert the values into the formula

(a). We will compute the coefficient of permeability

Applying the formula for permeability

Where, Q=discharge

l = length

A = cross-sectional area

h=constant head that causes flow

Plugging the value into the formula

The coefficient of permeability measures as .

(c). To ascertain the discharge velocity during the testing phase

Utilizing the discharge velocity formula

Substituting the values into the equation

The discharge velocity during the test measures 0.0187 cm/s.

Thus, (a). The coefficient of permeability is .

(b). The seepage velocity computes at 0.0330 cm/s.

(c). The observed discharge velocity during the test equals 0.0187 cm/s.

Answer:

50.2 cm

Explanation:

We have the following data:

Height, h=3.5 m

Initial horizontal velocity,

Time, t=0.32 s

We need to determine how far the ball is from the ground after 0.32 s.

Initial vertical velocity,

Where

The complete question is;

A ski jumper descends a ramp and exits the ski track at a horizontal speed of 24 m/s. The slope at the landing site angles downwards at θ = 59◦. The acceleration due to gravity is 9.8 m/s².

What is the value of the relative angle φ at which the ski jumper makes contact with the slope? Provide the answer in degrees.

Answer:

14.08°

Explanation:

The time taken can be calculated using the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Next, the slope of the trajectory at the impact point is determined by;

tan α = V_y/V_x

Given V_x = 24 m/s

We will find V_y using;

v = gt

Therefore;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Consequently;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

La imagen del objeto distante se formará en el punto lejano o a 52.5 cm, así que la distancia del objeto u = infinito, la distancia de la imagen v = -52.5 cm, la longitud focal requerida = f. Usando la fórmula de la lente 1 / v - 1 / u = 1 / f, obtenemos f = -52.5 cm = -0.525 m. La potencia P = 1 / f = -1 / 0.525 = -1.90. Ahora, para el ojo con gafas, debemos encontrar el nuevo punto cercano.