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Pavlova-9 [17]
3 years ago
15

A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end

is allowed to fall. What is the angular acceleration as it is released
Physics
1 answer:
Natalka [10]3 years ago
5 0

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod  and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

Substituting the value of the variables, we have

mL²α/12 = mgL/2

Simplifying by dividing through by mL, we have

mL²α/12mL = mgL/2mL

Lα/12 = g/2

multiplying both sides by 12, we have

Lα/12 × 12 = g/2 × 12

αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

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Se necesita subir una carga de 500 kg (4900 N) a una altura de 1.5 m deslizándola sobre una rampa inclinada. ¿Qué longitud debe
marusya05 [52]

Answer:

4.22 m

Explanation:

Una rampa es una máquina que se utiliza para levantar un objeto con una fuerza menor a la que realmente necesitarías. Cuanto mayor sea la longitud de la rampa, menor será la magnitud de la fuerza necesaria para levantar el objeto.

Dado que:

altura de la rampa = 1.5 m, carga = 4900 N, fuerza aplicada = 1633.33 N.

La fórmula de la rampa se da como:

fuerza aplicada * longitud de la rampa = peso de la carga * altura de la rampa

1633.33 * longitud de la rampa = 4900 * 1.5

longitud de la rampa = 4900 * 1.5 / 1633.33

longitud de la rampa = 4.22 m

6 0
3 years ago
Ir
NemiM [27]

Answer:

what is your question because I seem to not see a question in this question....

3 0
2 years ago
You approach a railroad crossing with a cross buck sign that has no lights or gates. What should you do?
maxonik [38]
Stop and look to see if anything is coming 
6 0
2 years ago
A rotating light is located 13 feet from a wall. The light completes one rotation every 3 seconds. Find the rate at which the li
saveliy_v [14]

Answer:

29.2 ft/s

Explanation:

The distance of the light's projection on the wall

y = 13 tan θ

where θ is the light's angle from perpendicular to the wall.

The light completes one rotation every 3 seconds, that is, 2π in 3 seconds,

Angular speed = w = (2π/3)

w = (θ/t)

θ = wt = (2πt/3)

(dθ/dt) = (2π/3)

y = 13 tan θ

(dy/dt) = 13 sec² θ (dθ/dt)

(dy/dt) = 13 sec² θ (2π/3)

(dy/dt) = (26π/3) sec² θ

when θ = 15°

(dy/dt) = (26π/3) sec² (15°)

(dy/dt) = 29.2 ft/s

5 0
3 years ago
A barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a va
viktelen [127]

Answer:

(a). 14.4 lbf/in^2.

(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

Explanation:

So, from the question above we are given the following parameters which are going to help us in solving this particular Question;

=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"

=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."

With these knowledge, let us delve right into the solution;

(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].

The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.

(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.

Length of mercury = 27.8 in.

AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

5 0
3 years ago
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