A helicopter and a jet airplane must work against gravity in order to fly. Which one flys faster

A certain 48 V electric forklift can lift up to 7000 lb at a

Gwar [14]

Answer:

256.25 A

Explanation:

7 0

3 years ago

Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s

nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

$x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}$

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

$v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}$

with this value of vo you calculate the max time:

$t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s$

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0

3 years ago

The third one pls help

Nadya [2.5K]

Answer:

20 ms¯¹

Explanation:

3. Determination of the final velocity

From the question given above, the following data were obtained:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

Acceleration is simply defined as the change in velocity per unit time.

Mathematically, it can be expressed as:

Acceleration (a) = final velocity – Initial velocity / time

a = v – u / t

With the above formula, we can obtain the final velocity of the car as follow:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

a = v – u / t

5 = v – 0 / 4

5 = v / 4

Cross multiply

v = 5 × 4

v = 20 ms¯¹

Thus, the final velocity of the car is 20 ms¯¹

7 0

3 years ago

The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

sattari [20]

Answer:

k = 11,564 N / m, w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

let's apply the equilibrium condition at this point

Axis y

W_{y} - Fr = 0

Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

sin 46 = $W_{y}$ / W

W_{y} = W sin 46

we substitute

mg sin 46 = k y

k = mg / y sin 46

If the length of the bar is L

sin 46 = y / L

y = L sin46

we substitute

k = mg / L sin 46 sin 46

k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

k = 1.18 9.8 / 1

k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

Em₀ = U = mgy

end point. Point at 30º

$Em_{f}$ = K -Ke = ½ I w² - ½ k y²

em₀ = Em_{f}

mgy = ½ I w² - ½ k y²

w = √ (mgy + ½ ky²) 2 / I

the height by 30º

sin 30 = y / L

y = L sin 30

y = 0.5 m

the moment of inertia of a bar that rotates at one end is

I = ⅓ mL 2

I = ½ 1.18 12

I = 0.3933 kg m²

let's calculate

w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

w = 6.06 rad / s

7 0

3 years ago

A 3 mm inside diameter tube is placed in a fluid with a surface tension of 600 mN/m and density of 3.7 g/cm3. The contact angle

Aleks04 [339]

Answer: The height of the fluid rise is 0.01m

Explanation:

Using the equation

h = (2TcosѲ )/rpg

h= height of the fluid rise

diameter of the tube =3mm

radius of the tube= 3/2 =1.5mm=0.0015

T= surface tension = 600mN/m=0.6N/m

Ѳ = contact angle = $60^o$ C

p= density =3.7g/cm3= 3700kg/m3

g= acceleration due to gravity =9.8m/s2

h = ( 2*0.6*0.5)/(0.0015*3700*9.8)

h = 0.6/54.39

h= 0.01m

Therefore,the height of the fluid rise is 0.01m

8 0

3 years ago