What hormone does the pineal gland release

El tubo de entrada que suministra presión de aire para operar un gato hidráulico tiene 2 cm de diámetro. El pistón de salida es

Bess [88]

Answer:

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

Explanation:

Asumiendo que la presión ( $P$ ), medida en pascales, tiene una distribución uniforme sobre la superficie del pistón, se calcula a partir de la siguiente expresion:

$P = \frac{F}{A}$

Donde:

$F$ - Fuerza motriz, medida en newtons.

$A$ - Área del pistón, medida en metros cuadrados.

La fuerza motriz es equivalente al peso del automóvil. El área del pistón ( $A$ ), medido en metros cuadrados, es determinado por:

$A=\frac{\pi}{4}\cdot D^{2}$

Donde $D$ es el diámetro del pistón, medido en metros.

Si $D = 0.32\,m$ y $F =17,640\,N$ , entonces la presión neumática es:

$A = \frac{\pi}{4}\cdot (0.32\,m)^{2}$

$A \approx 0.080\,m^{2}$

$P = \frac{17,640\,N}{0.080\,m^{2}}$

$P = 220,500\,Pa$

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

8 0

3 years ago

What is the idea behind the law of multiple proportions?

Akimi4 [234]

Answer:

The law of multiple proportions states that when two elements can combine in different ratios to form different compounds, the masses of the element combining with the fixed mass of another element result in whole number ratios. This shows that the law of multiple proportions is followed

Explanation:

5 0

3 years ago

A tow truck drags a stalled car along a road. The chain makes an angle of 30???? with the road and the tension in the chain is 1

My name is Ann [436]

Answer: work = 1,305kJ

Explanation:

angle= 30°

force= 1,500N

distance= 1,000m

The formula for work is : Work= force x distance, however there is an angle of 30° between the direction of force applied and the direction of motion, therefore force must be decomposed to its value on the horizontal axis which is the direction of motion by using the cosine of the very angle.

W= F×cos(α)×D

W= 1,500×cos (30)×1,000

W= 1,305kJ ( kilojoules)

3 0

3 years ago

On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance

Tresset [83]

Answer B. 112 m

Step-by-Step Explanation

initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s
acceleration = (v - U) / t
= (36 - 20) / 4
a=4m/s2
from the formula
7-u2=2as , sis distance covered
putting the values
362-202=2×4×s
1296 - 400 = 8 x S
S= 112 m

4 0

2 years ago

Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r

cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

E_net = E_2 + E_4

E_2 = E_4

E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

E_net = 2*(99.518)*cos(27.12)

E_net = 177.151 N / C

5 0

3 years ago