Ask question

Ask question

All categories

3 years ago

12

¿conjunto de fuerzas aplicado sobre un objeto de manera que la linea de accion de todas las fuerzas convergan en un mismo punto?

, ¿ cual es el conjunto de esas fuerzas? . agradecería mucho si lo contestan porfis uwu

1 answer:

Naddika [18.5K]3 years ago

8 0

Answer:

Fuerzas Convergentes

Fuerza Resultante

You might be interested in

-34-r=-7(8r-3) what is the work and answer

NeX [460]

<span>-34-r=-7(8r-3)
- 34 - r = - 56r + 21
- r + 56r = 21 + 34
55r = 55
r = 55/55
r = 1</span>

5 0

3 years ago

The base of a rectangular fish tank measures 18 inches long by 10 inches wide. If it is filled with 1,080 cubic

Oxana [17]

Answer:

Volume = length x width x height

V = 18 x 8 x 12 = 1728 cm3 = 1728 ml = 1.729 liters

Step-by-step explanation:

5 0

3 years ago

The locations, given in polar coordinates, for two ships are (8 mi, 63º) and (8 mi, 123º). Find the distance between the two shi

PolarNik [594]

The arrangement forms an isosceles triangle with equal legs of 8 mi. The angle between the legs is,

123-63 = 60°. Therefore, the other two angles ares:

Angles = (180-60)/2 = 120/2 = 60°

It can therefore be noted that all angles are equal and thus the resulting triangle is actual an equilateral triangle and thus all the sides are equal.

It can be concluded that the ships are 8 miles apart.

8 0

3 years ago

Read 2 more answers

4xi + 5yi^8+ 6xi^3 + 2yi^4

Ad libitum [116K]

Answer:

The correct answer is,

$7y - 2ix$

5 0

3 years ago

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

$\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.$

3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

4 0

2 years ago