Cuantos CM son: 8 newtons

La resultante de dos fuerzas perpendiculares aplicadas a un mismo cuerpo es 11.18 N y el módulo de una de ellas es de 10 N. ¿Cuá

Dmitrij [34]

Answer:

English?

Explanation:

3 0

3 years ago

A bar magnet is doplar. This means is has ___ poles. A. Two B. Zero C. Three D. One

Mandarinka [93]

This means it has TWO poles

The 2 poles are NORTH and SOUTH

8 0

2 years ago

The acceleration due to gravity on Earth is 9. 8 m/s2. What is the weight of a 75 kg person on Earth? 9. 8 N 75 N 85 N 735 N.

Illusion [34]

The weight of a 75 kg person on Earth, which has an acceleration due to gravity of 9.8m/s², is 735N.

CALCULATE WEIGHT:

Weight is a force, hence, measured in Newton. It is a function of an object's mass in relation to gravity.

Weight of an object can be calculated using the following formula:

W = mg

Where;

W = weight (N)
m = mass (kg)
g = acceleration due to gravity (m/s²)

The weight of the person in this question can be calculated as follows:

W = 75 × 9.8

W = 735N

Therefore, the weight of a 75 kg person on Earth, which has an acceleration due to gravity of 9.8m/s², is 735N.

Learn more about weight at: brainly.com/question/18554478

3 0

2 years ago

A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the

Shalnov [3]

Answer:

x ’= 1,735 m, measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

x_{cm} = 1.2 -1

x_ {cm} = 0.2 m

Σ τ = 0

w₁ 1.2 + mg 0.2 - W₂ x = 0

x = $\frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}$

x = $\frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}$

let's calculate

x = $\frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}$ 2.9 1.2 + 4 0.2 / 8

x = 0.535 m

measured from the pivot point

measured from the far left is

x’= 1,2 + x

x'= 1.2 + 0.535

x ’= 1,735 m

8 0

2 years ago

Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such

Lyrx [107]

This question is incomplete, the complete question is;

- Calculate the difference in blood pressure between the feet and top of the head of a person who is 1.80m Tall

- Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head

Answer:

- the difference in blood pressure is 18698.4 Pa

- the additional outward force F is 4.86 N

Explanation:

Given the data in the question;

we know that the expression for difference in blood pressure is;

ΔP = pgh

where p is density = 1060 kg/m³

g is acceleration due to gravity = 9.8 m/s²

and h is height = 1.80 m

now we substitute

ΔP = 1060 × 9.8 × 1.80

ΔP = 18698.4 Pa

therefore the difference in blood pressure is 18698.4 Pa

Also given that;

diameter of blood vessel d = 3.10 mm

radius r = 3.10 mm / 2 = 1.55 mm = 0.00155 m

length l = 2.70 cm = 0.027 m

Surface area of the cylindrical segment of a blood vessel is

A = 2πrl

we substitute

A = 2 × π × 0.00155 × 0.027

A = 2.6 × 10⁻⁴ m²

so

the required for will be;

F = PA

we substitute

F = 18698.4 Pa × 2.6 × 10⁻⁴ m²

F = 4.86 N

Therefore, the additional outward force F is 4.86 N

5 0

3 years ago

Cuantos CM son:8 newtons