What element in the third period is a metalloid?

A 70 kg man's arm, including the hand, can be modeled as BIO 75-cm-long uniform rod with a mass of 3.5 kg. When the man raises b

Butoxors [25]

Answer:

Δy = 7.1 cm

Explanation:

The center of mass of a body is defined

$y_{cm}$ = 1 /M ∑ $m_{i} y_{i}$ i

Where M is the total mass of the body, m mass of each part and ‘y’ height

Let's apply this equation to our case

We locate the reference system on the shoulders

The height of the arms is at its midpoint

y = -75/2 = 37.5 cm

With arms down

$y_{cm}$ = 1/70 (63 y₀ - 3.5 37.5 - 3.5 37.5)

$y_{cm}$ = 1/70 (63 y)₀ - 7 37.5)

With arms up

$y_{cm}$ ’= 1/70 (63 y₀ + 3.5 y + 3.5 y)

$y_{cm}$ ’= 1/70 (63y₀ + 7 35.5)

let's subtract the two equations

$y_{cm}$ ’ - $y_{cm}$ = 1/70 2 (7 35.5)

Δy = $y_{cm}$ ’ - $y_{cm}$ = 2 7 35.5 / 70

ΔY = 7.1 cm

7 0

2 years ago

A 1-kg block is lifted vertically 1 m at constant speed by a boy. The work done by the boy is about:

Igoryamba

Answer:

$9.8\; {\rm J}$ , assuming that the gravitational field strength is $g = 9.8\; {\rm N \cdot kg^{-1}}$ .

Explanation:

Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.

By Newton's Second Law, the net force on this block would be $0$ . External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.

Let $m$ denote the mass of this block. It is given that $m = 1\; {\rm kg}$ . The weight of this block would be:

$\begin{aligned}\text{weight} &= m\, g \\ &= 1\; {\rm kg} \times 9.8\; {\rm N \cdot kg^{-1}} \\ &= 9.8\; {\rm N}\end{aligned}$ .

Hence, the force that the boy applies on this block would be upward with a magnitude of $F = 9.8\; {\rm N}$ .

The mechanical work that a force did is equal to the product of:

the magnitude of the force, and
the displacement of the object in the direction of the force.

The displacement of this block (upward by $s = 1\; {\rm m}$ ) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:

the magnitude of the force that this boy exerted, $F = 9.8\; {\rm N}$ , and
the displacement of this block in the direction, $s = 1\; {\rm m}$ .

$\begin{aligned}\text{work} &= F\, s \\ &= 9.8\; {\rm N} \times 1\; {\rm m} \\ &= 9.8\; {\rm J}\end{aligned}$ .

5 0

1 year ago

Which statement is true for objects in dynamic equilibrium?

lakkis [162]

Answer:

The correct answer to the question is objects have zero acceleration.

Explanation:

Before answering the question, first we have to understand dynamic equilibrium .

A body moving with uniform velocity is said to be in dynamic equilibrium if the net external forces acting on the body is zero. Hence, the body is under balanced forces.

If the external forces acting on a body is not balanced, then the body will accelerate which will destroy its equilibrium condition. Hence, the necessary and sufficient condition for a body to be in dynamic equilibrium is that the forces are balanced.

When a body is in dynamic equilibrium, the body moves with uniform velocity along a straight line unless and until it is compelled by some external unbalanced forces.

Hence, the rate of change of velocity or acceleration of the body will be zero.

5 0

3 years ago

Read 2 more answers

The Moon orbits Earth in an average of p = 27.3 days at an average distance of a =384,000 kilometers. Using Newton’s version of

Korvikt [17]

Answer:

The mass of the earth, $M=6.023\times 10^{24}\ kg$