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2 years ago

Which processes occur during the second stage of technological design? Check all that apply.

Physics

1 answer:

Tema [17]2 years ago

7 0

Answer:

designing a solution

Explanation:

after knowing & identifying a problem we start putting a solution

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A 56.0kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.50 turns each

Arada [10]

Answer:

Net force = 129.4, Force as multiple of weight of her hand = 18.84

Explanation:

Given Data:

Total body weight = 56.0 kg ;

no. of turns = 2.5/second ;

hand to hand distance = 1.5m ;

weight of hand = 1.25% of body weight ;

Solution:

mass of hand = $\frac{1.25}{100}$ *56 = 0.7kg ;

radius = d/2 = 1.5/2 = 0.75m ;

Now we need to find velocity, as we know that velocity can be calculated by dividing distance by time

v = d/t = $\frac{2.5*2*3.14*0.75}{1}$ = 11.775 m/s or 12 m/s;

The formula to calculate force is given as

F = mv²/r = (0.7*11.775²)/0.75 = 129.4 N

To calculate force as multiple of weight on her hand, we need to calculate the gravitational force W on her hand first.

W = gm = 9.81 * 0.7 = 6.867 N

Now the wieght on her hand can be represented by

$\frac{Force_{net} }{weight of hand}$ = 129.4 / 6.867 = 18.84

3 0

4 years ago

Which law is described by the equation p1v1 = p2v2 ?

kupik [55]

Boyle's law is modeled by the equation p1v1=p2v2.

5 0

3 years ago

Read 2 more answers

Falling objects drop with an average acceleration of 9.8 m/s2. An arrow is shot with a velocity of 11.76 m/s straight down from

In-s [12.5K]

Answer:

3.8 secs

Explanation:

Parameters given:

Acceleration due to gravity, g = 9.8 $m/s^2$

Initial velocity, u = 11.76 m/s

Final velocity, v = 49 m/s

Using one of Newton's equations of linear motion, we have that:

$v = u + gt$

where t = time of flight of arrow

The sign is positive because the arrow is moving downward, in the same direction as gravitational force.

Therefore:

$49 = 11.76 + 9.8*t\\\\\\\49 - 11.76 = 9.8t\\\\=> 9.8t = 37.24\\\\\\t = \frac{37.24}{9.8} \\\\\\t = 3.8 secs$

The arrow was in flight for 3.8 secs

6 0

4 years ago

What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on

Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is $252.52\ m/s^2$ .

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = $\frac{2500}{9}\ m/s$

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

$\therefore 0=\frac{2500}{9} + a(1.1)$

$\Rightarrow a=-\frac{2500}{9(1.1)}$

$\therefore a = - 252.52 \ m/s^2$

Hence , the deceleration of the rocket is $252.52\ m/s^2$ .

To learn more about Attention here:

brainly.com/question/28500124

#SPJ4

6 0

2 years ago

An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass

gizmo_the_mogwai [7]

Answer:

Explanation:

First of all we shall find the velocity at equilibrium point of mass 1.2 kg .

It will be ω A , where ω is angular frequency and A is amplitude .

ω = √ ( k / m )

= √ (170 / 1.2 )

= 11.90 rad /s

amplitude A = .045 m

velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s

= .5355 m /s

At middle point , no force acts so we can apply law of conservation of momentum

m₁ v₁ = ( m₁ + m₂ ) v

1.2 x .5355 = ( 1.2 + .48 ) x v

v = .3825 m /s

= 38.25 cm /s

Let new amplitude be A₁ .

1/2 m v² = 1/2 k A₁²

( 1.2 + .48 ) x v² = 170 x A₁²

( 1.2 + .48 ) x .3825² = 170 x A₁²

A₁ = .0379 m

New amplitude is .0379 m

7 0

3 years ago