All categories

4 years ago

Which statement is true regarding the "relationship merry-go-round"?

2 answers:

7 0

A common finding with couples who can’t connect or resolve their problems is that they reason out or feel that it’s the other individual who can’t connect, not them. It’s the other individual that has the problem and they remain to think "I’m right and you’re wrong." The answer is letter a. <span>In 40 percent of new weddings, one or both partners has been wedded no less than once before. Men are more probable to re-wed than women who've been wedded before.</span>

s344n2d4d5 [400]4 years ago

5 0

What was the correct answer?

You might be interested in

What parts are found in an electric generator

Len [333]

Here are a list of items I found.

some brushes

a armature

a permanent magnet

some slip rings

3 0

3 years ago

Read 2 more answers

a trolley of mass 20kg was originally at rest on a smoth horizontal surface. bu how much will it accelerate if a pulling force o

DochEvi [55]

Answer:

just divide 22 N by 20 kg to get the acceleration in m/s2

Explanation:

I hope this is right-

6 0

3 years ago

Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!

DedPeter [7]

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

v² - u² = 2a ∆x

where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

6 0

3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago

A skier is accelerating down a 30.0-degree hill at 3.80 m/s^2.

Bond [772]

Answer:

ax = -3.29[m/s²]

ay = -1.9[m/s²]

Explanation:

We must remember that acceleration is a vector and therefore has magnitude and direction.

In this case, it is accelerating downwards, therefore for a greater understanding we will make a diagram of said vector, this diagram is attached.

$a_{x}=-3.8*cos(30) = -3.29 [m/s^{2}]\\ a_{y}=-3.8*sin(30) = -1.9 [m/s^{2}]$

3 0

3 years ago