All categories

3 years ago

HELP I NEED THIS ANSWERED AS FAST AS POSSIBLE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics

1 answer:

WITCHER [35]3 years ago

8 0

Displacement is d

Vf² = Vi² + 2 g d

(-20²) = (+10²) + 2 (-9.8) d

-19.6 d = 300

d = -15.3 m

negative means lower

time is t

d = Vi t + 1/2 g t²

-15.3 = 10 t + (-4.9) t²

4.9 t² - 10 t -15.3 = 0

t = 3.06 s

Hope this helps -John

You might be interested in

Mariah trained for months leading up to the marathon, and she won. After seeing this, her friends and family would describe her

alexandr402 [8]

Answer:

Mariah trained for months leading up to the marathon and won. Her sole motivation was that after seeing her winning the marathon, her friends and family would call her as motivated and athletic. It means that Mariah wanted to fulfill her esteem needs.

Explanation:

According to Abraham Maslow:

Safety needs includes the personal security, the safety of health, resources and property. etc.

Physiological needs falls at the lowest level of basic needs. It includes food, water, rest. etc which are necessary for an individual's survival.

Esteem includes the need of respect and self-confidence.

Cognitive needs includes the desire of knowledge, to know things, to know what is happening and why is it happening around you.

In Mariah's case, she needed respect and motivation and thus she was trying to fulfill her esteem needs by winning the marathon.

8 0

3 years ago

Read 2 more answers

A person walks first at a constant speed of 5.50 m/s along a straight line from point A to point B and then back along the line

Sav [38]

Answer:

4.25 m/s

Explanation:

They walked the first distance at 5.50 m/s, then the same distance at 3 m/s.

Since the distances are equal, the average speed is simply the average of 5.50 and 3.

(5.50 + 3) / 2 = 4.25

Her average speed over the entire trip is 4.25 m/s.

8 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

How is energy transferred during the water cycle? Question 1 options: Water gains energy during evaporation and releases it duri

Margaret [11]

Answer:

Water gains energy during evaporation and releases it during condensation in the atmosphere

Explanation:

In the water cycle, heat energy is gained or lost by water as it undergoes various processes in the cycle.

In evaporation, water molecules gains energy because the molecules of water vibrate faster and become more energetic. Hence they are able to escape into the atmosphere from the surface of the liquid.

In condensation, the molecules of gaseous water looses energy and becomes liquid.

Hence, water gains energy during evaporation and releases it during condensation in the atmosphere.

8 0

2 years ago

Read 2 more answers

a 0.40kg soccer ball approaches a player horizontally with a velocity of 18m/s north. the player strikes the ball and causes it

denis23 [38]

Impulse = Ft = (m)(delta v)
delta v = change in velocity = velocity final - velocity initial.
= -22m/s - +18m/s = -40m/s.
mdeltav = (0.40kg)(-40m/s) = -16kgm/s or -16Ns.

4 0

3 years ago