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3 years ago

10

describe what is meant by "a constant change of direction".Identify whether the examples provided show a constant change of dire

ction when moving

1 answer:

liraira [26]3 years ago

7 0

"Constant change of direction" simply means on a curve.

The examples provided don't show a durn thing.

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

3 years ago

A student of weight 659 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude

Misha Larkins [42]

Answer:

a) The student feel light

b) Nbottom = 758 N

c) N'top= 236 N

d) N'bottom= 1055 N

Explanation:

a) W= 659N , Ntop= 560N

W > Ntop ---> Student feel less weight

b) Top:

∑F= W - Ntop = m.v²/R

m.v²/R = 659N - 560 N = 99 N

Bottom:

∑F= Nbottom- W = m.v²/R

Nbottom= W + m.v²/R = 659N + 99 N = 758N

c) W= 659 N , Ntop= 560 N , v'=2.v

N'top= ?

∑F= W - N'top = m.v'²/R

N'top= W - 4.m.v²/R

N'top = 659 N - 4. 99 N = 263 N

d) N'bottom = ?

∑Fbottom= N'bottom- W = m.v'²/R

N'bottom = W + 4.m.v²/R = 659 N + 4. 99 N = 1055 N

4 0

2 years ago

Water is traveling into what sphere during infiltration?

o-na [289]

When precipitation hits the ground and hydrosphere

5 0

3 years ago

The chemical substances in food that help maintain our body are known as what?

rodikova [14]

Answer:

Nutrients

Explanation:

4 0

3 years ago

Read 2 more answers

Two cars having different weights are traveling on a level surface at different constant velocities. Within the same time interv

GuDViN [60]

I think you're saying that once you start pushing on the cars, you want to be able to stop each one in the same time.

This is sneaky. At first, I thought it must be both 'c' and 'd'. But it's not
kinetic energy, for reasons I'm not ambitious enough to go into.
(And besides, there's no great honor awarded around here for explaining
why any given choice is NOT the answer.)

The answer is momentum.

Momentum is (mass x speed). Change in momentum is (force x time).

No matter the weight (mass) or speed of the car, the one with the greater
momentum is always the one that will require the greater (force x time)
to stop it. If the time is the same for any car, then more momentum
will always require more force.

4 0

3 years ago