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2 years ago

In the graph, which region shows nonuniform positive acceleration?

Physics

2 answers:

cricket20 [7]2 years ago

3 0

Answer: A.AB

Explanation:

This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.

As we can see in the attached image, the graph can be divided in four segments:

OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.

AB: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the <u>acceleration is nonuniform and positive.</u>

BC: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.

CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.

From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:

Segment AB

lesya692 [45]2 years ago

3 0

Answer:

AB plato users ;)

Explanation:

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2 years ago

A skateboarder shoots off a ramp with a velocity of 5.1 m/s, directed at an angle of 55° above the horizontal. The end of the ra

ExtremeBDS [4]

Answer

given,

initial velocity of skateboard = 5.1 m/s

angle above the horizontal = 55°

height of the ramp = 1 m

a) maximum height of projectile

$H = \dfrac{u^2sin^2 \theta}{2g}$

$H = \dfrac{5.1^2\times sin^2 55^0}{2\times 9.81}$

H = 0.889 m

the maximum height of the skateboard above the ground

= 1 + 0.889

= 1.889 m

b) time to reach the height

$t = \dfrac{u\ sin\theta}{g}$

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t = 0.426 s

horizontal distance = u cos θ × t

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3 years ago

Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and

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A) Acceleration: $a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2$

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

Acceleration is defined as:

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

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

$a_1 = \frac{11-8}{4}=0.75 m/s^2$ (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: $a_2 = 0$

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

$a=\frac{0-11}{7}=-1.57 m/s^2$

And the negative sign means the acceleration is south, opposite to the direction of motion.

In a uniformly accelerated motion, the displacement can be calculated as:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

$u = 0\\a = 0.75 m/s^2\\t=4 s$

So the displacement is

$s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m$

- For the second segment, we have

$u = 11 m/s\\a = 0\\t=15 s$

So the displacement is

$s_2 = (11)(15)+0=165 m$

- For the third segment, we have

$u = 11\\a = -1.57 m/s^2\\t=7 s$

So the displacement is

$s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m$

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

$v=\frac{209.5}{26}=8.06 m/s$ (north)

Learn more about accelerated motion:

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Answer:

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Explanation:

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Using Snell's Law we can write:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

$\implies \sin(\theta_1)=\frac{n_2}{n_1}\sin(\theta_2)$ ,

Let's approximate the index of refraction of the air (medium 1 in the picture) to 1.

We thus have:

$\sin(\theta_1)=n_2\sin(\theta_2)=1.333\sin(47)$

$\implies\theta_1=\arcsin[n_2\sin(\theta_2)]=\arcsin[1.333\sin(47)]\approx 77.13$ . Calling $\alpha$ the actual angle of elevation, we get from the picture that $\alpha=90-77.13=12.97$

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2 years ago

Why pressure above the surface is greater than in the air

nexus9112 [7]

Answer:

At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels. ... Since most of the atmosphere's molecules are held close to the earth's surface by the force of gravity, air pressure decreases rapidly at first, then more slowly at higher levels.

Explanation:

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