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3 years ago

At higher elevations, air pressure.. O increases decreases O stays the same

Physics

1 answer:

taurus [48]3 years ago

3 0

Answer:

decreases because there is not much air. when you are at a lower elevation, the air preassure is high and heavy.

Explanation:

hope I helped you. If I did please mark as brainliest

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What is the object’s velocity, in meters per second, at time t = 2.9? Calculate the object’s acceleration, in meters per second

madreJ [45]

Answer:

the question is incomplete, below is the complete question

"An object is undergoing simple harmonic motion along the x-axis. Its position is described as a function of time by x(t) = 5.5 cos(4.4t - 1.8), where x is in meters, the time, t, is in seconds, and the argument of the cosine is in radians.

a.What is the object's velocity, in meters per second, at time t = 2.9?

b.Calculate the object's acceleration, in meters per second squared, at time t = 2.9.

c. What is the magnitude of the object's maximum acceleration, in meters per second squared?

d.What is the magnitude of the object's maximum velocity, in meters per second?"

a. $v(t)==24.1m/s$

b. $a(t)=3.79m/s^{2}$

c. $a_{max}=106.48m/s^{2}$

d. $v_{max}=24.2m/s$

Explanation:

the gneral expression for the displacement of object in simple harmonic motion is represented by

$x(t)=Acos(wt- \alpha)\\$

while the velocity is express as

$v(t)=-Awcos(4.4t-1.8)\\$

and the acceleration is

$a(t)=-aw^{2}cos(wt- \alpha )\\$

Note: the angle is in radians

The expression for the displacement from the question is $x(t)=5.5cos(4.4t-1.8)\\$

comparing, A=5.5, w=4.4,α=1.8

a.To determine the object velocity at t=2.9secs,

we substitute for t in the velocity equation

$v(t)=-5.5*4.4sin(4.4*2.9-1.8)\\v(t)=-24.2sin(10.96)\\$

$v(t)=-24.2*(-0.9993)\\v(t)==24.1m/s$

b.To determine the object acceleration at t=2.9secs,

we substitute for t in the acceleration equation

$a(t)=-5.5*4.4^{2} cos(4.4*2.9-1.8)\\a(t)=-106.48cos(10.96)\\$

$a(t)=-106.48*0.0356\\a(t)=3.79m/s^{2}$

c. The acceleration is maximum when the displacement equals the amplitude. hence magnitude of the object acceleration is

$a_{max}=-w^{2}A\\ a_{max}=-4.4^{2}*5.5\\ a_{max}=106.48m/s^{2}$

d.The maximum velocity is expressed as

$v_{max}=wA\\v_{max}=4.4*5.5\\v_{max}=24.2m/s$

5 0

3 years ago

The physical quantity represented as rate of change of change in position in a

Oduvanchick [21]

Answer:

rate of change of its position with respect to a frame of reference, and is a function of time.

3 0

3 years ago

Two forces are acting on a 0.250 kg hockey puck as it slides along the ice. The first force has a magnitude of 0.340 N and point

jasenka [17]

Answer:

the answer is self explanition

Explanation:

The answer is _________

8 0

3 years ago

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt

Volgvan

Answer:

It takes you 32.27 seconds to travel 121 m using the speed ramp

Explanation:

Lets explain how to solve the problem

- The speed ramp has a length of 121 m and is moving at a speed of

2.2 m/s relative to the ground

- That means the speed of the ramp is 2.2 m/s

- You can cover the same distance in 78 seconds when walking on

the ground

Lets find your speed on the ground

Speed = Distance ÷ Time

The distance is 121 meters

The time is 78 seconds

Your speed on the ground = 121 ÷ 78 = 1.55 m/s

If you walk at the same rate with respect to the speed ramp that

you walk on the ground

That means you walk with speed 1.55 m/s and the ramp moves by

speed 2.2 m/s

So your speed using the ramp = 2.2 + 1.55 = 3.75 m/s

Now we want to find the time you will take to travel 121 meters using

the speed ramp

Time = Distance ÷ speed

Distance = 121 meters

Speed 3.75 m/s

Time = 121 ÷ 3.75 = 32.27 seconds

It takes you 32.27 seconds to travel 121 m using the speed ramp

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3 years ago

An object’s speed is determined from the object’s

shusha [124]

shape and yeah so you must find objects speed to determine the shape

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4 years ago