1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Volgvan
3 years ago
8

What is the average speed of particles of atoms at room temperature?

Physics
1 answer:
Paha777 [63]3 years ago
8 0

Answer:

300 meters per second. That's equal to about 670 miles per hour.

Explanation:

Not only are air particles incredibly small, they are always moving. And they move fast. At room temperature, they are going about 300 meters per second. That's equal to about 670 miles per hour.

You might be interested in
A proton is a subatomic particle that carries a ___________ charge
Dovator [93]

The answer would be a positive charge

4 0
3 years ago
The ______ is between the mesosphere and the exosphere. troposphere stratosphere thermosphere exosphere
Aliun [14]

Answer:

The thermosphere

4 0
3 years ago
Read 2 more answers
Dr. John Paul Stapp was U.S. Air Force officer whostudied the effects of extreme deceleration on thehuman body. On December 10,
Ira Lisetskai [31]

Answer:

(a) a = 56.4 m/s², his acceleration a, in multiples of gravity g, is 5.76 g

(b) a = -201.43 m/s², his deceleration -a, in multiples of gravity g, is -20.56 g

Explanation:

(a)

When moving upwards, the initial velocity, u = 0 (he accelerated from rest)

When moving upwards, the final velocity, v = 282 m/s

time of  motion during this acceleration, t = 5 s

His acceleration is calculated as;

v = u + at

282 = 0 + 5a

a = 282 / 5

a = 56.4 m/s²

Ratio of his acceleration, a to gravity, g = a/g = 56.4 / 9.8 = 5.76

a = 5.76 g

(b)

When moving downwards, the initial velocity, u = 282 m/s

When moving downwards, the final velocity, v = 0 (he was brought to rest)

time of  motion during this deceleration, t = 1.4 s

His deceleration is calculated as;

v = u + at

0 = 282 + 1.4a

1.4a = -282

a = -282 / 1.4

a = -201.43 m/s²

Ratio of his deceleration, -a to gravity, g = -a/g = 201.43 / 9.8 = 20.56

a = -20.56 g

3 0
3 years ago
The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 50 km/
velikii [3]

To solve this problem, you must figure out (in vector form) both the wind vector and plane vector

w⃗ = wind vector

P⃗ = plane vector

 To get the  true course of the plane, you need to add the plane and wind vectors, the formula would be

w⃗ +P⃗ ,

 which will result to the ground speed.

ground speed=||w⃗ +P⃗ ||

 

Using the planar representation of your situation, this will help you understand the equation, use this to make the equation more understandable.

w⃗ =AB¯¯¯¯¯¯¯¯,   P⃗ =AC¯¯¯¯¯¯¯¯

 

the smaller circle is of radius 50 (similar to the wind speed) and the larger circle is of radius 200 (similar to the plane vector.  To get the coordinates of these two vectors,  use polar coordinates.

Let East be 0 degrees, so since the wind vector is on the circle of radius 50, we have:

w⃗ =⟨50cos(135),50sin(135)⟩=⟨−252√, 252√⟩.

P⃗ =⟨200cos(60),200sin(60)⟩=⟨100,\1003√⟩.

w⃗ +P⃗ =⟨100−252√ , 1003√+252√⟩

||w⃗ +P⃗ ||=(100−252√)2+(1003√+252√)2

√≈218.349218.

 

<span> </span>

8 0
3 years ago
A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif
Serggg [28]

Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

v(t)=-gt-v_e\times \ln \frac{m-rt}{m}

v = velocity of rocket at time t

g = Acceleration due to gravity =9.8 m/s^2

v_e = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})

v(60) = 668.97 m/s

Height of the rocket = h

Velocity=\frac{Displacement}{time}

668.97 m/s=\frac{h}{60 s}

h=668.97 m/s\times 60 s=40,138.2 m = 40.1382 km

Height of the rocket be one minute after liftoff is 40.1382 km.

4 0
3 years ago
Other questions:
  • Calculate the linear acceleration (in m/s2) of a car, the 0.330 m radius tires of which have an angular acceleration of 13.0 rad
    14·1 answer
  • Which planet orbits in a different plane than all of the others?
    6·1 answer
  • 5. Two negative charges that are both - 3.8 C push each other apart with a force of 19.0 N. How far apart
    5·1 answer
  • Which is a warning sign that someone needs the help of a mental health professional
    11·2 answers
  • Explain constant and changing velocity.
    5·2 answers
  • 17. If a good is a luxury, what type of response will it cause?
    10·1 answer
  • Is the system in the image moving? Explain. If it is moving, find its acceleration.
    6·1 answer
  • 1. Using the data table below, model (draw) the orbital motion of the two planets (D and
    7·1 answer
  • A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,
    14·1 answer
  • . A girl of mass 35 kg and boy of mass 25 kg are connected by a light rope. They move horizontally in a skating rink. A second r
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!