What will happen if our atmosphere was oxygen

How does velocity differ from speed? What changes in motion can result in a change in velocity?

olga55 [171]

-Velocity is the speed of any moving object in a given direction, whilst Speed is the rate of an object's ability to move.
-Velocity can change if the direction or time is changed, the basic equation of velocity is: v = d/t
v - velocity
d - distance
t - time
If one of these factors change, it affects the other.

Hope this answers the question!

4 0

3 years ago

Read 2 more answers

ustapha Jones is speeding on the interstate in his Ferrari at 231km/hr when he passes a police car at rest . If the cop accelera

Lena [83]

Answer:

461 km/h

Explanation:

In order to solve this problem we must first sketch a drawing of what the situation looks like so we can better visualize it. (See attached picture).

We have two situations there, the first one is Mustapha's car that is traveling at a constant speed of 231km/hr.

The second situation is the police that is accelerating from rest until he reaches Mustapha. (We are going to suppose the acceleration is constant and that he will not stop accelerating until he reaches Mustapha). He has an acceleration of 15m/ $s^{2}$ .

We want to find what the final velocity of the police is at the time he reaches Mustapha. From this we can imply that the displacement x will be the same for both particles.

So let's model the first situation.

The displacement of Mustapha can be found by using the following equation:

$V_{M}=\frac{x}{t}$

when solving the equation for the displacement x we get that it will be:

$x=V_{M}t$

Now let's model the displacement of the cop. Since the cop has a constant acceleration, we can model his displacement with the following formula.

$x=V_{0}t+\frac{1}{2}at^{2}$

Since the initial velocity of the cop is zero, we can get rid of that part of the equation leaving us with:

$x=\frac{1}{2}at^{2}$

We can now set both equations equal to each other so we get:

$\frac{1}{2}at^{2}=V_{M}t$

When solving this for t, we get that:

$t=\frac{2V_{M}}{a}$ (let's call this equation 1)

Now, we know the cop has constant acceleration, so we can model it with the following formula too:

$a=\frac{V_{f}-V_{0}}{t}$

since the initial velocity of the cop is zero, we can get rid of that here too, so we get the following formula:

$a=\frac{V_{f}}{t}$

when solving for the final velocity, we get that:

$V_{f}=at$ (let's call this equation 2)

when substituting equation 1 into equation 2 we get:

$V_{f}=a(\frac{2V_{M}}{a})$

we can now cancel a leaving us with:

$V_{f}=2V_{M}$

This tells us that the final velocity of the cop will not depend on his acceleration. (This is only if the acceleration is constant all the time) So we get that the final velocity of the cop is:

$V_{f}=2(231km/hr)$

so

$V_{f}=461km/hr$

which is our answer.

8 0

3 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

4 years ago

An object is placed 96.5 cm from a glass lens (n = 1.51) with one concave surface of radius 24 cm and one convex surface of radi

poizon [28]

Answer:

image is vertical at distance -203.62 cm

magnification is 2.110

Explanation:

given data

n = 1.51

distance u = 96.5 cm

concave radius r1 = 24 cm

convex radius r2 = 19.1 cm

to find out

final image distance and magnification

solution

we will apply here lens formula to find focal length f

1/f = n-1 ( 1/r1 - 1/r2) .......................1

put here all value

1/f = 1.51 -1 ( -1/24 + 1/19.1)

f = 183.43

so from lens formula

1/f = 1/v + 1/u .............................2

put here all value and find v

1/183.43 = 1/v + 1/96.5

so

v = −203.62 cm

so here image is vertical at distance -203.62 cm

and

magnification are = -v /u

magnification = 203.62 / 96.5

magnification is 2.110

3 0

4 years ago

20 kg who is running at a speed of 4.0 m/s jumps onto a stationary sled of mass 5.0 kg on a frozen lake. the speed at which the

Step2247 [10]

Here we can use momentum conservation as there is no external force on sled and child while he jump on sled

by momentum conservation equation

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v{2f}$

$20*4 + 5*0 = 20*v + 5 *v$

since sled and child both moves with same speed so here they both will have same final speed "v"

by solving above equation we will have

25 v = 80

v = 3.2 m/s

So they will move together with speed 3.2 m/s

6 0

3 years ago