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Elan Coil [88]
3 years ago
10

4. How does a molecule differ from an atom

Physics
2 answers:
tia_tia [17]3 years ago
8 0

Answer:

Explanation:

In a molecule, atoms are bonded together by single, double, or triple bonds. An atom has a nucleus surrounded by electrons. ... So another difference between atoms and molecules is that when similar atoms combine together in varying numbers, molecules of different properties can be formed.

Fudgin [204]3 years ago
7 0
An atom is much smaller than a molecule
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What energy is directly dependent upon velocity and mass?
erastova [34]

Answer:

Kinetic energy

Explanation:

3 0
2 years ago
Why it is important to have exact standards of measurement
Tamiku [17]

Answer:

reference against doubt

Explanation:

exact standards of measurement are a reference point for situations in doubt , a line against which to test and observe.

6 0
3 years ago
The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T
zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than -0.25 m/s^2

f) The two trains avoid collision if the acceleration of the freight train is at least 0.35 m/s^2

Explanation:

a)

We can describe the position of the passenger train at time t with the equation

x_p(t)=u_p t + \frac{1}{2}at^2

where

u_p = 25.0 m/s is the initial velocity of the passenger train

a=-0.100 m/s^2 is the deceleration of the train

On the other hand, the position of the freight train is given by

x_f(t)=x_0 + v_f t

where

x_0=200 m is the initial position of the freight train

v_f = 15.0 m/s is the constant velocity of the train

The collision occurs if the two trains meet, so

x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

b)

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

x_f(t)=x_0 +v_ft

And substituting t = 22.5 s, we find:

x_f(22.5)=200+(15)(22.5)=537.5 m

We can verify that the passenger train is at the same position at the time of the collision:

x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m

So, the two trains collide at x = 537.5 m.

c)

In the graph in attachment, the position-time graph of each train is represented. We have:

  • The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
  • The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

d)

In order to avoid the collision, the freight train must have a initial position of

x_0'

such that the two trains never meet.

We said that the two trains meet if:

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t

Re-arranging,

\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0

Substituting the values for the acceleration and the velocity,

0.05t^2-10t+x_0'=0

The solution of this equation is given by the formula

t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m

Therefore, the freight train must have a head start of 500 m.

e)

In this case, we want to find the acceleration a' of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t

Re-arranging

\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0

Substituting,

-0.5at^2-10t+200=0

The solution to this equation is

t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}

Again, the two trains never meet if the discriminant is negative, so

10^2-4\cdot (-0.5a') \cdot (200)

So, the deceleration must be smaller (towards negative value) than -0.25 m/s^2

f)

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2

where a_f is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2

Re-arranging,

\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0

And the solution is

t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}

Again, the two trains avoid collision if the discriminant is negative, so

10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
2 years ago
A ball is dropped from 8.5 meters above the ground. If it begins at rest, how long does it take to hit the ground?
Anna35 [415]

Answer:

Explanation:

Givens

d = 8.5 meters

vi = 0

a = 9.81

t = ?

Formula

d = vi * t + 1/2 a t^2

Solution

8.5 = 0 + 1/2 9.81 * t^2       multiply both sides by 2

8.5 = 4.095 t^2                  Divide both sides by 4.095

8.5/4.095 = t^2

1.7329 = t^2                       Take the square root of both sides

t = 1.316

It takes 1.316 seconds to hit the ground.

6 0
3 years ago
How would you describe the difference between momentum and angular momentum?
zheka24 [161]

Answer:

Analogous

Explanation:

Momentum is linear

Angular momentum is spinning

momentum is based on the mass and has no geometric component.

Angular momentum is based on moment of inertia and greatly depends on how far the mass is from the center of spin.

3 0
2 years ago
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