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3 years ago

What’s the difference between atoms and molecules in a substance?

Physics

2 answers:

Lerok [7]3 years ago

8 0

Answer:

Atoms are single particles while molecules are particles made of two or more atoms bonded together.

Explanation:

Goryan [66]3 years ago

5 0

Answer:

Atoms are single neutral particles.

example: Ne, O

Molecules are neutral particles made of two or more atoms bonded together.

example:O2,HCl

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A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.

Kobotan [32]

Explanation:

The given data is as follows.

F = $5.75 \times 10^{-16} N$

q = $1.6 \times 10^{-19} C$

v = 385 m/s

$sin (63.9^{o})$ = 0.876

Now, we will calculate the magnitude of magnetic field as follows.

B = $\frac{F}{qv sin (\theta)}$

= $\frac{5.75 \times 10^{-16} N}{1.6 \times 10^{-19} C \times 385 m/s \times 0.876}$

= $0.01065 \times 10^{3}$ T

= 10.65 T

Thus, we can conclude that magnitude of the magnetic field is 10.65 T.

4 0

3 years ago

Science please help!<br>

IceJOKER [234]

Explanation:

spherical lenses which are curved outward are CONVEX lenses

7 0

3 years ago

Two competing models attempt to explain the motions and changing brightness of the planets: Ptolemy's geocentric model and Coper

AlekseyPX

Answer:

A. Geocentric: This model is Earth Centered . Retrograde motion is explained by epicycles .

B. Heliocentric: This model is Sun centered. Retrograde motion is explained by the orbital speeds of planets

C. Both geocentric and heliocentric: Epicycles and deferents help explain planetary motion . Planets move in circular orbits and with uniform motion . The brightness of a planet increases when the planet is closest to Earth.

Explanation:

The principle of the Ptolemy's geocentric model was developed on the assumption that the center of the universe is the Earth. On the other hand, the principle of the Copernicus' heliocentric model was based on the assumption that the center of the universe is the sun. However, both models have a common ideology on uniform circular motion and epicycles.

6 0

3 years ago

For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant

myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4

Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0

3 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

2 years ago