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umka2103 [35]
3 years ago
14

Look at the diagrams. Each model the arrangement of particles in a substance.

Physics
1 answer:
Kisachek [45]3 years ago
5 0

Answer:

a

Explanation:

because it has  compact  molecules  

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A beam of sunlight falling on a prism refracts and forms seven color bands. This illustrates that
ExtremeBDS [4]

That isn"t the right answer the correct answer is B.

4 0
3 years ago
Read 2 more answers
How to find the velocity of an object in a circular path?
Luda [366]
First of all, let's just talk about the speed, and not get wound up
in the velocity. OK ?

If a fly is sitting on the rim of the wheel and the wheel is rotating, then for
each full revolution of the wheel, the fly travels the circumference of the
wheel, which is (2 π) x (radius of the wheel).

In 'N' revolutions, the fly travels (2 N π) x (the radius). and so on.

So if the wheel is going, let's say 71 revs per minute (RPM), a point
on the rim is moving at (2 π times 71) x (the radius) per minute.

Another way to say it:

Speed of a point on the circle = (2 π) x (rotation frequency) x (radius).

The 'rotation frequency' takes care of the unit of time, and the 'radius'
takes care of the unit of length, so the result is a speed.
7 0
3 years ago
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During the last shot of the game, the basketball goes from rest to 15 m/s and reaches the backboard in 0.41 s. What was the acce
Anna [14]

Answer: a= 37m

Explanation: V= 15 m/s (Velocity) t= 0.41s (time) formula: a= v/t

15 m/s / 0.41 (15 divided by 0.41) = 36.583m

There are 2 significant digits, 36, you look at the third digit, either round up or down in this case up to 36. a= 37m

5 0
3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin
Mashcka [7]

Answer:

r_{cm}=[12.73,12.73]cm

Explanation:

The general equation to calculate the center of mass is:

r_{cm}=1/M*\int\limits {r} \, dm

Any differential of mass can be calculated as:

dm = \lambda*a*d\theta  Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}

So, the differential of mass is:

dm = \frac{2M}{a\pi}*a*d\theta

dm = \frac{2M}{\pi}*d\theta

Now we proceed to calculate X and Y coordinates of the center of mass separately:

X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta

Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta

Solving both integrals, we get:

X_{cm}=2*a/\pi=12.73cm

Y_{cm}=2*a/\pi=12.73cm

Therefore, the position of the center of mass is:

r_{cm}=[12.73,12.73]cm

5 0
3 years ago
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