Which be behavior of light makes it possible for you to see a spectrum of colors in a spray of water on a sunny day?

Zina [86]

Answer:

The speed of the 8-ball is 2.125 m/s after the collision.

Explanation:

Law Of Conservation Of Linear Momentum

The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:

P=mv

If we have a system of masses, then the total momentum is the sum of all the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

When a collision occurs, the velocities change to v' and the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

In a system of two masses, the law of conservation of linear momentum is simplified to:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.

After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.

The above equation is solved for v1':

$\displaystyle v'_1=\frac{m_1v_1+m_2v_2-m_2v'_2}{m_1}$

$\displaystyle v'_1=\frac{0.16*0+0.17*2-0.17*0}{0.16}$

$\displaystyle v'_1=\frac{0.34}{0.16}$

$v'_1=2.125\ m/s$

The speed of the 8-ball is 2.125 m/s after the collision.

8 0

2 years ago

I dont know how to do question 3，4

dlinn [17]

Prior to the Cambrian, organisms with hard parts (eg. shells and bones) weren't present for most the time, and it's generally such hard parts that get to fossilize. There's also been more time for Precambrian deposits to get destroyed by natural processes, 

The supply of Precambrian fossils has actually greatly improved over the last few decades. The so called Ediacaran faunas have been found at many parts of the world, and would be worth looking into.hope it helps

7 0

3 years ago

Where can electromagnetic waves travel that mechanical waves do not

Slav-nsk [51]

Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate . this means that electromagnetic waves can travel not only through air and solid materials bit also through vacuum of space.

7 0

3 years ago

In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.43 km mark a

Novay_Z [31]

Answer:

Acceleration, a = 0.101 m/s²

Explanation:

Average speed = total distance / total time.

At the 7.43km mark, total distance = 7.43km or 7430m

Total time = 25 * 60 s = 1500s

Average speed = 7430m/1500s = 4.95m/s

He then covers (10 - 7.43)km = 2.57 km = 2570 m

in t = 27m43.6s - 25min = 2m43.6s = 163.6 s

Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.

From V = u + at; V = 4.95m/s + a *60s

Distance covered while accelerating is

s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²

Distance covered while at constant velocity, v after accelerating is

D = velocity * time

Where v = 4.95m/s + a*60s

D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²

Total distance covered after initial 7.43 km, S + D = 2570 m, so

2570 m = 297m + a*1800s² + 512.82m + a*6216s²

2570 = 809.82 + a*8016

a = 809.82m / 8016s² = 0.101 m/s²

8 0

2 years ago

The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the

xz_007 [3.2K]

Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First, convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

5 0

2 years ago