2.3c. What is the frequency of a wave with a length of 12 m that travels at a speed of 36 m/s?

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

a)

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago

Max ran 100 feet in 10 seconds.

Ivanshal [37]

Answer:

for max :

100 feet in 10 secs

for molly :

60 feet in 5 secs = 120 feet in 10 secs

so, molly ran farther in the same time interval i.e. covered 120 feet where as Max covered 100 feet

Explanation:

brainliest plz

8 0

2 years ago

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Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light

professor190 [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$\theta_{max} =18.38^o$

b

New $n_{cladding} =1.491$

Explanation:

From the question we are told that

The refractive index of the core is $n_{core} = 1.497$

The refractive index of the cladding is $n_{cladding} = 1.421$

Generally according to Snell's law

$n_{core} * sin(90- \theta) = n_{cladding} * sin (90)$

Where $\theta_{max}$ is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection

$\theta_{max} = 90 - sin^{-1} [\frac{n_{cladding}}{n_{core}} ]$

$\theta_{max} = 90 - sin^{-1} [\frac{1.421}{1.497}} ]$

$\theta_{max} =18.38^o$

Given from the question the the largest angle is 5°

Generally the refraction index of the cladding is mathematically represented as

$n_{cladding} = n_{core} * sin (90 - 5)$

$n_{cladding} =1.491$

5 0

3 years ago

You and your friend are pushing a box at the same constant speed. You pushed 20 feet in 10 minutes, your friend was pushing for

anzhelika [568]

Answer:

60 N.C

Explanation:

The box will move in the direction of the push and pull with a force of 60 N. C.

7 0

2 years ago

QUICK PLZ I NEED TO TURN IT IN

Anni [7]

The answer i got is 133.3 but i’m not completely sure . I got that because to get velocity i’m pretty sure it’s distance over time. So in between 8 and 10 minutes it’s 9 minutes. I pretty much just did 1200 divided by 9.

8 0

3 years ago

Read 2 more answers