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Leni [432]
3 years ago
9

The carpal bones in the hands are an example of __________.

Physics
2 answers:
Olenka [21]3 years ago
6 0
The carpal bones in the hands are an example of __________.
Answer: gliding joints

<span>A gliding joint means a freely moving joint in which the articulations allow only gliding motions</span>
sasho [114]3 years ago
4 0

Answer:

A gliding joint means a freely moving joint in which the articulations allow only gliding motions

Explanation:

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the rock of 10 kg is falling near the Earth's surface.assume that g =10N/kg and the is no air resistance. what is the accelerati
Ivanshal [37]

Answer:

kya faltu sawal h repetitive g=10N/kg

5 0
3 years ago
a car travels at 15 m/s for 10 s. It then speeds up with a constant acceleration of 2.0 m/s? for 15 s. At the end of this time,
Firlakuza [10]

V = u + at where u is initial velocity (15 m/s), a is acceleration (2m/s^2) and t is time (15 seconds)

V = 15 + 2 X 15

V = 45 m/s

6 0
3 years ago
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
Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
jenyasd209 [6]

The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.

2. This is immediate from (2.2.7).

3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =

3 0
3 years ago
A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration
Rufina [12.5K]

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

F_f = \mu N

Where,

\mu = Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894\\

By condition of Balance the friction force must be equal to the total net force, that is to say

F_{net} = F_f

m_{total}a = \mu m_hg

(m_h+m_c)a = \mu*m_h*g

Re-arrange to find acceleration,

a= \frac{\mu*m_h*g}{(m_h+m_c)}

a = \frac{0.894*242*9.8}{(242+224)}

a = 4.54 m/s^2

Therefore the acceleration the horse can give is 4.54m/s^2

3 0
3 years ago
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