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3 years ago

As Clinton walks he pushes his shoe against the track.

Physics

1 answer:

Valentin [98]3 years ago

4 0

Answer:A:The track pushes back on Clinton's shoe with the same force.

Explanation:According to Newton's third law of motion, for every action force there is an equal and opposite reaction force. In this case, the action force is Clinton's shoe pushing on the track. As this happens, there is an equal and opposite reaction force in which the track pushes back on Clinton's shoe with the same force.

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All simple machines are types of

taurus [48]

Answer: Metal. There are six of them. inclined plane, the wedge, the screw, the lever, the wheel and axle, and the pulley.

Hope this helps!

~Jarvis

Explanation:

6 0

3 years ago

The density of table sugar is 1.59 g/cm3. What is the volume of 7.85 g of sugar?

Likurg_2 [28]

$\frac{7.85g }{1.59 \frac{g}{ cm^{3} } }$ = 4.937 $cm^{3}$

4 0

3 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

D is diameter of the eye

$D = \frac{1.22 (539nm)}{5.11 mm}$

$D= 1.287*10^{-4}m$

The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

Therefore,

$L = \frac{d}{sin\theta}$

$L = \frac{0.691m}{1.287*10^{-4}}$

$L = 5367m$

Therefore the distance is 5.367km.

4 0

4 years ago

One object (m1 = 0.220 kg) is moving to the right with a speed of 2.10 m/s when it is struck from behind by another object (m2 =

blagie [28]

Answer:

vf₁ = 6.86 m/s , to the right

vf₂ = 2.96 m/s, to the right

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

p=m*v

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ :Initial linear momentum quantity

Pf : Final linear momentum quantity

Data

m₁= 0.220 kg : mass of object₁

m₂= 0.345 kg : mass of object₂

v₀₁ = 2.1 m/s ₁ , to the right : initial velocity of m₁

v₀₂= 6 m/s, to the right i :initial velocity of m₂

Problem development

We appy the formula (1):

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:

(0.22)*(2.1) + (0.345)*(6) = (0.22)*vf₁ +(0.345)*vf₂

2.532 = (0.22)*vf₁ +(0.345)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

$e= \frac{v_{f2}-v_{f1} }{v_{o1} -v_{o2} }$

1*(v₀₁ - v₀₂ ) = (vf₂ -vf₁)

(2.1 - 6 ) = (vf₂ -vf₁)

-3.9 = (vf₂ -vf₁)

vf₂ = vf₁ - 3.9

vf₂ = vf₁ - 3.9 Equation (2)

We replace Equation (2) in the Equation (1)

2.532 = (0.22)*vf₁ +(0.345)*( vf₁ - 3.9)

2.532 = (0.22)*vf₁ +(0.345)* (vf₁) -(0.345)( 3.9)

2.532 + 1.3455 = (0.565)*vf₁

3.8775 = (0.565)*vf₁

vf₁ = (3.8775) / (0.565)

vf₁ = 6.86 m/s, to the right

We replace vf₁ = 6.86 m/s in the Equation (2)

vf₂ = 6.86 - 3.9

vf₂ = 2.96 m/s, to the right

8 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:

$\theta_{max} =18.38^o$

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