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sleet_krkn [62]
3 years ago
9

Use the following free body diagram to answer questions 8 and 9:

Physics
2 answers:
Nat2105 [25]3 years ago
5 0

Answer:

8 is c and 9 is b it is shown though the practice as god lol

nalin [4]3 years ago
3 0

Question 8

The net force acting on the cart is:

F = 150 N - 90 N

F = 60 N

Question 9

By the Newtons second law:

F = ma

Solving for a:

a = F/m

a = 60 N / 30 kg

a = 2 m/s²

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A Physicist on the riverbank, watching the canoe drift by at 1 m/s, will say that its kinetic energy is 9 Joules.

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Explanation:

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A certain thin lens is made of glass with refraction index ????lens=1.500. In air, where the index of refraction is 1.000, the l
son4ous [18]

Answer:

The focal length of the lens in ethyl alcohol is 41.07 cm.

Explanation:

Given that,

Refractive index of glass= 1.500

Refractive index of air= 1.000

Refractive index of ethyl alcohol = 1.360

Focal length = 11.5 cm

We need to calculate the focal length of the lens in ethyl alcohol

Using formula of focal length for glass air system

\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})

Using formula of focal length for glass ethyl alcohol system

\dfrac{1}{f'}=(n_{g}-n_{ethyl})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})

Divided equation (II) by (I)

\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}

Where, n_{g} = refractive index of glass

n_{a} = refractive index of air

n_{ethyl} = refractive index of ethyl

Put the value into the formula

\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}

\dfrac{f'}{11.5}=\dfrac{25}{7}

f'=\dfrac{25}{7}\times11.5

f'=41.07\ cm

Hence, The focal length of the lens in ethyl alcohol is 41.07 cm.

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