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2 years ago

12

The following figure shows a Ferris wheel that rotates 6 times each minute and has a diameter of 17.8. What is the centripetal a

cceleration of a rider?

1 answer:

Leokris [45]2 years ago

8 0

Answer:

106.8

Explanation:

If you multiply 6 and 17.8, you'll get 106.8.

If that's wrong, my apology's !

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What part of the hammer acts as the fulcrum when the hammer is used to remove a nail

Natali5045456 [20]

The end of it with the dent

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4 years ago

Tectonic plates are large segments of the earth's crust that move slowly. Suppose one such plate has an average speed of 6.0 cm

faltersainse [42]

Answer:

1.35×10⁻⁷ m

37.278 mi/My

Explanation:

Speed of the tectonic plate= 6 cm/yr

Converting to seconds

$6=\frac{6}{365.25\times 24\times 60\times 60}$

So in one second it will move

$\frac{6}{365.25\times 24\times 60\times 60}$

In 71 seconds

$71\times \frac{6}{365.25\times 24\times 60\times 60}=1.35\times 10^{-5}\ cm$

The tectonic plate will move 1.35×10⁻⁵ cm or 1.35×10⁻⁷ m

Convert to mi/My

1 cm = 6.213×10⁻⁶ mi

1 M = 10⁶ years

$6\times 6.213\times 10^{-6}\times 10^6=37.278\ mi/My$

Speed of the tectonic plate is 37.278 mi/My

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3 years ago

One earth day equals 24 hours.

Artyom0805 [142]

Answer:

Correct, is there another part to the question?

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3 years ago

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Which action will increase the gravitational force between two objects?

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The answer is D have a nice day!

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3 years ago

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A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago