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Blizzard [7]
3 years ago
12

The shanghai maglev train is capable of reaching speeds of up to 350km h-1 what is the speed in m/h

Physics
1 answer:
LenaWriter [7]3 years ago
8 0

Answer:

The shanghai maglev train is capable of reaching speeds of up to 217.48 in mph

217.48 mph

Explanation:

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xxMikexx [17]
The requirement is to change the °F to °C scale. That is;

T(°C) = [T(°F) - 32]/1.8

Substituting;

T(°C) = [428 - 32]/1.8 = 396/1.8 = 220 °C.

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4 years ago
023 (part 1 of 2) 10.0 points
Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

\Delta KE_{rotational} = \dfrac{1}{2}  \cdot I \cdot \omega ^2 = \dfrac{1}{2}  \cdot m \cdot v ^2

I  = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

\Delta KE_{rotational} = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

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What conversion takes place in a motor?
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C should be the right answer! hopefully this helps!
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The correct answer is Reverberation (C)
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The answer is B- the process of cells making new cells
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