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katrin [286]
2 years ago
13

What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan

ts Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.22 Hz, and the acceleration of the top of the building can reach 1.5% of the free-fall acceleration, enough to cause discomfort for occupants. What is the total distance, side to side, that the top of the...
Physics
1 answer:
kramer2 years ago
5 0
I think the question should be the below:

<span>What is the total distance, side to side, that the top of the building moves during such an oscillation?
</span>
Answer is the below:

 <span>Acceleration .. a = (-) ω² x </span>
<span>(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) </span>

<span>x (max) = a(max) /ω² </span>

<span>x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)</span>
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A Nichrome wire 44 cm long and 0.30 mm in diameter is connected to a 3.1 V flashlight battery. What is the electric field inside
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Answer:

7.05 Volts/m

Explanation:

L = length of the Nichrome wire = 44 cm = 0.44 m

V = Potential difference across the end of the wire = battery voltage = 3.1 Volts

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2 years ago
A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a rope that goes over an ideal pulley. If the masses are
irga5000 [103]

Answer:

The time taken will be 0.553 seconds.

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We should start off by finding the force exerted by the rope on the 3kg weight in this case.

Weight of 5kg mass = 5 * 9.81 = 49.05 N

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The force acting upward on the 3kg mass will equal the weight of the 5kg mass. Thus the resultant force acting on the 3kg mass is:

Total force = 49.05 - 29.43 = 19.62 N (upwards)

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F = m * a

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4 0
3 years ago
You are in charge of designing a movie stunt for a new Fast and Furious film. In this scene a stunt person has to jump from a br
blondinia [14]

Answer:

103.5 meters

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Given that a stunt person has to jump from a bridge and land on a boat in the water 22.5 m below. The boat is cruising at a constant velocity of 48.3 m/s towards the bridge. The stunt person will jump up at 6.45 m/s as they leave the bridge.

The time the person will jump to a certain spot under the bridge can be calculated by using the formula below:

h = Ut + 1/2gt^2

since the person will fall under gravity, g = 9.8 m/s^2

Also, let assume that the person jump from rest, then, U = 0

Substitute h, U and g into the formula above

22.5 = 1/2 * 9.8 * t^2

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t^2 = 4.59

t = \sqrt{4.592}

t = 2.143 seconds

From definition of speed,

speed = distance /time

Given that the boat is cruising at a constant velocity of 48.3 m/s towards the bridge, substitute the speed and the time to get the distance.

48.3 = distance / 2.143

distance = 48.3 * 2.143

distance = 103.5 m

Therefore,  the boat should be 103.5m away from the bridge at the moment the stunt person jumps?

3 0
3 years ago
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