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katrin [286]
3 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:
kramer3 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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sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting ahorizontal force on it. The coefficient
lara31 [8.8K]

Answer:

2.5 x 10^{5} J

Explanation:

weight = 5,000 N

coefficient of friction = 0.05

distance = 1000 m

how much work is done by the dogs pulling the sledge

work done = force x coefficient of friction x distance

work done = 5000 x 0.05 x 1000 = 2.5 x 10^{5} J

6 0
3 years ago
If your weight is 120 pounds and your mass is 54 kilograms how would those values change if you were on the moon
BlackZzzverrR [31]
The gravitation acceleration on the moon is different than on Earth. It is 1.6 m/s^2. If you weigh 120 lbs, then you would multiply 120 pounds by the gravitational acceleration on the moon and then divide by the acceleration on Earth.

(120 lbs * 1.6) / 9.8 = 20 pounds.

The mass will always be the same no matter what planet you’re on, so it’s still 54 kg.
4 0
3 years ago
If you walk a distance of 8 blocks and then 3 blocks south from home, what is your position compared to home? What distance did
Harman [31]

The position compared to that of home is a reference to displacement, I believe.

Displacement = x total - x initial

So I believe the answer is 5 blocks due north (if you’re walking linearly from your home), unless the questions is referring to relative displacement, in which then you’d need to use the Pythagorean theorem to find the hypotenuse between both positions. And then you’d have to find theta for the degrees between the south direction and the other unmentioned direction. But I don’t think that’s the case.

Distance refers to x total and doesn’t care for direction, as this refers to a scalar quantity opposed to a vector. Thus the equation is just

d = x

So 8 blocks + 3 blocks = a distance of eleven blocks walked total

7 0
3 years ago
Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

3 0
3 years ago
Special relativity can be used to study an object in which frame of reference?
Tresset [83]
Special relativity led the path for general relativity; special relativity is in a sense a special application of the rules of general relativity. While general relativity is in position to tackle all of these problems, special relativity can tackle only problems in inertial frames. Inertial frame means that the frame of reference is inot accelerating. So, we disqualify answers A and D. However, remember that moving in a circle means that there is an acceleration, the centrifugal one, even if the speed does not change. Hence C is also incorrect.
The correct answer is B, since if there is no change in velocity, the frame does not accelerate and it is inertial.
3 0
3 years ago
Read 2 more answers
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