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sdas [7]
3 years ago
6

The New England Merchants Bank Building in Boston is 152 mm high. On windy days it sways with a frequency of 0.20 HzHz , and the

acceleration of the top of the building can reach 2.5 %% of the free-fall acceleration, enough to cause discomfort for occupants.What is the total distance, side to side, that the top of the building moves during such an oscillation?
Physics
1 answer:
kkurt [141]3 years ago
3 0

Answer:

The total distance, side to side, that the top of the building moves during such an oscillation = 31 cm

Explanation:

Let the total side to side motion be 2A. Where A is maximum acceleration.

Now, we know know that equation for maximum acceleration is;

A = α(max) / [(2πf)^(2)]

So 2A = 2[α(max) / [(2πf)^(2)] ]

α(max) = (0.025 x 9.81) while frequency(f) from the question is 0.2Hz.

Therefore 2A = 2 [(0.025 x 9.81) / [((2π(0.2)) ^(2)] ] = 2( 0.245 / 1.58) = 0.31m or 31cm

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Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot
Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² =  2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ =  m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁  = 4.427 - 2v₂  ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂  = 8.854

v₂ = 8.854 / 3

v₂  = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁  = - 1.477 m/s

v₁  = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

mgh_{max} = \frac{1}{2}mv_{max}^2

(a) The maximum height achieved by the first ball (v₁  = 1.477 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max}  = 0.11 \ m

(b) The maximum height achieved by the second ball (v₂  = 2.95 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max}  = 0.44 \ m

6 0
3 years ago
During an earthquake, you should do all of the following EXCEPT
krok68 [10]
Run inside if you are outdoors .
7 0
4 years ago
A battery has an electric potential of 1.5V and transfers 10.0 C between the two terminals. How much work was done?
bogdanovich [222]

Answer:

15 Joules

Explanation:

work = charge x potential difference

= 10 x 1.5

= 15

8 0
3 years ago
When looking at a food label, to be considered a good source of a nutrient, the nutrient must be? A.More than 20% of your daily
aleksklad [387]

Answer:

A. More than 20% of your daily recommended amount.

Explanation:

Reading food labels can be tricky. The percent daily value listed on the right of all food labels lets you know what percent out of the recommended daily intake of each nutrient you are consuming in that specific food.

To check if the food you're consuming is a good source of that nutrient you need in higher amount, the nutrient must be labeled 20% or higher.

The rule used here is called the 5/20 rule. According to this rule, A nutrient that is 5% or below is considered less and a nutrient which is labeled 20% or higher is considered good enough in that food source.

8 0
3 years ago
A drone flies 8 m/s due East with respects to the wind. The wind is blowing 6 m/s due North with respects to the ground. What is
vekshin1

Answer:

B. 10m/s

Explanation:

If a drone flies 8 m/s due East with respects to the wind and the wind is blowing 6 m/s due North, the speed of the drone with respect to the ground is its displacement.

Displacement is calculated using Pythagoras theorem.

d² = 8²+6²

d² = 64+36

d² = 100

Square root both sides

√d² = √100

d = 10m/s

Hence the distance of the drone with respect to the ground is 10m/s

Option B is correct

8 0
3 years ago
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