(PLEASE answer ASAP, Will give brainliest)

2000, the Millennium Bridge, a new footbridge over the River Thames in London, England, was opened to the public. However, after

sweet [91]

Answer:

n = 1810

A = 25 mm

Explanation:

Given:

Lateral force amplitude, F = 25 N

Frequency, f = 1 Hz

mass of the bridge, m = 2000 kg/m

Span, L = 144 m

Amplitude of the oscillation, A = 75 mm = 0.075 m

time, t = 6T

now,

Amplitude as a function of time is given as:

$A(t)=A_oe^{\frac{-bt}{2m}}$

or amplitude for unforce oscillation

$\frac{A_o}{e}=A_oe^{\frac{-b(6T)}{2m}}$

or

$\frac{6bt}{2m}=1$

or

$b=\frac{m}{3T}$

Now, provided in the question Amplitude of the driven oscillation

$A=\frac{F_{max}}{\sqrt{(k-m\omega_d^2)+(b\omega_d^2)}}$

the value of the maximum amplitude is obtained $(k=m\omega_d^2)$

thus,

$A=\frac{F_{max}}{(b\omega_d}$

Now, for n people on the bridge

Fmax = nF

thus,

max amplitude

$0.075=\frac{nF}{((\frac{m}{3T})2\pi}$

or

n = 1810

hence, there were 1810 people on the bridge

b) $A=\frac{F_{max}}{(b\omega_d}$

since the effect of damping in the millenium bridge is 3 times

thus,

b=3b

therefore,

$A=\frac{F_{max}}{(3b\omega_d}$

or

$A=\frac{1}{(3}A_o$

or

$A=\frac{1}{(3}0.075$

or

A = 0.025 m = 25 mm

6 0

3 years ago

A 91 kg zebra is traveling 7 m/s east. What is the zebra’s momentum? kg-m/s

frez [133]

To find momentum you multiply the mass and velocity.
91 × 7 = 637 kg-m/s

5 0

3 years ago

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

4 years ago

Air at 207 kPa and 200◦C enters a 2.5-cm-ID tube at 6 m/s. The tube is constructed ofcopper with a thickness of 0.8 mm and a len

Serga [27]

Answer:

Temperature of air at exit = 24.32 C, After reducing hot air the temperature of the exit air becomes = 20.11 C

Explanation:

ρ = P/R(Ti) where ρ is the density of air at the entry, P is pressure of air at entrance, R is the gas constant, Ti is the temperature at entry

ρ = (2.07 x 10⁵)/(287)(473) = 1.525 kg/m³

Calculate the mass flow rate given by

m (flow rate) = (ρ x u(i) x A(i)) where u(i) is the speed of air, A(i) is the area of the tube (πr²) of the tube

m (flow rate) = 1.525 x (π x 0.0125²) x 6 = 4.491 x 10⁻³ kg/s

The Reynold's Number for the air inside the tube is given by

R(i) = (ρ x u(i) x d)/μ where d is the inner diameter of the tube and μ is the dynamic viscosity of air (found from the table at Temp = 473 K)

R(i) = (1.525) x (6) x 0.025/2.58 x 10⁻⁵ = 8866

Calculate the convection heat transfer Coefficient as

h(i) = (k/d)(R(i)^0.8)(Pr^0.3) where k is the thermal conductivity constant known from table and Pr is the Prandtl's Number which can also be found from the table at Temperature = 473 K

h(i) = (0.0383/0.025) x (8866^0.8) x (0.681^0.3) = 1965.1 W/m². C

The fluid temperature is given by T(f) = (T(i) + T(o))/2 where T(i) is the temperature of entry and T(o) is the temperature of air at exit

T(f) = (200 + 20)/2 = 110 C = 383 K

Now calculate the Reynold's Number and the Convection heat transfer Coefficient for the outside

R(o) = (μ∞ x do)/V(f) where μ∞ is the speed of the air outside, do is the outer diameter of the tube and V(f) is the kinematic viscosity which can be known from the table at temperature = 383 K

R(o) = (12 x 0.0266)/(25.15 x 10⁻⁶) = 12692

h(o) = K(f)/d(o)(0.193 x Ro^0.618)(∛Pr) where K(f) is the Thermal conductivity of air on the outside known from the table along with the Prandtl's Number (Pr) from the table at temperature = 383 K

h(o) = (0.0324/0.0266) x (0.193 x 12692^0.618) x (0.69^1/3) = 71.36 W/m². C

Calculate the overall heat transfer coefficient given by

U = 1/{(1/h(i)) + A(i)/(A(o) x h(o))} simplifying the equation we get

U = 1/{(1/h(i) + (πd(i)L)/(πd(o)L) x h(o)} = 1/{(1/h(i) + di/(d(o) x h(o))}

U = 1/{(1/1965.1) + 0.025/(0.0266 x 71.36)} = 73.1 W/m². C

Find out the minimum capacity rate by

C(min) = m (flow rate) x C(a) where C(a) is the specific heat of air known from the table at temperature = 473 K

C(min) = (4.491 x 10⁻³) x (1030) = 4.626 W/ C

hence the Number of Units Transferred may be calculated by

NTU = U x A(i)/C(min) = (73.1 x π x 0.025 x 3)/4.626 = 3.723

Calculate the effectiveness of heat ex-changer using

∈ = 1 - е^(-NTU) = 1 - e^(-3.723) = 0.976

Use the following equation to find the exit temperature of the air

(Ti - Te) = ∈(Ti - To) where Te is the exit temperature

(200 - Te) = (0.976) x (200 - 20)

Te = 24.32 C

The effect of reducing the hot air flow by half, we need to calculate a new value of Number of Units transferred followed by the new Effectiveness of heat ex-changer and finally the exit temperature under these new conditions.

Since the new NTU is half of the previous NTU we can say that

NTU (new) = 2 x NTU = 2 x 3.723 = 7.446

∈(new) = 1 - e^(-7.446) = 0.999

(200 - Te (new)) = (0.999) x (200 - 20)

Te (new) = 20.11 C

5 0

3 years ago

An Indy car is pushed down on the track by the Bernoulli effect. Where does the air flowing around the moving car travel the fas

GalinKa [24]

D
Reason: cause im built different way

6 0

3 years ago