Ask question

Ask question

All categories

3 years ago

9

The displacement of a certain object is described by y(t) = 23 sin 5t, where t is measured in seconds. Compute its period and it

s oscillation frequency in rad/sec and in hertz

1 answer:

serious [3.7K]3 years ago

7 0

Answer:

5 rad/sec

0.796 Hz

1.256 seconds

Explanation:

Y(t) = 23sin5t ----1

Y(t) = aSin(w)t----2

w = 5 rad/sec

Then we get the Oscillation frequency from the formula below

w = 2πf

We make f to be the subject of this formula

f = w/2π

f = 5/2*3.24

f = 5/6.28

f = 0.796Hz

From the calculated frequency, we get the period = 1/f

= 1/0.796

= 1.256 seconds

You might be interested in

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0

3 years ago

A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det

Vinil7 [7]

Answer:

a) $T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)$ , b) $\dot Q_{H} = 26.875\,kW$

Explanation:

a) The Coefficient of Performance of the Carnot Heat Pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

After some algebraic handling, the temperature of the cold reservoir is determined:

$T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}$

$T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}} \right)$

$T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)$

$T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)$

b) The heating load provided by the heat pump is:

$\dot Q_{H} = COP_{HP}\cdot \dot W$

$\dot Q_{H} = (12.5)\cdot (2.15\,kW)$

$\dot Q_{H} = 26.875\,kW$

4 0

3 years ago

What type of foundation do engineers use for a small and light building and when the load of the building is borne by columns? A

ikadub [295]

Answer:

A.

Explanation:

Individual footings are the commonest, and they are often used if the load of the building is borne by columns. Typically, every column will have an own footing. The footing is usually only a rectangular or square pad of concrete on which the column is erected

8 0

3 years ago

A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f

dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14, pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, $head\ loss = \frac{f L V^2}{( 2 g D)}$

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

$V = \frac{Q}{A}$

$= \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s$

obtained Darcy friction factor

calculate Reynold number (Re) ,

$Re = \frac{\rho V D}{\mu}$

where, $\rho$ = density of water

$\mu$ = Dynamic viscosity of water at 15 degree C = 0.001 Ns/m2

so reynold number is

$Re = \frac{1000\times 2.015\times 0.318}{0.001}$

= 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is 0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

$HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}$

HL = 1.64 m

7 0

4 years ago

Consider a cubic workpiece of rigid perfect plastic material with side length lo. The cube is deformed plastically to the shape

Taya2010 [7]

Answer: ε₁+ε₂+ε₃ = 0

Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-

l₀l₀l₀=l₁l₂l₃

$\frac{lo*lo*lo}{l1*l2*l3}=1.0$

taking natural log on both sides

$ln(\frac{(lo*lo*lo)}{l1*l2*l3})=ln(1)$

Considering the logarithmic Laws of division and multiplication :

ln(AB) = ln(A)+ln(B)

ln(A/B) = ln(A)-ln(B)

$ln(\frac{(l1)}{lo})*ln(\frac{(l2)}{lo})*ln(\frac{(l3)}{lo}) = 0$

Use the image attached to see the definition of true strain defined as

ln(l1/1o)= ε₁

which then proves that ε₁+ε₂+ε₃ = 0

8 0

3 years ago