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3 years ago

True or False? Mechanical waves do NOT need a medium to travel through.

Physics

2 answers:

motikmotik3 years ago

8 0

Answer:

True

Explanation:

Mechanical waves require a medium in order to transport their energy from one location to another.

stellarik [79]3 years ago

6 0

Answer:

False.

Explanation:

Mechanical waves require a medium in order to transport their energy from one location to another.

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A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with neglig

Blababa [14]

Answer:

<u>Please Mark As Brainliest!!</u>

a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a) If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf ⇒ I₁ * ω₁ = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)

W = 7.03 J

4 0

3 years ago

A student is throwing a ball. She throws the ball up with her left hand, it passes over her head, and she catches it with her ri

pychu [463]

Answer:

(c) at point 2, the ball is at its highest height do its PE is max. Also at ms height, velocity is zero therefore KE is zero.

8 0

3 years ago

A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformati

Slav-nsk [51]

The question is incomplete. The complete question is :

A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformation cycle. At a time, t = 0, a tensile stress of 20 MPa is applied instantaneously and maintained for 100 s. The stress is then removed at a rate of 0.2 MPa s−1 until the polymer is unloaded. If the creep compliance of the material is given by:

J(t) = Jo (1 - exp (-t/to))

Where,

Jo= 3m^2/ GPA

to= 200s

Determine

a) the strain after 100's (before stress is reversed)

b) the residual strain when stress falls to zero.

Answer:

a)-60GPA

b) 0

Explanation:

Given t= 0,

σ = 20Mpa

Change in σ= 0.2Mpas^-1

For creep compliance material,

J(t) = Jo (1 - exp (-t/to))

J(t) = 3 (1 - exp (-0/100))= 3m^2/Gpa

a) t= 100s

E(t)= ΔσJ (t - Jo)

= 0.2 × 3 ( 100 - 200 )

= 0.6 (-100)

= - 60 GPA

Residual strain, σ= 0

E(t)= Jσ (Jo) ∫t (t - Jo) dt

3 × 0 × 200 ∫t (t - Jo) dt

E(t) = 0

5 0

3 years ago

How does the structure of the stigma aid in pollination

natali 33 [55]

<span>Pollination is the process by which pollen is transferred to the female part of a plant. The stigma is the central part of the flower, that is supported by a style and is part of the female reproductive organ within plants. Its structure is optimised to promote pollination by having hairs, sticky surfaces and three-dimensional sculptures that capture and trap pollen.</span>

5 0

3 years ago

An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency $\zeta$ = 3%

where;

$\zeta = \dfrac{W_{out}}{Q_{supplied }}$

$Q_{supplied } = \dfrac{2}{0.03} \ MW$

$Q_{supplied } = 66.66 \ MW$

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

$LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}$

Also;

$\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K$

$\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K$

$LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}$

$LMTD = \dfrac{8}{In (5)}$

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

$Q_H = UA (LMTD)$

where;

U = overall heat coefficient given as 1200 W/m².K

$66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}$

The mass flow rate:

$Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}$

3 0

3 years ago