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2 years ago

Identify the vibrating media in three different types of musical instruments.

Physics

1 answer:

mart [117]2 years ago

3 0

Explanation:

Hole. Hole. Different notes can be played on the flute by blocking holes. ...

Drum skin. Drum skin. Hitting the bongo drum makes its tight elastic skin vibrate.

String. String. ...

Sound. hole. ...

Bow. Bow.

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Identical forces act for the same length of time on two different masses. The change in momentum of the smaller mass is

xenn [34]

Answer:

Equal to change in momentum of larger mass.

Explanation:

We are given that

Two difference masses .

Force act on both masses for the same length of time.

We have to find the change in momentum of the smaller mass.

Let M and m are two masses

M>m

We know that

Change in momentum for large mass= $F\Delta t$

Change in momentum for small mass= $F\Delta t$

Because Force and length of time are same for both masses .

Hence, the change in momentum of smaller mass is equal to change in momentum of larger mass.

7 0

2 years ago

Formula:

s2008m [1.1K]

Answer:

55N

Explanation:

Using Newton's second law of motion:

F=ma

Force=mass × acceleration

F=25×2.2

F=55N

So 55 Newtons are needed

8 0

2 years ago

A 4-kilogram ball moving at 8 m/sec to the right collides with a 1-kilogram ball at rest. After the collision,

elena-14-01-66 [18.8K]

Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32

Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32

Therefore β = 12.8 m/s

4 0

2 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

2 years ago

Lasers emit light of a certain frequency in one, precise direction. The light that a laser emits can be tuned to have a high fre

makkiz [27]

Answer:

An ultra intense laser is one with which intensities greater than 1015 W cm-2 can be achieved.

Explanation:

This intensity, which was the upper limit of lasers until the invention of the Chirped Pulse Amplification, CPA technique, is the value around which nonlinear effects on the transport of radiation in materials begin to appear.

Currently, the most powerful lasers reach intensities of the order of 1021W cm-2 and powers of Petawatts, PW, in each pulse. This range of intensities has opened the door for lasers to a multitude of disciplines and scientific areas traditionally reserved for accelerators and nuclear reactors, applying as generators of high-energy electron, ion, neutron and photon beams, without the need for expensive infrastructure.

8 0

2 years ago

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