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3 years ago

10

Ilocano dancesshow inward movement and are closer to the center and their musical notes stay close to each other. what character

istics or value does this show?
a. fright
b. reservation
c. proudness
d. poorness

1 answer:

love history [14]3 years ago

7 0

Answer:

letter c

Explanation:

i hope it helps at pa brainly naman po pls

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What measures the distance between two consecutive crests of a wave?

Marina86 [1]

Answer A is incorrect
A crest is just one point. It is not the distance between 2 crests.

B is incorrect
A trough is just 1 point. It is not the distance between 2 troughs.

C is incorrect.
the amplitude measures the height of a crest from the middle of the wave to the crest (or trough).

D is the correct answer. That is the distance between 2 crests or 2 troughs

8 0

3 years ago

11. A sensitive thermometer is one which

Marta_Voda [28]

Answer:

A. ls sensitive to heat

this is correct answer

I hope it's helpful for you.......

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3 years ago

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

$\theta= \omega t$

and substituting t = 75 seconds, we find

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, it is

$15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

$v=\omega r$

where we have

$\omega=0.20 rad/s$

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

$v=(0.20 rad/s)(13.5 m)=2.7 m/s$

6 0

3 years ago

Thomas and John are carrying a 43kg cylinder head on a 510cm X 510mm board. The cylinder head with dimensions of 43cm X 250mm li

tatyana61 [14]

John carry the heaviest load.

<h3>How to find out who is carrying the heavy load?</h3>

Write down given data from questions:

Board=510cm X 510mm.

Cylinder head with dimensions=43cm X 250mm.

Cylinder lies across the board 210cm from john.

Find out: Who is carry the heaviest load?

Calculation:

We assume that mass of cylinder head = x kg

Then weight=x x 9*81

W=9.81x Newton.

Weight per unit length= Weight/Total leanth

Weight per unit length= 9.81x/43

(w/l)=0.23x N/cm

From equation contition: $(F_{J} +F_{T} =9.81x n)$

$F_{T} (510)=9.81x$ (210+21.5)

$F_{T} (510)=9.81 x (231.5)$

$F_{T} =4.452x N.$

$F_{J} =9.81x-4.452x$

$F_{J} =5.358 xN$

Therefore $F_{J} \geq F_{T}$

To learn more about mass per unit length, refer to:

brainly.com/question/24180692

#SPJ9

4 0

1 year ago

A book rest on a table which it's face having a sides 30cm by 25cm . if it exerts apressure of 200pa then determine the mass of

Simora [160]

Answer:

Approximately $1.5\; \rm kg$ . (Assuming that this table is level, and that the gravitational field strength is $g = 9.8\; \rm N \cdot kg^{-1}$ .

Explanation:

Convert the dimensions of this book to standard units:

$\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.30\; \rm m$ .

$\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.25\; \rm m$ .

Calculate the surface area of this book:

$0.30\; \rm m \times 0.25\; \rm m = 0.075\; \rm m^{2}$ .

Pressure is the ratio between normal force and the area over which this force is applied.

$\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}$ .

Equivalently:

$(\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area})$ .

In this question, $\text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^{-2}$ .

It was found that $(\text{contact Area}) = 0.075\; \rm m^{2}$ (assuming that the entire side of this book is in contact with the table.

Hence:

$\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^{-2} \times 0.075\; \rm m^{2} \\ &= 15\; \rm N \end{aligned}$ .

If that the table is level, this normal force would be equal to the weight of this book:

$\text{weight} = 15\; \rm N$ .

Assuming that the gravitational field strength is $g = 9.8\; \rm N \cdot kg^{-1}$ . The mass of this book would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{15\; \rm N}{9.8\; \rm N \cdot kg^{-1}}\approx 1.5\; \rm kg\end{aligned}$ .

4 0

2 years ago