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Viktor [21]
3 years ago
15

1. What coefficients would balance the following equation?

Physics
2 answers:
Dovator [93]3 years ago
4 0

Answer:

B

Explanation:

Stolb23 [73]3 years ago
3 0

Answer:

no question matee

Explanation:

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Two boats - Boat A and Boat B - are anchored a distance of 24 meters apart. The incoming water waves force the boats to oscillat
ozzi

Answer:

wavelength = 24 m

Period = 10 s

f = 0.1 Hz

Amplitude = 4 m

Explanation:

Wavelength:

Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:

<u>wavelength = 24 m</u>  

Period:

The period is given as:

Period = \frac{time}{no.\ of\ cycles} \\\\Period = \frac{10\ s}{1}\\\\

<u>Period = 10 s</u>

<u></u>

Frequency:

The frequency is given as:

f = \frac{1}{time\ period}\\\\f = \frac{1}{10\ s}\\\\

<u>f = 0.1 Hz</u>

<u></u>

Amplitude:

Amplitude will be half the distance between extreme points, that is, crest and trough:

Amplitude = 8 m/2

<u>Amplitude = 4 m</u>

5 0
3 years ago
While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find th
Dovator [93]

Answer:

Explanation:

From the given information:

radius = 15 m

Time T = 23 s

a) Speed (v) = \dfrac{2 \pi r}{T}

v = \dfrac{2\times \pi \times 15}{23}

v = 4.10 m/s

b) The magnitude of the acceleration is:

a = \dfrac{v^2}{r} \\ \\ a = \dfrac{(4.10)^2}{15}

a = 1.12 m/s²

c) True weight = mg

Apparent weight = normal force

From the top;

the normal force = upward direction,

weight is downward as well as the acceleration.

true weight - normal force = ma  

apparent weight =mg - ma  

\dfrac{apparent \ weight}{true \  weight} = \dfrac{(mg - ma)}{(mg)}

=1- \dfrac{1.12}{9.8}

= 0.886 m/s²

d)

From the bottom;

acceleration is upward, so:

apparent weight - true weight = ma

apparent weight = true weight + ma

\dfrac{apparent \ weight }{true \ weight} =\dfrac{ mg + ma}{mg}

\dfrac{apparent \ weight }{true \ weight} =1+ \dfrac{a}{g} \\ \\ = 1 + \dfrac{1.12}{9.8}

= 1 + \dfrac{1.12}{9.8} \\ \\

= 1.114 m/s²

6 0
3 years ago
Mechanical locks can accept a variety of inputs as keys, including magnetic strips on ID cards, radio signals from name badges,
lidiya [134]

Answer:

The answer is B) False

3 0
4 years ago
If the load is 13 cm from the fulcrum, how much effort is needed to lifth the load ? Help me.
snow_tiger [21]
That depends on a few things that you haven't told us about the setup.
So I'm going to assume one of them, and then give you the answer
in terms of another one:

-- Assume a Class-I lever . . . the fulcrum is between the load and the effort.

-- Then the effort needed to lift the load is

(the weight of the load) x (13 / the distance between the fulcrum and the effort)
7 0
3 years ago
State the relative position for the sun and earth during the lunar eclipse (ASAP)
Mademuasel [1]

Answer:

Solar eclipses result from the Moon blocking the Sun relative to the Earth; thus Earth, Moon and Sun all lie on a line. Lunar eclipses work the same way in a different order: Moon, Earth and Sun all on a line. In this case the Earth's shadow hides the Moon from view.

8 0
3 years ago
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