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3 years ago

Which of the following is an instantaneous speed?

Physics

1 answer:

Brilliant_brown [7]3 years ago

5 0

Answer:

A: All of the above

Explanation:

The instantaneous speed of an object is simply the current seed of the object at any given time. The SI unit is m/S and it is a vector quantity.

Therefore, according to the given options, they all have SI units that are consistent with distance and time which makes them all an example of instantaneous speed.

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What happens when sediment builds up over time

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When sediment has built up over time layers of rock start to form, starting with sedimentry rocks, then metamorphic rocks

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3 years ago

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In a lightning bolt, a large amount of charge flows during a time of 1.2 x 10-3 s. Assume that the bolt can be represented as a

ohaa [14]

Answer: 10.58 C has flowed during the lightning bolt

Explanation:

Given that;

Time of flow t = 1.2 × 10⁻³

perpendicular distance r = 21 m

Magnetic field B = 8.4 x 10⁻⁵ T

Now lets consider the expression for magnetic field;

B = u₀I / 2πr

the current flow is;

I = ( B × 2πr ) / u₀

so we substitute

I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷

= 0.01107792 / 0.000001256

= 8820 A

Hence the charge flows during lightning bolt will be;

q = It

so we substitute

q = 8820 × 1.2 × 10⁻³

q = 10.58 C

therefore 10.58 C has flowed during the lightning bolt

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3 years ago

What kind of relationship does a human have with the bacteria that lives in her/his digestive tract?

Misha Larkins [42]

The one that both benefits each other is the one I think it's mutalistic

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3 years ago

A 4.00 kg ball is swung in a circle on the edge of a 1.50 m rope. The time it takes for the ball to complete one rotation is 3.4

lorasvet [3.4K]

Answer:

The answer is below

Explanation:

The length of the rope is equal to the radius of the circle formed by the complete rotation of the rope. Therefore the radius = 1.50 m.

a) The distance covered by the rope when completing one rotation is the same as the perimeter of the circle. Hence:

Distance covered in one rotation = 2π * radius = 2π * 1.5 = 3π meters

The velocity of the ball = Distance / time = 3π meters / 3.4 seconds = 2.77 m/s

b) The initial velocity (u) is 0 m/s, the final velocity is 2.77 m/s during time (t) = 3.4 s. Hence acceleration (a):

v = u + at

2.77 = 3.4a

a = 0.82 m/s²

c) Force on ball = mass * acceleration = 4 * 0.82 = 3.28 N

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2 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$

Then,

$f_3 = 3f_1$

Rearranging to find the fundamental frequency

$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

7 0

2 years ago