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3 years ago

Topic: Chapter 19: Some wiggle room

Physics

1 answer:

Nana76 [90]3 years ago

3 0

Answer:

4.86 m

Explanation:

Given that,

The frequency produced by a humming bird, f = 70 Hz

The speed of sound, v = 340 m/s

We need to find how far does the sound travel between wing flaps. Let the distance is equal to its wavelength. So,

$v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{340}{70}\\\\\lambda=4.86\ m$

So, the sound travel 4.86 m between wings flaps.

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What is the equivalent resistance of a 7ohm resistor and a 93 ohm resistor connected in parallel? PLEASE HELP ASAP

lukranit [14]

When you have several resistors in parallel, their equivalent resistance is the reciprocal of the sum of their individual reciprocals.

When there are only two of them, it gets a lot easier. In that case, their equivalent resistance is equal to

(their product) divided by (their sum).

Equivalent = (7 x 93) / (7 + 93)

Equivalent = (651) / (100)

Equivalent = 6.51 ohms .

7 0

4 years ago

Find the current flowing through the following circuit

kap26 [50]

Answer:

0.5

Explanation:

calculating the total resistance for the both the parallel and the series resistors then later using the ohm's law formulae V=IR to calculate current.

4 0

3 years ago

The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca

larisa86 [58]

Answer:

The value is $E_i = 1.5596 *10^{-18} \ J$

Explanation:

From the question we are told that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$

The velocity is $v = 2.371*10^6 \ m/s$

The mass of electron is $m_e = 9.109*10^{-31} \ kg$

Generally the energy of the incident light is mathematically represented as

$E = \frac{h * c}{\lambda}$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

h is the Planck constant with value $h = 6.62607015 * 10^{-34 } J\cdot s$

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

=> $E = 4.12 *10^{-18} \ J$

Generally the kinetic energy is mathematically represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$

=> $E_k = 2.56 *0^{-18} \ J$

Generally the ionization energy is mathematically represented as

$E_i = 4.12 *10^{-18} - 2.56 *0^{-18}$

=> $E_i = 1.5596 *10^{-18} \ J$

8 0

3 years ago

If the Sun were to collapse into a black hole, the point of no return for an investigator would be approximately 3 km from the c

Marysya12 [62]

Answer:

1.97*10⁴N .He cannot survive.

Explanation:

The Tidal force can be defined as the difference in force between center of two bodies like center of earth and other bodies.This force stretches the body too and fro from the center of mass of another body due to the difference in the gravitational force.

The tidal force is responsible for various phenomena like ocean tides,celestial body tearing up.

The tidal force is the effect of massive body gravitationally affecting another body.

The tidal force is the difference between the gravitational force of two bodies and directly proportional to the mass of body affecting another body and inversely proportional to the square of radius of the distance between center and the object.

Here let us consider height of person as 2m and d=1m

ΔF= $\frac{GMm}{(R-d)^{2} }$ - $\frac{GMm}{(R+d)^{2} }$

=GMm[ $\frac{1}{(R-d)^{2} }$ - $\frac{1}{(R+d)^{2} }$ ]

=(6.67*10⁻¹¹)(1.99*10³⁰)(1.0)[ $\frac{1}{(300*10^3-1)^{2} }$ - $\frac{1}{(300*10^3+1)^{2} }$ }

ΔF=1.97*10⁴N.

Thus he cannot survive

4 0

3 years ago

Use the chart to determine the half-life of carbon-14

lubasha [3.4K]

The correct answer to the question is B) 5,700 years.

EXPLANATION:

Before going to answer this question, first we have to understand half life period of a radio active substance.

The half life period $[\ T_{1/2}\ ]$ of a radio active substance is defined as the time in which half of the radio active specimen has undergone decay.

As per the question, the initial concentration of the radioactive carbon-14 is given as 10,000.

Hence, initial concentration $N_{0} =\ 10,000$ .

After 5,700 years, the remaining radio active specimen is 5000. It means that half of the radio active specimen has undergone decay in 5,700 years.

Hence, half life period of the radio active carbon-14 specimen is 5,700 years.

7 0

3 years ago

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