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2 years ago

What are action-reaction force pairs?

2 answers:

8 0

Exerted by two objects on each other are often called an action- reaction force pair. Either force can be considered the action force or the reaction force.

Cloud [144]2 years ago

7 0

Two objects on each other

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If a force of 25 N is applied to an object with a mass of 8 kg, the object will accelerate at

Blizzard [7]

3.13 m/s2
.
the formula for acceleration is as follows:
force/mass = acceleration
-
so 25/8 = 3.13

3 0

3 years ago

1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.

marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

$B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB$

<h3>Speed of sound at the given temperature</h3>

$v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s$

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

$f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )$

where;

-v₀ is velocity of the observer moving away from the source
-vs is the velocity of the source moving towards the observer
fs is the source frequency
fo is the observed frequency
v is speed of sound

$f_0 = f_s(\frac{v-v_0}{v- v_s} )$

$f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz$

Learn more about intensity of sound here: brainly.com/question/17062836

3 0

1 year ago

Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h

Gelneren [198K]

Answer: W = $1.04.10^{-20}$ J

Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.

The work to transport an ion from a lower potential side to a higher potential side is calculated by

$W=q.\Delta V$

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = $1.6.10^{-19}$ C

To determine work in joules, potential has to be in Volts, so:

$\Delta V=65.10^{-3}V$

Then, work is

$W=1.6.10^{-19}.65.10^{-3}$

$W=1.04.10^{-20}$

To move a potassium ion from the exterior to the interior of the cell, it is required $W=1.04.10^{-20}$ J of energy.

8 0

3 years ago

If you wish to observe features that are around the size of atoms, say 1 .5 x 100 m, with electromagnetic radiation, the radiati

chubhunter [2.5K]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

If you wish to observe features that around the size of atoms, say 1.5×10⁻¹⁰ m, with electromagnetic radiation, the radiation must have a wavelength about the size of the atom itself.

a) If you had a microscope which was capable of doing this, what would the frequency of electromagnetic radiation be, in hertz that you would have to use?

b) What type of electromagnetic radiation would this be?

Given Information:

Wavelength = λ = 1.5×10⁻¹⁰ m

Required Information:

a) Frequency = f = ?

b) Type of electromagnetic radiation = ?

Answer:

a) Frequency = f = 2×10¹⁸ Hz

b) Type of electromagnetic radiation = X-rays

Explanation:

a) The frequency of the electromagnetic radiation is given by

f = c/ λ

Where λ is the wavelength of the electromagnetic radiation and c is the speed of light and its value is 3×10⁸ m/s

f = 3×10⁸/1.5×10⁻¹⁰

f = 2×10¹⁸ Hz

Therefore, the frequency of the electromagnetic radiation would be 2×10¹⁸ Hz.

The frequency range of X-rays is 3×10¹⁶ Hz to 3×10¹⁹ Hz

The frequency 2×10¹⁸ lies in that range, therefore, the type of electromagnetic radiation is X-rays

5 0

3 years ago

Two identical satellites are in orbit about the earth. One orbit has a radius r and the other 2r. The centripetal force on the s

velikii [3]

Answer:

the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

Explanation:

Mass of satellite, m

orbit radius of first, r1 = r

orbit radius of second, r2 = 2r

Centripetal force is given by

$F= \frac{mv^{2}}{r}$

Where v be the orbital velocity, which is given by

$v=\sqrt{gr}$

So, the centripetal force is given by

$F= \frac{mgr}}{r}}=mg$

where, g bet the acceleration due to gravity

$g=\frac{GM}{r^{2}}$

So, the centripetal force

$F= \frac{GMm}}{r^{2}}}$

Gravitational force on the satellite having larger orbit

$F= \frac{GMm}{4r^{2}}$ .... (1)

Gravitational force on the satellite having smaller orbit

$F'= \frac{GMm}{r^{2}}$ .... (2)

Comparing (1) and (2),

F' = 4 F

So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

8 0

3 years ago