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3 years ago

Is there any difference between reflection and echo???

Physics

1 answer:

Fofino [41]3 years ago

6 0

Answer:

reflection is the act of reflecting or the state of being reflected while echo is

reflection of sound waves from a surface back to the initial listener

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The sound level of one person talking at a certain distance from you is 61 dB. If she is joined by 5 more friends, and all of th

Lady_Fox [76]

Answer:

Explanation:

For sound level in decibel scale the relation is

dB = 10 log I / I₀ where I₀ = 10⁻¹² and I is intensity of sound whose decibel scale is to be calculated .

Putting the given values

61 = 10 log I / 10⁻¹²

log I / 10⁻¹² = 6.1

I = 10⁻¹² x 10⁶°¹

$=10^{-5.9}$

intensity of sound of 5 persons

$I=5\times 10^{-5.9}$

$dB=10log\frac{5 X 10^{-5.9}}{10^{-12}}$

= 10log 5 x 10⁶°¹

= 10( 6.1 + log 5 )

= 67.98

sound level will be 67.98 dB .

4 0

3 years ago

A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial

Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

$qvB=m\frac{v^2}{r}$

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

$r=\frac{mv}{qB}$

For the ion of oxygen-16, we have:

$m_A=2.66\cdot 10^{-26}kg$

$q_A = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_A=2.90\cdot 10^6 m/s$

$B_A=1.30 T$

So the radius of its orbit is

$r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m$

For the ion of oxygen-18, we have:

$m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg$

$q_B = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_B=2.90\cdot 10^6 m/s$

$B_B=1.30 T$

So the radius of its orbit is

$r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m$

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

$d=2(r_B-r_A)=2(0.417-0.371)=0.092 m$

3 0

3 years ago

How to tackle questions on resultant force

Solnce55 [7]

Answer:

February. we will be in the future types, I

Explanation:

Collins the best year fixed come in handy when but it at a very nice relaxing. it will not have the option?

3 0

3 years ago

A mass weight of 120N is hung from two strings. what is the tension?

kramer

The weight should be shared between the two string equally. Therefore, tension in each string, T is;

T = 120 N/2 = 60 N

7 0

4 years ago

Read 2 more answers

An archer puts a 0.30-kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m. Assuming th

vovikov84 [41]

Answer:

41.74 m/s

Explanation:

The energy used to draw the bowstring = the kinetic energy of the arrow.

Fd = 1/2mv²................................ Equation 1

Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.

make v the subject of the equation

v = √(2Fd/m)...................... Equation 2

Given: F = 201 N, m = 0.3 kg, d = 1.3 m.

Substitute into equation 2

v = √(2×201×1.3/0.3)

v = √(1742)

v = 41.74 m/s.

Hence the arrow leave the bow with a speed of 41.74 m/s

3 0

3 years ago

Read 2 more answers