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4 years ago

What is magnification

Physics

1 answer:

il63 [147K]4 years ago

5 0

The action or process of magnifying something or being magnified, especially visually. Hope this helped

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What is velocity and it’s SI Unit?

svp [43]

Answer:

Velocity is the displacement (change In position of an an object it is similar to distance ) divided by the time taken. SI unit m/s (metre per second)

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3 years ago

What quantity is represented by a unit called newton (N) ?

bixtya [17]

Newton is the SI unit of force . Newton is the name of a British scientist and the name of unit is to honour him. The unit is actually Kg.m/s 2 The unit can be derived by the formula. Take the example of weight. It's formula is W = mg . We know that the unit of mass is kg and gravity is m/s 2 so the unit of weight becomes kg.m/s 2 This unit is known as a Newton. It is always given a capital letter because it is someone's name. Other units that are always capitalised (upper case) are Ampere (Amp), Watt, Volt, Coulomb, Kelvin, Celsius, Fahrenheit, Curie, Roentgen because they are also people's names.

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3 years ago

Mass is measured in ____________, and weight is measured in ______________.

chubhunter [2.5K]

Answer:

MASS is measured in KILOGRAMS

Weight is measured in NEWTONS

Explanation:

7 0

3 years ago

Read 2 more answers

Two strings on a musical instrument are tuned to play at 196 hz (g) and 523 hz (c). (a) what are the first two overtones for eac

Tems11 [23]

(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
$f_n = n f_1$
where f1 is the fundamental frequency.

So, the first overtone (2nd harmonic) of the string is
$f_2 = 2 f_1 = 2 \cdot 196 Hz = 392 Hz$
while the second overtone (3rd harmonic) is
$f_3 = 3 f_1 = 3 \cdot 196 Hz = 588 Hz$

Similarly, for the second string with fundamental frequency $f_1 = 523 Hz$ , the first overtone is
$f_2 = 2 f_1 = 2 \cdot 523 Hz = 1046 Hz$
and the second overtone is
$f_3 = 3 f_1 = 3 \cdot 523 Hz = 1569 Hz$

(b) The fundamental frequency of a string is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where L is the string length, T the tension, and $\mu = m/L$ is the mass per unit of length. This part of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them $m_{196}$ , the mass of the string of frequency 196 Hz, and $m_{523}$ , the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
$\frac{523 Hz}{196 Hz} = \frac{ \frac{1}{2L} \sqrt{ \frac{T}{m_{523}/L} } }{\frac{1}{2L} \sqrt{ \frac{T}{m_{196}/L} }}$
and since L and T simplify in the equation, we can find the ratio between the two masses:
$\frac{m_{196}}{m_{523}}=( \frac{523 Hz}{196 Hz} )^2 = 7.1$

(c) Now the tension T and the mass per unit of length $\mu$ is the same for the strings, while the lengths are different (let's call them $L_{196}$ and $L_{523}$ ). Let's write again the ratio between the two fundamental frequencies
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L_{523}} \sqrt{ \frac{T}{\mu} } }{\frac{1}{2L_{196}} \sqrt{ \frac{T}{\mu} }}$
And since T and $\mu$ simplify, we get the ratio between the two lengths:
$\frac{L_{196}}{L_{523}}= \frac{523 Hz}{196 Hz}=2.67$

(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them $T_{196}$ and $T_{523}$ . Let's write again the ratio of the frequencies:
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L} \sqrt{ \frac{T_{523}}{m/L} } }{\frac{1}{2L} \sqrt{ \frac{T_{196}}{m/L} }}$
Now m and L simplify, and we get the ratio between the two tensions:
$\frac{T_{196}}{T_{523}}=( \frac{196 Hz}{523 Hz} )^2=0.14$

7 0

3 years ago

Explain what is happening in this<br> induction example.

Semmy [17]

Answer:

An electrical field duh

Explanation:

None needed self explanitory

4 0

2 years ago