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3 years ago

Explain the relationship between force, mass and acceleration in Newton's second law

Physics

1 answer:

Nikitich [7]3 years ago

7 0

Newton's second law of motion describes what happens to a body when an external force is applied to it.

Newton's second law of motion states that the force acting on an object is equal to the mass of that object times its acceleration. In mathematical form this is written as

F = ma

Where F is force , m is mass and a is acceleration. The math or logic behind this is that if you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half.

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Gunther needs a 75% concentrated solution. When making his solution at room temperature, he could only make a solution that was

jolli1 [7]

Answer:

Heat the solution, dissolve the solute, and let it cool verifying nothing settled out.

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

Name the two scales that are used to measure earth intensity

nikklg [1K]

Answer:

Although several scales have been developed over the years, the two commonly used today in the United States are the moment magnitude scale, which measures magnitude (M), or size, and the Modified Mercalli scale, which measures intensity.

3 0

4 years ago

If the sign of work is negative,

aivan3 [116]

If the sign of work is negative, that means the force and the motion are in opposite directions.

Let's say you see something roll off of the shelf. You catch it, and you let it down slowly and gently.

Gravity exerted down-force on it and it moved down. Gravity did positive work on it.

YOU exerted UP-force on it and it moved down. YOU did negative work on it.

(Also, the falling object exerted down-force on your hand, and your hand moved down. The falling object did positive work on your hand ! Where did THAT energy come from ? It came from the potential energy that the object had while it was on the shelf. Your hand absorbed that energy on the way down, doing negative work. So the object didn't have any kinetic energy when it reached the floor, and it did NOT splinter the floor or shatter in smithereens. It had barely enough energy left to make a sound when it hit the floor.)

5 0

3 years ago

A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train the

sergij07 [2.7K]

Answer:

c) 0.16 m/s2

Explanation:

The computation of the acceleration during the first km of travel is shown below

Given that

Final velocity = v = 42 m/s

Initial velocity = u = 0 m/s

Distance = 5.6km

Based on the above information, we need to apply the following formula

As we know that

$v^2 - u^2 = 2as$

$a = \frac{v^2 - u^2}{2s}$

$= \dfrac{42^2 - 0^2}{2 \times 5.6 \times \frac{1000\ m}{1\ km} }$

= 0.1575 m/s ^2

hence, the correct option is c. 0.16 m.s^2

4 0

3 years ago