What line of code will call force with a value of 10 for mass and a value of 9.81 for acceleration?

1 answer:

7 0

Line of code will call force with a value of 10 for mass and a value of 9.81 for acceleration is force(10, 9.81).

<h3 /><h3>Line of code for force and acceleration</h3>

In mechanics, acceleration refers to the rate at which an object's velocity with respect to time varies.
Acceleration is a vector quantity (in that they have magnitude and direction).
The direction of an object's acceleration is determined by the direction of the net force acting on it.
Newton's Second Law states that the combined effect of two factors determines how much an item accelerates.
The size of the net balance of all external forces acting on the object is, in accordance with the materials used to create it.
It inversely proportional to its mass, whereas the magnitude of the net resultant force is directly proportional to the net force.

def force(mass, acceleration):

force_val = mass*acceleration

return force_val

10 is assigned to mass and 9.81 is assigned to acceleration

def force(10, 9.81)

So, Line of code will call force with a value of 10 for mass and a value of 9.81 for acceleration is force(10, 9.81).

Learn more about acceleration here:

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Two microwave signals of nearly equal wavelengths can generate a beat frequency if both are directed onto the same microwave det

alekssr [168]

Answer:

1.5106 cm

Explanation:

The beat frequency is equal to the absolute value of the difference between the frequencies of the two signals:

$f_B = |f_1 - f_2|$

using the wave equation, we can re-write each frequency as

$f=\frac{c}{\lambda}$

where c is the speed of light and $\lambda$ is the wavelength. Therefore,

$f_B = |\frac{c}{\lambda_1}-\frac{c}{\lambda_2}|$

where:

$f_B = 140 MHz = 140\cdot 10^6 Hz$ is the beat frequency

$\lambda_1 = 1.50 cm = 0.015 m$ is the wavelength of the first generator

$\lambda_2$ is the wavelength of the second generator

We also know that the second generator emits the longer wavelength, so we already know that the term inside the module is positive. Therefore, we can now solve for $\lambda_2$ :

$f_B = c(\frac{1}{\lambda_1}-\frac{1}{\lambda_2})\\\lambda_2=(\frac{1}{\lambda_1}-\frac{f_B}{c})^{-1}=(\frac{1}{0.015}-\frac{140\cdot 10^6}{3\cdot 10^8})^{-1}=0.015106 m = 1.5106 cm$