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Ivan
1 year ago
6

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

Physics
1 answer:
Kaylis [27]1 year ago
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:

brainly.com/question/460763

#SPJ4

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A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0
MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

\omega=\frac{2\pi}{T}

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

\omega=\frac{2\pi}{1 s}=6.28 rad/s

Now we can find the linear speed of the ladybug, which is given by

v=\omega r

where:

\omega=6.28 rad/s is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

v=(6.28)(0.65)=4.1 m/s

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7 0
3 years ago
Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl
ollegr [7]
<h2>Answer:</h2>

(a) standard deviation = σ = 4.9996

(b) variance = σ² = 24.996

<h2>Explanation:</h2><h2 />

<em>Given frequency table (find attached as Table 1);</em>

<u></u>

(a) To find the sample standard deviation and sample variance, follow these steps;

<em>i. Calculate the mid-point c for each group by using the mid-point formula;</em>

c = (lower bound + upper bound) / 2

=> c = (6.51 + 8.50) / 2 = 7.505

=> c = (8.51 + 10.50) / 2 = 9.505

=> c = (10.51 + 12.50) / 2 = 11.505

=> c = (12.51 + 14.50) / 2 = 13.505

=> c = (14.51 + 16.50) / 2 = 15.505

<em>So the new table becomes (find attached as Table 2);</em>

<em>ii. Calculate the total number of samples (n) which is the sum of all the frequencies.</em>

n = 50+18+42+20+46

n = 176

<em>iii. Calculate the mean (M)</em>

This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).

M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176

M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176

M = [2012.88] / 176

M = 11.44

<em>iv. Find the variance (σ²);</em>

The variance is calculated using the following formula

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)                ------------(i)

Where;

f = frequency of each boundary data point

<em>=>  Let's first calculate </em>Σ(f x c²).

This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)

Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]

Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]

Σ(f x c²) = 24708.1244

<em>=> Now calculate (n x M²)</em>

n x M² = 176 x 11.44²

n x M² = 23033.7536

<em>=> Now substitute these values into equation (i) to calculate the variance</em>

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)

σ² = [24708.1244 - 23033.7536] / (176 - 1)

σ² = [4374.3708] / (175)

σ² = 24.996

Therefore, the variance is 24.996

<em>v. Find the standard deviation (σ)</em>

The standard deviation is the square root of the variance. i.e

σ = √σ²

σ = √24.996

σ = 4.9996

Therefore, the standard deviation is 4.9996

4 0
2 years ago
The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that th
Mars2501 [29]

Answer:

According to Newton's 2nd law

The force acting on a body produces acceleration in its direction which is directly propotional to the force but inversly propotinal to the mass of tbe body.

Explanation:

a = F/m

F = ma

Where( F) is force (m) is mass and (a) is acceleration.

6 0
3 years ago
Convert the following to relative uncertainties <br>a) 2.70 ± 0.05cm<br>b) 12.02 ± 0.08cm
DENIUS [597]

data which is expressed in form of following way

a = a_o + \Delta a

here in above expression

a_o = true value

\Delta a = uncertainty in the value

now the relative uncertainty is given as

\frac{\Delta a}{a_o}

now by above formula we can say

a) 2.70 ± 0.05cm

here

True value = 2.70

uncertainty = 0.05

Relative uncertainty = \frac{0.05}{2.70} = 0.0185

b) 12.02 ± 0.08cm

here

True value = 12.02

uncertainty = 0.08

Relative uncertainty = \frac{0.08}{12.02} = 0.00665

4 0
3 years ago
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