A poodle racing across a lawn is an example of ____

26. Keenan found the mass of a book to be 4.56*10^ -2 kg . What is the mass of the book in milligrams?

vagabundo [1.1K]

Taking into account the rule of three for the change of units, the mass of the book is 45600 miligrams.

First of all, the rule of three is a mathematical tool that helps you quickly solve proportionality problems.

Having three known values and one unknown, a proportional relationship is established between all of them in order to find the fourth term of the proportion.

If the relationship between the magnitudes is direct (when one magnitude increases, so does the other; or when one magnitude decreases, so does the other), the rule of three is applied as follows, where a, b and c are known values and x is the unknown to calculate:

a → b

c → x

So: $x=\frac{cxb}{a}$

Being 1 kg equivalent to 1000000 milligrams, In this case the rule of three is applied as follows: if 1 kg equals 1000000 milligrams, 4.56×10⁻² kg equals how many milligrams?

1 kg → 1000000 milligrams

4.56×10⁻² kg → x

So:

$x=\frac{4.56x10^{-2} kg x1000000 miligrams }{1 kg}$

<u><em>x=45600 miligrams</em></u>

In summary, the mass of the book is 45600 miligrams.

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7 0

2 years ago

If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, ass

mixas84 [53]

Answer: $3.874 m/s^2$

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

$60mi/h \approx 26.8224m/s$

$34mi/h \approx 15.1994 m/s$

we know acceleration is given by $=\frac{velocity}{Time}$

$a=\frac{15.1994-26.8224}{3}$

$a=-3.874 m/s^2$

negative indicates that it is stopping the car

Distance traveled

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

$\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s$

$s=\frac{488.419}{2\times 3.874}$

s=63.038 m

7 0

2 years ago

In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21

Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

$z = 20km$