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4 years ago

The maximum speed of the Tupolev Tu-144 airline is 694 m/s. What is this speed in kilometers per hour???

1 answer:

3 0

There are 60 seconds in 1 minute and 60 minutes in 1 hour.

and there are 1.609km in 1 mile, so, let's use those ratios.

$\bf \cfrac{694~\underline{mile}}{\underline{sec}}\cdot \cfrac{60~\underline{sec}}{\underline{min}}\cdot \cfrac{60~\underline{min}}{hr}\cdot \cfrac{1.609~km}{\underline{mile}}\implies \cfrac{694\cdot 60\cdot 60\cdot 1.609~km}{hr}$

and I'm sure you know how much that is, since is just a product.

You might be interested in

The three squares that will form a right triangle are ...?

MAVERICK [17]

Answer:

Three squares that form right triangle:

A, E and C

Step-by-step explanation:

Area of E = 1^2 = 1

Area of A = 2.4^2 = 5.76

Area of C = 2.6^2 = 6.76

Area A + Area E = Area C

5 0

3 years ago

Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou

mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

$\frac{dQ}{dt} = -kQ$

In which k is the decay rate.

The solution is:

$Q(t) = Q(0)e^{-kt}$

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that $Q(0) = 8000$

500 counts per minute 13 days later.

This means that $Q(13) = 500$ . We use this to find k.

$Q(t) = Q(0)e^{-kt}$

$500 = 8000e^{-13k}$

$e^{-13k} = \frac{500}{8000}$

$\ln{e^{-13k}} = \ln{\frac{500}{8000}}$

$-13k = \ln{\frac{500}{8000}}$

$k = -\frac{\ln{\frac{500}{8000}}}{13}$

$k = 0.2133$

$Q(t) = Q(0)e^{-0.2133t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.2133t}$

$0.5Q(0) = Q(0)e^{-0.2133t}$

$e^{-0.2133t} = 0.5$

$\ln{e^{-0.2133t}} = \ln{0.5}$

$-0.2133t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.2133}$

$t = 3.25$

The half-life of the radioactive substance is of 3.25 days.

7 0

3 years ago

I need help with #3 & #4

lina2011 [118]

Step-by-step explanation:

Q3, B

Q4, I'm not too sure about this one but I think its D

5 0

3 years ago

What is the value of 3 in the figure shown below?

NeX [460]

9514 1404 393

Answer:

Step-by-step explanation:

Angle E is congruent to angle C, so is the value required to make the sum of angles in triangle ABC be 180°.

∠E = 180° -118° -30° = 32°

The value of x is 32.

7 0

3 years ago

What is the parallel line for (0,4) and y=3/2x+3

pochemuha

The parallel line to y = $\frac{3}{2}$ x + 3 and passes through

the point (0 , 4) is y = $\frac{3}{2}$ x + 4

Step-by-step explanation:

Parallel lines have:

1. The same slopes

2. The different y-intercept

The slope-intercept form of the linear equation is y = m x + c, where

m is the slope of the line and c is the y-intercept

We need to find the equation of the line that is parallel to

y = $\frac{3}{2}$ x + 3 and passes through the point (0 , 4)

∵ The two lines are parallel

∴ Their slopes are equal

∵ The equation of the given line is y = $\frac{3}{2}$ x + 3

∵ The form of the equation is y = m x + c

∴ m = $\frac{3}{2}$

∴ The slope of the line is $\frac{3}{2}$

∵ The equation of the line is y = mx + c

∵ m = $\frac{3}{2}$

∴ The equation of the line is y = $\frac{3}{2}$ x + c

- To find c substitute x and y in the equation by the coordinates of

a point lies on the line

∵ The line passes through the point (0 , 4)

∴ x = 0 , y = 4

∴ 4 = $\frac{3}{2}$ (0) + c

∴ c = 4

∴ The equation of the line is y = $\frac{3}{2}$ x + 4

The parallel line to y = $\frac{3}{2}$ x + 3 and passes through

the point (0 , 4) is y = $\frac{3}{2}$ x + 4

Learn more:

You can learn more about equations of parallel lines in brainly.com/question/8628615

#LearnwithBrainly

8 0

3 years ago