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A car traveling at 14 m/s accelerates at 3.5 m/s² for 5 seconds. How much distance does it travel during that time?

Rainbow [258]

Answer: 113.75

Explanation:

You know

acceleration = a = 3.5 m/s²

time = t = 5 seconds

initial velocity = u = 14 m/s

Unknown is distance = s = ?

Use equation: s = ut + $\frac{1}{2}$ at²

Substitute all the known values inside the equation:

s = (14*5) + 0.5 * 3.5 * 5²

s = 70 + 43.75 = 113.75 m

The car travels 113.75 metres.

3 0

3 years ago

Please help meeeeeeeeeeeeeeee

maxonik [38]

probabilityAnswer:

2/27

Explanation:

The elk can not be eaten so we remove that from the probablity

so we have x/18

songbird = 4/18

mice = 6/18

4/18*6/18 = 2/27

5 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,

$z = 20km$

And the velocity can be written as,

$V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$V = 980.55m/s$

From the properties of standard atmosphere at altitude z = 20km temperature is

$T = 216.66K$

$k = 1.4$

$R = 287 J/kg$

Velocity of sound at this altitude is

$a = \sqrt{kRT}$

$a = \sqrt{(1.4)(287)(216.66)}$

$a = 295.049m/s$

Then the Mach number

$Ma = \frac{V}{a}$

$Ma = \frac{980.55}{296.049}$

$Ma = 3.312$

So front stagnation temperature

$T_0 = T(1+\frac{k-1}{2}Ma^2)$

$T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)$

$T_0 = 689.87K$

Therefore the temperature at its front stagnation point is 689.87K

6 0

3 years ago

Two rams run toward each other. One ram has a mass of 44 kg and runs south with a speed of 6 m/s, while the other has a mass of

gladu [14]

Answer = D. 114 kg-m/s north

3 0

3 years ago

Which transition represents a time when water molecules are moving closer together?

oee [108]

The best transition between the four options presented to represent a time when water molecules are moving closer together would be A. Frost forms on a window pane.

The closest distance that the water molecules can do is when the water is in the state of being solid. It is known that the solid state of matter has the closest distance from molecule to molecule that when a molecule tries to move, the others move as well creating a vibration and thus producing heat in the process. When they are in a liquid state, they are quite far from each other. In a gas state, they really are far from each. This explains the difference in their characteristics.

4 0

3 years ago

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Pictures below :) Have a nice day!!