Ask question

Ask question

All categories

2 years ago

12

A drag racer starts her car from rest and accelerates at 11.9 m/s2 for the entire distance of 400 m. What is the speed of the ra

ce car at the end of the run?

1 answer:

miv72 [106K]2 years ago

6 0

<h2>Speed of the race car at the end of the run is 97.57 m/s</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s

Acceleration, a = 11.9 m/s²

Final velocity, v = ?

Displacement,s = 400 m

Substituting

v² = u² + 2as

v² = 0² + 2 x 11.9 x 400

v = 97.57 m /s

Speed of the race car at the end of the run is 97.57 m/s

You might be interested in

What is an isotope ?

Step2247 [10]

<span>An isotope is a form of a chemical element whose atomic nucleus contains a specific number of neutrons in addition to the number of protons that distinctively defines the element. The nuclei of most atoms have neutrons as well as protons.</span>

8 0

3 years ago

Which image shows the correct way of lining up vectors to add them

Marina CMI [18]

Answer:

it's A

Explanation:

wen aligning the vectors the head and the tail should meet

4 0

11 months ago

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh

Alinara [238K]

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c) $k_{mol}=3.74\times 10^{3}J/mol$

d) $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ( $k_{avg}$ ) is given as

$k_{avg}=\frac{3}{2}\times kT$ ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$

$k_{avg}=6.22\times 10^{-21}J$

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$

$k_{avg}=8.61\times 10^{-21}J$

c ) The translational kinetic energy per mole of an ideal gas is given as:

$k_{mol}=A_{v}\times k_{avg}$

here $A_{v}$ = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

$k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}$

$k_{mol}=3.74\times 10^{3}J/mol$

d) now at T = 143° C

$k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

$k_{mol}=5.1\times 10^{3}J/mol$

8 0

3 years ago

What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 43

kiruha [24]

Answer:

The magnification is $m = 0.3674$

Explanation:

From the question we are told that

The power of the lens is $P = -4.00 D(dioptre)$

Generally $1 dioptre = 1 \ meter$

The object distance is $u = -43 \ cm$ the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )

Generally the focal length is mathematically represented as

$f = \frac{1}{P}$

=> $f = \frac{1}{4.00 }$

=> $f = 0.25 \ m$

converting to cm

=> $f = 0.25 \ m = 0.25 * 100 = 25 \ cm$

Generally from lens equation we have that

$\frac{1}{f} +\frac{1}{v} -\frac{1}{u}$

=> $\frac{1}{25} +\frac{1}{v} -\frac{1}{-43}$

=> $v = -15.8 \ cm$

Generally the magnification is mathematically represented as

$m = \frac{v}{u}$

=> $m = \frac{- 15.8}{-43}$

=> $m = 0.3674$

6 0

2 years ago

Which sentence in the passage describes retrograde motion of a planet

Soloha48 [4]

Answer:

what is the passage? I don’t see it

Explanation:

4 0

3 years ago