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2 years ago

A car speeds up from 18.54 m/s to 29.52 m/s in 13.84 s. The acceleration of the car is:

Physics

2 answers:

valkas [14]2 years ago

5 0

Answer:

.7934 $m/s^{2}$

Explanation:

Acceleration = change in velocity / change in time

A = 10.98 $m/s$ / 13.84 $s$

A = .7934 $m/s^{2}$

raketka [301]2 years ago

3 0

Answer:0.8 m/s^2

Explanation:

initial velocity(u)=18.54m/s

Final velocity(v)=29.52m/s

Time(t)=13.84 sec

Acceleration =(v-u)/t

acceleration =(29.52-18.54)/13.84

Acceleration =10.98/13.34

Acceleration=0.8 m/s^2

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Describe a light wave and explain how light wave travel through solids liquids and gasses

marin [14]

Through refraction , it bends as it passes into a solid object

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2 years ago

Determine the mass of Ar in 1.00 liter of a gas mixture at 25oC which contains 0.300 atm of Ne and has a total pressure of 4.00

sergejj [24]

Answer:

The mass of Ar is 36.91g

Explanation:

The gas mixture consist of Neon(Ne) and Argon(Ar)

Partial pressure of Ar = total pressure of mixture - partial pressure = 4 - 0.3 = 3.7 atm

Mole fraction of Ar = partial pressure of Ar ÷ total pressure of mixture = 3.7/4 = 0.925

Mass of Ar = 0.925 × molecular weight of Ar = 0.925 × 39.9 = 36.91g

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3 years ago

If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average

Nina [5.8K]

Answer:

Average accelation = -4V

Explanation:

$a=\frac{V-V0}{t}$

V=0 m/s (because the frog stopped)

V0 = V (average velocity)

t= 0,25 s

So;

$a=\frac{V-V0}{t}=\frac{0-V}{0.25}=-4V$

4 0

2 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

2 years ago

Two cars are initially separated by 2500 m and traveling towards each other. One car travels at 4.5 m/s and the second car trave

mylen [45]

Answer:

714.285s

Explanation:

use relative velocity

8-4.5 = 3.5m/s

x = 2500m

2500/3.5 = 714.285s = 700s (with sig figs)

7 0

2 years ago