According to the kinetic-molecular theory, gases condense into liquids because of

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

A Car with drives into a solid object with 70 km/h, from which height would it fall if it was in free fall?

Ilya [14]

Answer:

18.9 m.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 70 km/h

Height (h) =?

Next, we shall convert 70 km/h to m/s. This can be obtained as follow:

3.6 km/h = 1 m/s

Therefore,

70 km/h = 70 km/h × 1 m/s / 3.6 km/h

70 km/h = 19.44 m/s

Finally, we shall determine the height. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 19.44 m/s

Acceleration due to gravity (g) = 10 m/s²

Height (h) =?

v² = u² + 2gh

19.44² = 0² + (2 × 10 × h)

377.9136 = 0 + 20h

377.9136 = 20h

Divide both side by 20

h = 377.9136 / 20

h = 18.9 m

Thus, the car will fall from a height of 18.9 m

8 0

3 years ago

An egg is dropped from the top of the band hall. if the band hall is 25 m tall, determine the time it takes the egg to hit the f

Vera_Pavlovna [14]

Answer:

2.26 s

Explanation:

The following data were obtained from the question:

Height (h) = 25 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =..?

The time taken for the egg to hit the floor can be obtained as illustrated below:

h = ½gt²

25 = ½ × 9.8 × t²

25 = 4.9 × t²

Divide both side by 4.9

t² = 25 / 4.9

Take the square root of both side

t = √(25 / 4.9)

t = 2.26 s

Thus, it will take 2.26 s for the egg to hit the floor.

7 0

3 years ago

A car weighing 12,000 N is parked on a 36° slope. The car starts to roll down the hill. What is the acceleration of the car?

Delicious77 [7]

Answer: 5.8 m/s squared

Explanation: just got that question lol

4 0

2 years ago

Where are you on Earth if you experience each of the following? (Refer to the discussion in Observing the Sky: The Birth of Astr

Aloiza [94]

Explanation:

We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.

Condition 1: The stars rise and set perpendicular to the horizon

The observer is at the equator

Condition 2: The stars circle the sky parallel to the horizon

The observer is at the Pole of the Earth

Condition 3: The celestial equator passes through the zenith

The observer is at the equator

Condition 4: In the course of a year, all stars are visible

The observer is at the equator

Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)

The observer is at North Pole

7 0

3 years ago