The ray diagram for the given object consists of 2 cm height of object, 4 cm object distance and 3 cm focal length.
<h3>Image formed by a diverging lens</h3>
Diverging lens is called a concave lens. The working of the lens is dependent on the refraction of the light rays as they pass through the lens.
Image formed by a diverging lens is always virtual, erect and diminished; smaller than the object and is located on the same side of the lens as the object.
The ray diagram for the given object is presented in the image in the diagram.
- Object height = 2 cm
- Focal length = 3 cm
- Object distance = 4 cm
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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
Calculation of the electric field at point P due to q1
Ep₁x=0
Calculation of the electric field at point P due to q2
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
This question apparently wants you to get comfortable
with E = m c² . But I must say, this question is a lame
way to do it.
c = 3 x 10⁸ m/s
E = m c²
1.03 x 10⁻¹³ joule = (m) (3 x 10⁸ m/s)²
Divide each side by (3 x 10⁸ m/s)²:
Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)
= (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)
= 1.144 x 10⁻³⁰ kg . (choice-1)
This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature. The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________
Something like this could have been much more impressive:
The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?
Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)
= (2,242 x 10⁶ x 86,400 x 365) joules
= 7.0704 x 10¹⁶ joules .
How much converted mass is that ?
E = m c²
Divide each side by c² : Mass = E / c² .
c = 3 x 10⁸ m/s
Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)
= 0.786 kilogram ! ! !
THAT should impress us ! If I've done the arithmetic correctly,
then roughly (1 pound 11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !
</span>
Answer:
Fc = 19.2 N
Explanation:
In this case, the force of the Honda over the rock, is a centripetal force. Then, you have:
m: mass of the rock = 600g = 0.6 kg
v: tangential velocity of the Honda = 4m/s
r: radius of the Honda = 50cm = 0.5m
You replace the values of m, r and v in the equation for Fc:
hence, the force has a magnitude of 19.2 N
If the rock would have more mass the centripetal force would be higher
