The increase the magnitude of the electric force ;
1) The charges on the objects must be increased
2) The distance of separation between the objects must be reduced.
<h3>
What is electric field?</h3>
The term electric field has to do with the region where the electric force on an object is expirieced. We must the note thate electric force is a result of the influence of a charge on another.
The increase the magnitude of the electric force ;
1) The charges on the objects must be increased
2) The distance of separation between the objects must be reduced.
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The discovery of it was so important to the study of geology and the study of Evolution because before that, researchers could only determine the relative orders of rock unit and fossils. They can only estimate the length of time involved , but failed to provide any further information
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Given Information:
Mass of electron = m = 9x10⁻³¹ kg
initial speed of electron = v₁ = 0.92c
Force = F = 1.4x10⁻¹³ J
Distance = d = 3 m
Required Information:
Final speed of electron = v₂ = ?
Answer:
Final speed of electron = v₂ = 2.974x10⁸ m/s
Explanation:
As we know from the conservation of energy,
E₂ - E₁ = W
E₂ = E₁ + W
Where E₂ is the final energy of electron and E₁ is the initial energy of electron
The above equation can be written in the form of particle energy
γ₂mc² = γ₁mc² + W
where γ₁ and γ₂ are given by
γ₁ = 1/√1 - (v₁/c)²
γ₂ = 1/√1 - (v₂/c)²
First calculate γ₁
γ₁ = 1/√1 - (0.92c/c)²
γ₁ = 2.55 m
Now calculate γ₂
γ₂ = (γ₁mc² + W)/mc²
First we need to find the work done
W = F*d
W = 1.4x10⁻¹³*3
W = 4.2x10⁻¹³ J
so γ₂ is
γ₂ = (2.55*9x10⁻³¹*(3x10⁸)² + 4.2x10⁻¹³)/9x10⁻³¹*(3x10⁸)²
γ₂ = 7.73
Now we can find the new speed of the electron
γ₂ = 1/√1 - (v₂/c)²
Re-arranging the above equation results in
v₂ = c*√(1 - 1/γ₂²)
v₂ = 3x10⁸*√(1 - 1/7.73²)
v₂ = 2.974x10⁸ m/s
Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance
∴ Area of each plate is 7.9060 m²
Voltage
∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb
Answer:
Explanation:
To calculate the period we need the formula:
Where 
is the radius of the moon, 
is the universal constant of gravitation and 
is the mass of mars.
The period of Phobos:
The period of Deimos:
The ratio of the period of Phobos and Deimos:
Most terms get canceled and we have:
According to the problem
so the ratio will be:
≈
the ratio of the period of revolution of Phobos to that of Deimos is 0.2528
