Answer:
It will increase by a factor of 8.
Explanation:
By Ohm's Law, we can relate current, voltage and resistance. It is expressed as V=IR. That is, there is a direct relationship between voltage and resistance and voltage and current.
V = IR
V1/2V1 = I1R1 / I2R1/4
1/2 = 4I1/I2
I2 = 8I1
The Electric field is E=23400 V/m.
What is electric potential ?
The amount of labor required to convey a unit of electric charge from a reference point to a given place in an electric field is known as the electric potential (also known as the electric field potential, potential drop, or the electrostatic potential). More specifically, it is the energy per unit charge for a test charge that is negligibly disruptive to the field under discussion.
The electric field of a spherical conductor is given by:
Here 'E' is the electric field and 'Q' is the electric charge.
therefore putting the values we get,
E=23400 V/m
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Answer:
ΔV=0.484mV
Explanation:
The potential difference across the end of conductor that obeys Ohms law:
ΔV=IR
Where I is current
R is resistance
The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A
R=(pL)/A
Given data
Length L=3.87 cm =0.0387m
Diameter d=2.11 cm =0.0211 m
Current I=165 A
Resistivity of aluminum p=2.65×10⁻⁸ ohms
So
ΔV=IR
ΔV=0.484mV
Answer:
v = 2.22 m/s
Explanation:
First we apply the second equation of motion to the vertical motion of the body:
s = Vi t + (1/2)gt²
where,
s = y = vertical distance covered = 200 m - 100 m = 100 m
Vi = V₀y = Vertical Component of Initial Velocity = 0 m/s (because spider man jumps horizontally, thus his velocity has no vertical component initially)
t = Time Taken to Land on 100 m high building = ?
g = 9.8 m/s²
Therefore,
100 m = (0 m/s)t + (0.5)(9.8 m/s²)t²
t² = (100 m)/(4.9 m/s²)
t = √(20.4 s²)
t = 4.5 s
Now, we analyze the horizontal motion. Neglecting air friction, the horizontal motion is uniform with uniform velocity. Therefore,
s = vt
where,
s = x = horizontal distance covered = 10 m
v = V₀ₓ = Horizontal Component of Initial Velocity = Initial Velocity = ?
Therefore,
10 m = v(4.5 s)
v = 10 m/4.5 s
<u>v = 2.22 m/s</u>
The net force acting on test charge is resultant of two force i.e., √2 F
The magnitude of force on test charge due to one charge Q is:
F = k qQ / r²
where k = 9 × 10^9
Similarly the force due to second charge will be:
F = k qQ / r²
Now both the charges of magnitude Q is present and it is pointed at right angle to each other then the force will be resultant i.e.,
F = √ F² + F² = F√2
Hence, net force acting on test charge is √2 F.
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