The strength of the electric field on the point charge at this distance will be 4000 V/m.
<h3>What is the strength of the electric field?</h3>
The strength of the electric field is the ratio of electric force per unit charge.
The given data in the problem is;
Qis the unit charge = 4.0 × 10⁻⁶ C
E is the strength of the electric field
R is the distance from point charge = 3 m
The strength of the electric field is;
Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.
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This means acceleration a is constant.
Let
a) vo be the initial speed, at t=0
b) v be the final speed after time t
c) d distance travelled in time t
Then we have:
a) v=vo+a×t
b) v²=vo²+2×a×d (Galilei's equation)
c) d=vo×t+a×t²/2
d) average speed vm=(vo+v)/2
Answer:
radiation heat.
Explanation:
That is any heat that passes through the air, like the sun.
Answer:
T=5797.8 K
Explanation:
Given that
Power P = 3.9 x 10²⁶ W
Radius ,r= 6.96 x 10⁸ m
We know that ,From Plank's law
P = σ A T⁴
σ = 5.67 x 10 ⁻⁸
A= Area ,T= Temperature ( in Kelvin)
Now by putting the values
T=5797.8 K
The temperature of surface will be 5797.8 K
