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:
Lauch velocity (u) = 26.15 m/s
Lauch Angle (θ) = 35°
Explanation:
From the question given above, the following data were obtained:
Range (R) = 65 m
Time of flight (T) = 3 s
Acceleration due to gravity (g) = 10 m/s²
Lauch velocity (u) =?
Lauch Angle (θ) =?
R = u²Sin2θ /g
65 = u² × Sin2θ /10
Recall:
Sin2θ = 2SinθCosθ
65 = u² × 2SinθCosθ / 10
65 = u² × SinθCosθ / 5
Cross multiply
65 × 5 = u² × SinθCosθ
325 = u² × SinθCosθ .....(1)
T = 2uSinθ / g
3 = 2uSinθ / 10
3 = uSinθ / 5
Cross multiply
3 × 5 = uSinθ
15 = u × Sinθ
Divide both side by Sinθ
u = 15 / Sinθ....... (2)
Substitute the value of u in equation (2) into equation (1)
325 = u² × SinθCosθ
u = 15 / Sinθ
325 = (15 / Sinθ)² × SinθCosθ
325 = 225 / Sin²θ × SinθCosθ
325 = 225 × SinθCosθ / Sin²θ
325 = 225 × Cosθ / Sinθ
Cross multiply
325 × Sineθ = 225 × Cosθ
Divide both side by Cosθ
325 × Sineθ / Cosθ = 225
Divide both side by 325
Sineθ / Cosθ = 225 / 325
Sineθ / Cosθ = 0.6923
Recall:
Sineθ / Cosθ = Tanθ
Tanθ = 0.6923
Take the inverse of Tan
θ = Tan¯¹ 0.6923
θ = 35°
Substitute the value of θ into equation (2) to obtain the value of u.
u = 15 / Sinθ
θ = 35°
u = 15 / Sin 35
u = 15 / 0.5736
u = 26.15 m/s
Summary:
Lauch velocity (u) = 26.15 m/s
Lauch Angle (θ) = 35°
That would be kinetic energy
Answer:
The larger star will burn out faster. as nuclear fuel is burned out due to higher percentage being consumed.
Explanation:
Gravity on the other hand is only affected a little during this process and mass use. Solar mass loss is caused by mass that is turned into energy (E=mc²), but also by the material that is lost to the solar wind.
