Determine the thrust that a boat with a volume of 1.2m³ receives when it is stranded at sea. The density of seawater is 1020kg /
1 answer:
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Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;
E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C
Answer:
2.87 km/s
Explanation:
radius of planet, R = 1.74 x 10^6 m
Mass of planet, M = 7.35 x 10^22 kg
height, h = 2.55 x 10^6 m
G = 6.67 x 106-11 Nm^2/kg^2
Use teh formula for acceleration due to gravity
g = 1.62 m/s^2
initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0
Use third equation of motion
0 = v² - 2 x 1.62 x 2.55 x 10^6
v² = 8262000
v = 2874.37 m/s
v = 2.87 km/s
Thus, the initial speed should be 2.87 km/s.