Answer:
8.9*10^6 V/m
Explanation:
The expression for electric field strength E is given as
where V= voltage
d= distance of separation
Given data
substituting our given data into the electric field strength formula we have
Answer:
Mass of the rocket, m = 20,000 kg Initial acceleration, a = 5 m/s Acceleration due to gravity, g = 10 m/s Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation: F – mg = ma F = m ( g + a ) = 20000 × (10 + 5)
Explanation:
Answer:A i think or D but its not c or b
Explanation:
Answer:
3) D: 31 m/s
4) D: 84.84 metres
Explanation:
3) Initial velocity along the x-axis is;
v_x = v_o•cos θ
Initial velocity along the y-axis is;
v_y = v_o•sin θ
Plugging in the relevant values, we have;
v_x = 31 cos 60
v_x = 31 × 0.5
v_x = 15.5 m/s
Similarly,
v_y = 31 sin 60
v_y = 31 × 0.8660
v_y = 26.85 m/s
Thus, magnitude of the initial velocity is;
v = √(15.5² + 26.85²)
v ≈ 31 m/s
4) Formula for horizontal range is;
R = (v² sin 2θ)/g
R = (31² × sin (2 × 60))/9.81
R = 84.84 m
Answer:
is uniquely defined.
Explanation:
The gravitational potential energy gained by an object being lifted is equal to the work done on the object by an applied force, which exactly cancels gravity. The applied force does positive work. When the object is being lifted, the gravitational force does negative work.
