Solar Radiation, The Solar Wind, and Gamma Ray Bursts
Answer:
Distance, d = 192 meters
Explanation:
We have,
Initial velocity of an object is 10 m/s
Acceleration of the object is 3.5 m/s²
Time, t = 8 s
We need to find the distance travelled by the object during that time. Second equation of motion gives the distance travelled by the object. It is given by :


So, the distance travelled by the object is 192 meters.
Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
Answer:
B. IT should have a very broad focus with many variables.