<h2><u>Question</u><u>:</u><u>-</u></h2>
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?
<h2><u>Answer:</u><u>-</u></h2>
<h3>Given,</h3>
=> Force applied by Ryan = 10N
=> Distance covered by the book after applying force = 30 cm
<h3>And,</h3>
30 cm = 0.3 m (distance)
<h3>So,</h3>
=> Work done = Force × Distance
=> 10 × 0.3
=> 3 Joules
Answer:
19.1 deg
Explanation:
v = speed of the proton = 8 x 10⁶ m/s
B = magnitude of the magnetic field = 1.72 T
q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C
F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N
θ = Angle between proton's velocity and magnetic field
magnitude of magnetic force on the proton is given as
F = q v B Sinθ
7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ
Sinθ = 0.327
θ = 19.1 deg
The lithosphere because it includes the outer region of the earth including the crust and outer mantle
Answer:
D. "The net force is zero, so the acceleration is zero"
Explanation:
edge 2020
