Define displacement and I'll help you
Te direction of the magnetic force for the velocity of the proton in the
-ve y direction will be +ve z direction.
As we know that the right-hand rule is based on the relation of magnetic fields and the forces that they exert on moving charges.When a charged particle moves under a magnetic field, it exerts a force on the particle, which is not in the same direction but different than the direction of the magnetic field.Under the right-hand rule, if we point our pointer finger in the direction of the charged particle is moving and the middle finger is representing the direction of the magnetic field then our thumb depicts the direction of the magnetic force which is exerted on the charged particle.
So, we are given that the direction of the velocity of the proton is in the negative y direction and the direction of the magnetic field is in the positive x direction, so the magnetic force is acting in the positive z direction.
To know more about the right-hand rule refer to the link brainly.com/question/9750730?referrer=searchResults.
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Answer:
The work done to get you safely away from the test is 2.47 X 10⁴ J.
Explanation:
Given;
length of the rope, L = 70 ft
mass per unit length of the rope, μ = 2 lb/ft
your mass, W = 120 lbs
mass of the 70 ft rope = 2 lb/ft x 70 ft
= 140 lbs.
Total mass to be pulled to the helicopter, M = 120 lbs + 140 lbs
= 260 lbs
The work done is calculated from work-energy theorem as follows;
W = Mgh
where;
g is acceleration due gravity = 32.17 ft/s²
h is height the total mass is raised = length of the rope = 70 ft
W = 260 Lb x 32.17 ft/s² x 70 ft
W = 585494 lb.ft²/s²
1 lb.ft²/s² = 0.0421 J
W = 585494 lb.ft²/s² = 2.47 X 10⁴ J.
Therefore, the work done to get you safely away from the test is 2.47 X 10⁴ J.
Answer:
t₁ > t₂
Explanation:
A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂, t₁ = t₂, t₁ >> t₂ , t₂ > t₁
Solution:
Newton's law of motion is given by:
s = ut + (1/2)gt²;
where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.
u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.
hence:
When stationary:
Hence t₂ < t₁, this means that t₁ > t₂.
