Answer: 200 J
Explanation: In order to explain this we have consider that the work done in a electric field is given by:
Work= Q*ΔV=2*100=200J
Answer:
The magnitude of F1 is
The magnitude of F2 is
And the direction of F2 is
Explanation:
<u>Net Force
</u>
Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.
The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.
Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:
The sum of the vertical components of F1 and F2 must equal the total force Ft
Solving for in the first equation
The magnitude of F1 is
The magnitude of F2 is
And the direction of F2 is
Answer:
v₂=- 34 .85 m/s
v₁=0.14 m/s
Explanation:
Given that
m₁=70 kg ,u₁=0 m/s
m₂=0.15 kg ,u₂=35 m/s
Given that collision is elastic .We know that for elastic collision
Lets take their final speed is v₁ and v₂
From momentum conservation
m₁u₁+m₂u₂=m₁v₁+m₂v₂
70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂
70 x v₁ + 0.15 x v₂=5.25 --------1
v₂-v₁=u₁-u₂ ( e= 1)
v₂-v₁ = -35 --------2
By solving above equations
v₂=- 34 .85 m/s
v₁=0.14 m/s