The diagram shows the electric field lines around two charges. Based on the field lines, what is the best description of the cha
rges ?
A.) they are negative, and they repel
B.) the left is negative and the right is positive, so they attract.
C.) the left is positive and the right is negative, so they attract.
D.) they are positive, they repel
A.) they are negative, and they repel
B.) the left is negative and the right is positive, so they attract.
C.) the left is positive and the right is negative, so they attract.
D.) they are positive, they repel
2 answers:
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Answer:
Explanation:
We are asked to find the final velocity of the boat.
We are given the initial velocity, acceleration, and time. Therefore, we will use the following kinematic equation.
The initial velocity is 2.7 meters per second. The acceleration is 0.15 meters per second squared. The time is 12 seconds.
= 2.7 m/s
- a= 0.15 m/s²
- t= 12 s
Substitute the values into the formula.
Multiply the numbers in parentheses.
Add.
The final velocity of the boat is <u>4.5 meters per second in the positive direction.</u>
Answer:
4x
Explanation:
Use to do the question.
For first instance,
-------------------( 1 )
for second instance,
-----------------( 2 )
So by (1) and (2),
s = 4x
<u>Given data:</u>
m= 50 Kg,
W= m×g = 50 × 9.81 = 490.5 N
ramp angle (α) = 30 degrees,
coefficient of friction (μs) = 0.5773,
Push at an angle (Θ) = 40 degrees,
Determine: Push to get box move up (P)=?
From the figure,
Resolving the forces along the plane
W sinα + μs.R = P cos Θ --------------------- (i)
Resolving the forces perpendicular to the inclined plane
W cosα = R+Psin Θ => R= W cosα - Psin Θ -------------- (ii)
Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>
<em>Pmin = W sin( α +Θ )</em>
<em> = W[ sin α.Cos Θ + cos α.sin Θ]</em>
<em> = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>
<em> = 460.9 N</em>
<em>Minimum push required to move the box up the ramp is 460.9 N</em>
