Answer:
x ’= 1,735 m, measured from the far left
Explanation:
For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.
Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive
They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,
the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar
x_{cm} = 1.2 -1
x_ {cm} = 0.2 m
Σ τ = 0
w₁ 1.2 + mg 0.2 - W₂ x = 0
x = 
x = 
let's calculate
x = 
2.9 1.2 + 4 0.2 / 8
x = 0.535 m
measured from the pivot point
measured from the far left is
x’= 1,2 + x
x'= 1.2 + 0.535
x ’= 1,735 m
<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
Here is the answer of the given problem above.
Use this formula: <span>P = FV = ma*at = ma^2 t
</span><span>Substitute the values, and therefore, we got m(a0)^2t = m(x)^2 (2t)
then, solve for x which is the acceleration at 2t.
</span>The <span>answer would be a0/sqrt(2).
Hope this answers your question. Thanks for posting.
</span>
Answer:
The answer to your question is vo = 5.43 m/s
Explanation:
Data
distance = d= 5.8 m
height = 3 m
height 2 = 1.7 m
angle = 60°
vo = ?
g = 9.81 m/s²
Formula
hmax = vo²sinФ/ 2g
Solve for vo²
vo² = 2ghmax / sinФ
Substitution
vo² = 2(9.81)(3 - 1.7) / 0.866
Simplification
vo² = 19.62(1.3) / 0.866
vo² = 25.51 / 0.866
vo² = 29.45
Result
vo = 5.43 m/s
Answer:
<em>Answer: positive velocity & negative acceleration</em>
Explanation:
<u>Accelerated Motion</u>
Both the velocity and acceleration are vectors because they have magnitude and direction. When the motion is restricted to one dimension, i.e. left-right or up-down, the direction is marked with the sign according to some preset reference.
The locomotive is moving at a certain speed with a (so far) unknown sign but the acceleration has a negative sign. Since the locomotive comes to a complete stop it means the velocity and the acceleration are of opposite signs.
Thus the velocity is positive.
Answer: positive velocity & negative acceleration
