I dont see it ;-; REEEEEEEE
I think its a higher frequency
Answer:
The Acceleration of the object = 6.4 m/s²
<u>Explanation:</u>
Mass of block (m) = 5 kg
Action force on block, (F₁) = 40 N
<u>To Find:</u>
Acceleration of the object (a) = ?
<u>Required solution:</u>
Frictional force opposing the motion (F₂) = 8 N
Here in this question we have to find Acceleration of the object. So, firstly we have to find Net force of block after that we will find Acceleration of the object on the basis of conditions given above
⇒ Net force = Action force on block - Opposing friction force
⇒ F = F₁ - F₂
⇒ F = 40 - 8
⇒ F = 32 N
Now, we have to two elements that used in formula, Net force and Mass of block.
Net force of the block (F) = 32 N
Mass of block (m) = 5 kg
And we have to find Acceleration of the object.
We can find Acceleration of the object by using the Second law of Newton which says F = ma
Here,
F is the Force in N.
m is the Mass in kg.
a is the Acceleration in m/s².
So let's find Acceleration (a) !
† From second law of Newton
⇛ F = ma
⇛ a = F/m
⇛ a = 32/5
⇛ a = 6.4 m/s²
Answer:
1.424 μC
Explanation:
I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),
tension T (acting along the string - to the pivot point), and
F (electric force – acting along the line connecting the charges).
We then have something like this
x: T•sin α = F,
y: T•cosα = mg.
Dividing the first one by the second one we have
T•sin α/ T•cosα = F/mg, ultimately,
tan α = F/mg.
Since we already know that
q1=q2=q, and
r=2•L•sinα,
k=9•10^9 N•m²/C²
Remember,
F =k•q1•q2/r², if we substitute for r, we have
F = k•q²/(2•L•sinα)².
tan α = F/mg =
= k•q²/(2•L•sinα)² •mg.
q = (2•L•sinα) • √(m•g•tanα/k)=
=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =
q = 0.486 • √(8.61•10^-12)
q = 0.486 • 2.93•10^-6
q = 1.424•10^-6 C
q = 1.424 μC.
