Answer:
D. −F
Explanation:
the rest of the answers are
2/3F
The force is represented as a positive quantity and is repulsive.
Electrostatic force is inversely proportional to the square of the distance.
The direction of the force changes, and the magnitude of the force quadruples.
hope this helps sorry if i was too late! :)
Answer:
The arrow will leave the bow with a velocity of 10 m/s.
Explanation:
Hi there!
The potential energy stored in the bow can be calculated using the following equation:
U = 1/2 · k · d²
Where
U = elastic potential energy.
k = spring constant.
d = stretched distance of the bow
Then:
U = 1/2 · 112 N/m · (0.29 m)²
U = 4.7 J
When the bow is released, the potential energy is transformed into kinetic energy. Then, the kinetic energy of the arrow when it leaves the bow will be:
KE = 1/2 · m · v² = 4.7J
Where:
KE = kinetic energy.
m = mass of the arrow.
v = velocity of the arrow:
Then:
4.7 J = 1/2 ·0.094 kg · v²
2 · 4.7 J / 0.094 kg = v²
9.4 kg · m²/s² / 0.094 kg = v²
v = 10 m/s
The arrow will leave the bow with a velocity of 10 m/s.
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:
a) i) x = 0.25 m, ii) x = 0.10 m, iii) x = 0.050 m
b) i) x = 0.40 m
Explanation:
a) For this exercise we use the rotational equilibrium equation, where we assume that the anticlockwise rotations are positive.
1) L = 2W
we set our reference system in the center of the bar where the fulcrum is
∑τ = 0
W 0.50 - L x = 0
x = 0.50 W / L
we substitute the value
x = 0.50 W / 2W
x = 0.25 m
ii) L = 5W
we calculate
x = 0.50 W / 5W
x = 0.10 m
iii) L = 10 W
x = 0.50 W / 10W
x = 0.050 m
b) a new weight is placed at x₂ = 30 cm on the left side
W 0.50 + W 0.30 - L x = 0
x = (0.50 + 0.30) W / L
x = 0.80 W / L
we calculate
i) L = 2W
x = 0.80 w / 2w
x = 0.40 m