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kirill [66]
3 years ago
6

The forces acting on a falling leaf are.

Physics
1 answer:
mojhsa [17]3 years ago
3 0
B. Gravity and air resistance.
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Three arrows are shot horizontally. They have left the bow and are traveling parallel to the ground. Air resistance is negligibl
timurjin [86]

Answer:

F₁ = F₂ = F₃ = 0 N

Explanation:

given,

Arrow 1 mass = 80 g   speed = 10 m/s

Arrow 2 mass = 80 g   speed = 9 m/s

Arrow 3 mass = 90 g   speed = 9 m/s

Horizontal Force:- F₁ , F₂ and F₃

There is no air resistance.

If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.

We know,

According to newton's second law

        F = m a

If Acceleration is equal to zero

Then Force is also equal to zero.

Hence, F₁ = F₂ = F₃ = 0 N

4 0
3 years ago
Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.
kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees

ELECTRIC FORCE (F)

F = \frac{KQq}{d^{2} }

Where K = 9 x 10^{9} Nm^{2}/C^{2}

The distance between q_{1} and q_{3} can be calculated by using Pythagoras theorem.

d = \sqrt{33^{2} + 33^{2}  }

d = 46.7 cm = 0.467 m

For force F_{1}, substitute all the parameters into the formula above

F_{1} = (9 x 10^{9} x 3 x 1)/0.467^{2}

F_{1} = 2.7 x 10^{10}/0.218

F_{1} = 1.24 x 10^{11} N

For force F_{4}, substitute all the parameters into the formula above

F_{4} = (9 x 10^{9} x 3 x 4)/0.33^{2}

F_{4} = 1.08 x 10^{11}/0.1089

F_{4} = 9.92 x 10^{11} N

For force F_{2}, substitute all the parameters into the formula above

F_{2} = (9 x 10^{9} x 3 x 2)/0.33^{2}

F_{2} = 5.4 x 10^{10}/0.1089

F_{2} = 4.96 x 10^{11} N

Summation of forces on Y component will be

F_{y} = F_{4} - F_{1} Sin 45

F_{y} = 9.92 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{y} = 9.04 x 10^{11} N

Summation of forces on X component will be

F_{x} = F_{2} - F_{1} Cos 45

F_{x} = 4.96 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{x} = 4.08 x 10^{11} N

Net Force = \sqrt{F_{x} ^{2} + F_{y} ^{2}  } }

Net force = \sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2}  }

Net force = 9.9 x 10^{11} N

The direction will be

Tan ∅ = F_{y}/F_{x}

Tan ∅ = 9.04 x 10^{11} / 4.08 x 10^{11}

Tan ∅ = 2.216

∅ = Tan^{-1}(2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0
2 years ago
Read 2 more answers
A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have
DiKsa [7]

Answer:

4.4×10⁻⁷ Coulomb

Explanation:

V = Voltage = 5.8 kV

d = Potential distance = 2.8 mm = 0.0028 m

A = Area = 0.3×0.08 = 0.024 m²

ε₀ = permittivity constant in a Vacuum= 8.85×10⁻¹² F/m

\frac{Q}{V}=\frac{A\epsilon_0}{d}\\\Rightarrow \Q=V\frac{A\epsilon_0}{d}\\\Rightarrow Q=5.8\times 10^3\frac{0.024\times 8.85\times 10^{-12}}{0.0028}\\\Rightarrow Q=4.4\times 10^{-7}\ C

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

6 0
3 years ago
What's the degree to which molecules are able to pass through a membrane
lawyer [7]
The degree is 4678% to pass theough all membranes.
8 0
3 years ago
Traveling in circle requires a net force
seraphim [82]

Answer:

<u><em>Circular motion requires a net inward or "centripetal" force. Without a net centripetal force, an object cannot travel in circular motion. In fact, if the forces are balanced, then an object in motion continues in motion in a straight line at constant speed.</em></u>

Explanation:

3 0
3 years ago
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