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Minchanka [31]
3 years ago
11

A strong gust of wind blows an apple off a tree. The apple has a mass of 0.25 kg. The gust of wind pushes the apple with a force

of 1.05 N to the right. What is the net force on the apple?
5.5 N
- 2.2 N
1.78 N
2.66 N

Physics
2 answers:
LenKa [72]3 years ago
8 0

Answer:

2.66N

Explanation:

Nesterboy [21]3 years ago
3 0

For this case we must draw the free body diagram of the apple.

It is observed in the attached figure that we have the force of the wind acting to the right, in addition to the force of the weight that goes down.

Mg: It is the force exerted by the weight of the apple (M: It is the mass and g: It is the acceleration of gravity)

F_ {1}: It is the force exerted by the wind

Then, the net force will be given by:

F_ {n} = \sqrt {(Mg) ^ 2 + (F_ {1}) ^ 2}

We have to:

Mg = 0.25kg * 9.8 \frac {m} {s ^ 2} = 2.45N\\F_ {1} = 1.05N

Substituting we have:

F_ {n} = \sqrt {(2.45) ^ 2 + (1.05) ^ 2}\\F_ {n} = \sqrt {6.0025 + 1.1025}\\F_ {n} = \sqrt {7.105}\\F_ {n} = 2.66

Thus, the net force acting on the apple is 2.66N

Answer:

Option D

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Otion
Scorpion4ik [409]

Answer:

SKID

Explanation:

In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.

Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.

Y axis y

     N- W = 0

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        fr = m a

the acceleration in the curve is centripetal

         a = \frac{v^2}{r}

the friction force has the expression

        fr = μ N

we substitute

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       v = \sqrt{\mu g r}

we calculate

      v = \sqrt{0.1 \ 9.8 \ 3}

      v = 1,715 m / s

to compare with the cyclist's speed let's reduce to the SI system

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