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Darya [45]
3 years ago
10

A toy car is tied to a string and pulled across a table horizontally. Which is the

Physics
2 answers:
Arturiano [62]3 years ago
7 0

Answer:

Correct answer is option D

Explanation:

In this question we have given that toy car is tied to a string and pulled across the table

Therefore there must be three forces which will act on toy car, these are as follows

1. Force due to gravity

2. Normal reaction force

3. Tension in string

Force due to gravity F_{g} will act in downward direction therefore to balance it there must be a normal force which will act in upward direction F_{norm} since the toy car is not accelerating in the y direction.

Since the car is accelerating in horizontal direction therefore tension force will act in the horizontal direction therefore, F_{t} will act to the right.

DerKrebs [107]3 years ago
4 0

Answer:

D

Explanation:

With F_{g} acting as a downward force, there must be a normal force counteracting it, because the toy car is not accelerating in the y direction. This force is the Normal Force or F_{norm}.

The Tension force must be going in the horizontal direction because the toy car is being pulled horizontally, so F_{t} mist be going to the right.

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Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The
Rudik [331]

Answer:

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Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image, respectively

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2) for an object located at p = 25 cm

            \frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        \frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}

        \frac{1}{f_2} = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = \frac{1}{f}

where the focal length is in meters

           

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3 years ago
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Answer:

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And this wavelength corresponds to red light.

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