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Karolina [17]
2 years ago
13

The net force (unbalanced force) applied on an object is the product of_________and___________​

Physics
1 answer:
RSB [31]2 years ago
5 0

Answer:

of its <u>mass</u> and its <u>acceleration</u>

<u>Explanation:</u>

based on Newton's second law of motion

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A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar ec
Sonja [21]
1) <span>A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar eclipse or an -->
Answer: ANULAR ECLIPSE. Since the moon is too far, it will cover only a part of the sun, and only the external ring of the moon will be visible; this is called anular eclipse.

2) </span><span>anyone looking from the night side of earth can, in principle, see a -->
Answer: LUNAR ECLIPSE. If the moon is the right position, and the Earth's shadow covers partially or totally the moon, then a lunar eclipse occurs.

3) </span><span>during some lunar eclipses, the moon's appearance changes only slightly, because it passes only through the part of earth's shadow called the -->
Answer: PENUMBRA. 

4) </span><span>a ... can occur only when the moon is new and has an angular size larger than the sun in the sky -->
Answer: TOTAL SOLAR ECLIPSE. When the moon is new, it means it is between the sun and the Earth, and its dark side faces the Earth. If the moon's angular size is also larger than the sun angular size, than it will completely cover the sun, and a total solar eclipse occurs.

5) </span><span>a partial lunar eclipse begins when the moon first touches earth's --> 
Answer: SHADOW. The Earth's shadow will start to cover the moon, and partial lunar eclipse will start.

6) </span><span> a point at which the moon crosses earth's orbital plane is called a(n)  --> 
Answer: NODE. Eclipses occur only when the Moon is at or close to a node, otherwise sun, earth and moon are not "aligned".</span>
5 0
3 years ago
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:

1)   P₁ = -2 D,   2) P₂ = 6 D

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

1) to see a distant object it must be at infinity (p = ∞)

          \frac{1}{f_1} = \frac{1}{q}

           q = f₁

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

           

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

4 0
2 years ago
A spelunker is surveying a cave. She follows a passage that takes her a distance 184 m straight west, then a distance 220 m in a
Sever21 [200]
Refer to the diagram shown below.

Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.

The first distance is 184 m west. It is represented by
d₁ = -184 i

The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j

The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) =  102.42 i + 18.06 j

Let the fourth distance be 
d₄ = a i + b j

Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.

The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95

The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94

The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m

The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°

Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.

7 0
3 years ago
Read 2 more answers
A mass attached to a spring vibrates back and forth. At maximum displacement, the spring force and the
Svetradugi [14.3K]

ANSWER

Velocity of the mass reaches zero

EXPLANATION

We want to identify what hapens to a mass attached toa a spring at maximum displacement.

When a mass attached to a spring is at its maximum position of displacement, the direction of the mass begins to change. This implies that the velocity of the mass will reach zero.

Hence, at maximum displacement, the velocity of the mass reaches zero.

8 0
1 year ago
Question 1<br> 2 pts<br> Explain what causes a solution to be a strong acid.
lubasha [3.4K]

Answer:

Cuanto más fuerte es el ácido, más rápido se disocia para generar H +start superscript, plus, end superscript. Por ejemplo, el ácido clorhídrico (HCl) se disocia completamente en iones hidrógeno y cloruro cuando se mezcla con agua, por lo que se considera un ácido fuerte.

5 0
3 years ago
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