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3 years ago

Why is Vy negative and not positive because I solved it and it resulted in a positive number?

Physics

1 answer:

11Alexandr11 [23.1K]3 years ago

8 0

Explanation:

You are given the initial velocity, the displacement, and the acceleration. You're looking for the final velocity. So you use the equation:

v² = v₀² + 2aΔy

When you solve for v, you take the square root. Your calculator will return a positive answer, but there are actually two possible answers: positive and negative.

v = ±√(v₀² + 2aΔy)

You must use the physical context of the problem. If we take up to be the positive direction, then v must be negative, since the projectile is moving down.

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The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an

yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

$\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}$

$\frac{1}{f}=0.028$

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

$p=\frac{1}{f}$

Here, p is the power of the lens.

Put f= 35.71 cm.

$p=\frac{1}{35.71}$

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

3 0

3 years ago

S Four point charges each having charge Q are located at the corners of a square having sides of length a. Find expressions for(

Ket [755]

The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.

<h3>What do you mean by electric potential? </h3>

The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.

V = kq/r

Where V represents electric potential, K is coulomb constant, q is Charge and r is distance between any two around charge to the point charge.

Electric potential at O due to four charges is given by,

V = 4KQ/ r

where, r = √2a/2 = a/√2

V = 4k × Q√2/a

V = √2Q/πÈa

The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.

To learn more about electric potential refer to:

brainly.com/question/12645463

#SPJ4

3 0

2 years ago

Nerve impulses in the human body travel at a speed of about 100 m/s. A 1.6 m tall man accidentally drops a hammer on his toe. Ho