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3 years ago

Turning a vector to a scalar is possible ???

Physics

1 answer:

sdas [7]3 years ago

3 0

Answer: Yes it is possible to turn a vector to a scalar.

Explanation:

Vectors and scalars represent different types of physical quantities. However, it is sometimes necessary for them to interact. Adding a scalar to a vector is impossible because of their different dimensions in space. Although It is possible to multiply a vector by a scalar.

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2 years ago

A student walks 1.0 mi west and then 1.0 mi north. afterward, how far is she from her starting point?

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1.0 mi because as he has covered 1 the disrance of 1 m and then 1 m in north so he can go straight 1m north so the distance is actually 1m from the starting point

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3 years ago

A wave that is traveling fast can be said to have a high ___

S_A_V [24]

A wave that is traveling fast can be said to have a high speed.<em> (b) </em>

Just like a car, motorcycle, or freight train that is traveling fast.

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3 years ago

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

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3 years ago

An object is five focal lengths from a concave mirror.how do the object and image heights compare?

enot [183]

An object distance is presented as s = 5f and we know that the mirror equation relates the image distance to the object distance and the focal length.

The mirror equation is 1/f = 1/s + 1/s’ where the variable f stands for the focal length of the mirror. Variable (s) represents the distance between the mirror surface and the object and the variable <span>(s’) represents the distance between the mirror surface and the image. </span>

In addition, a concave mirror will have a positive focal length (f) and a convex mirror will have a negative focal length (f).

Now, we then have 1/f = 1/5f + 1/s’ which is s’ = 5f/4

Then we get the magnification ratio that expresses the size or amount of magnification or reduction of the object or image and to get the magnification, we use this equation: M= s’/s

M= 5f/4x5f

s’ = 1/4s

Therefore, the image height is one fourth of the object height

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3 years ago

Turning a vector to a scalar is possible ???​

Turning a vector to a scalar is possible ???