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FromTheMoon [43]
4 years ago
15

If a vector A has components A. 0, and Ay -0, then the magnitude of the vector is negative. Select one: True False

Physics
2 answers:
Dima020 [189]4 years ago
8 0

Answer:

False

Explanation:

The magnitude of any vector is given by,

||A||=\sqrt{A_x^2+A_y^2}

The magnitude of anything is never negative. It can be even seen from the formula that the components are squared. A squared value can never be negative. Even if the component is negative the square will be always positive.

So, magnitude of the vector is <u>not</u> negative.

Anika [276]4 years ago
7 0

Answer:

The given statement is false.

Explanation:

For a given vector \overrightarrow{A}=A_{x}\widehat{i}+A_{y}\widehat{j}

The magnitude is given by |A|=\sqrt{A_{x}^{2}+A_{y}^{2}}

As we can see that the quantity on the right side of equation is always positive since square root of a quantity is always positive thus we conclude that the magnitude of any vector and hence vector A is always positive.

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Electromagnetic waves can transfer energy without a(n) _____________ (medium/electric field).
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Medium is matter, and electromagnetic waves do not need matter to travel. Just like the sun, its plasma light travels across space to reach us.
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If you were creating a lamp by hand what type of material would you chose for the wiring and what material would you chose to co
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A 100 N force is applied to a 500 kg crate resting on frictionless wheels.
Alla [95]

Answer:

<h2>0.2 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{100}{500}  =  \frac{1}{5}  = 0.2 \\

We have the final answer as

<h3>0.2 m/s²</h3>

Hope this helps you

6 0
2 years ago
Read 2 more answers
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
For numbers 2 and 3, which one is compression and which one is rarefaction?
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