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Brilliant_brown [7]
3 years ago
5

Can the sum of the magnitudes of two vectors ever be equal to the magnitude of the sum of the same two vectors? If no, why not?

If yes, when?
a-No, because of the angle between the two vectors.
b-No, it is impossible for the magnitude of the sum to be equal to the sum of the magnitudes.
c-Yes, if the two vectors are in the same direction.
d-Yes, if the two vectors are perpendicular.
e. Yes, if one of the vectors is zero.
Physics
1 answer:
Vikki [24]3 years ago
4 0

Answer:

c-Yes, if the two vectors are in the same direction.

e. Yes, if one of the vectors is zero.

Explanation:

Lets take

The magnitude of the two vectors are A and B and they are at angle θ

Then the Sum of these two vectors

\bar{R}=\bar{A}+\bar{B}

Resultant R

R=\sqrt{A^2+B^2+2ABcos\theta}

if the angle between these vectors is zero.It means that they are in the same direction.

θ = 0

R=\sqrt{A^2+B^2+2ABcos0}

R=\sqrt{A^2+B^2+2AB}

R=\sqrt{(A+B)^2}

R=A+B

If the one vector is zero vector.

Therefore the answer will be C and e.

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Answer

given,

gauge pressure =   1.94 x 10⁵ Pa

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( b )

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P = ρ g h

1.94 x 10⁵  = h x(1000) x (9.8)

h = 19.79 m

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Explanation:

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2 years ago
You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
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<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

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3 years ago
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