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3 years ago

15

What relationship exists between the applied force and the green displacement vector for a constant k? Is it linear or quadratic

1 answer:

Oksana_A [137]3 years ago

3 0

Answer:

Explanation:

Using Hook's law

$F=-kx$

The relationship between applied force and green displacement is directly proportional.

It is linear relation.

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inysia [295]

Answer:

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2.Atomic bonds

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Explanation:

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3 years ago

Which objects would sink in honey, which has a density of 1.4 g/cm³? Check all that apply.

bazaltina [42]

Answer:

any object that has density more than 1.4

Explanation:

The object that has density more than 1.4 is denser than the honey

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4 years ago

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Kazeer [188]

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3 years ago

The power in an electric circuit varies inversely with the resistance. If the power is 2,200 watts when the resistance is 25 ohm

valentinak56 [21]

Answer:

The value of resistance when power is 1100 watts = $R_{2}$ = 50 ohms

Explanation:

Power $P_{1}$ = 2200 Watts

Resistance $R_{1}$ = 25 ohms

Power $P_{2}$ = 1100 Watts

Resistance $R_{2}$ = we have to calculate

Given that the power in an electric circuit varies inversely with the resistance

⇒ P ∝ $\frac{1}{R}$

⇒ $\frac{P_{2} }{P_{1} }$ = $\frac{R_{1} }{R_{2} }$

⇒ $\frac{1100}{2200}$ = $\frac{25}{R_{2} }$

⇒ $R_{2}$ = 50 ohms

This is the value of resistance when power is 1100 watts.

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4 years ago

A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax

Colt1911 [192]

Answer:

From the question we are told that

The length of the rod is $L_o$

The speed is v

The angle made by the rod is $\theta$

Generally the x-component of the rod's length is

$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

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3 years ago