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3 years ago

Which statement explains why a red rose does not appear to be orange?

Physics

1 answer:

navik [9.2K]3 years ago

7 0

Answer:

Explanation:

you see red because it's reflected and other colors are absorbed, light transmitted through something is when it travels through something like glass or a gem

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Brainliest!

USPshnik [31]

Answer:

C) 40,000 Joules

Explanation:

½(1000)10² - 10000 = 40000

3 0

3 years ago

The gravitational potential energy of a particle of mass m moving under the influence of a fixed mass M is given by - , where G

djverab [1.8K]

-GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.

Given

A particle of mass m moving under the influence of a fixed mass's M, gravitational potential energy of formula -GMm/r, where r is the separation between the masses and G is the gravitational constant of the universe.

As the Gravity Potential energy of particle = -GMm/r

Total energy of particle = Kinetic energy + Potential Energy

As we know that

Kinetic energy = 1/2mv²

Also, v is equals to square root of GM/r

v = √GM/r

Put the value of v in the formula of kinetic energy

We get,

Kinetic Energy = GMm/2r

Total Energy = GMm/2r + (-GMm/r)

= GMm/2r - GMm/r

= -GMm/2r

Hence, -GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.

Learn more about Gravitational Potential Energy here brainly.com/question/15896499

#SPJ4

3 0

1 year ago

Rearrange the equation to solve for the permeability of free space (μ0). Remember that the slope is the ratio of magnetic field

IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %

Answer:

The permeability of free space is $\mu_0 = 1.32*10^{-6} \ Tm/A$

The percentage error is % error = 5.25%

Explanation:

From the question we are told that

The slope is $s= 3.9\ G/A$ , and

Generally $1 \ gauss = 10^{-4 } tesla$

So $s = 3.9 *10^ {-4} T /A$

From the relation given in the question

$\mu_0 = \frac{2R}{N} [\frac{B}{I} ]$

Where R is the radius of the coil $=\frac{Diameter \ of \ coil }{2} = 0.017m$

N is the number of loops of the coil = 10

Now from the question we are told that

$s = \frac{B}{I}$

substituting into the equation above

$\mu_0 = \frac{2R }{N} s$

Substituting values

$\mu_0= \frac{2 * 0.017}{10 } * 3.9*10^{-4}$

$= 1.326 *10^ {-6} \ Tm/A$

Generally the % error is mathematically represented as

% $error = \frac{Measured \ value - accepted \ value}{accepted \ value } *100$

Given that the accepted value is $\mu_o_ {acc} = 1.26 *10 ^{-6} \ T \cdot m /A$

Hence substituting values

% $error = \frac{(1.326-1.26)*10^{-6}}{1.26 *10 ^ {-6}} *100$

$= 5.24$

6 0

3 years ago

The speed of light in vacuum is defined to be c = 299,792,458 m/s = 1 μ0ε0 . The permeability constant of vacuum is defined to b

Radda [10]

Explanation:

It is given that,

The speed of light in vacuum is, c = 299,792,458 m/s

The permeability constant of vacuum is, $\mu_o=4\pi\times 10^{-7}\ N.s^2/C^2$

Let $\epsilon_o$ is the permittivity of free space. The relation between $\mu_o,\epsilon_o\ and\ c$ is given by :

$c=\dfrac{1}{\sqrt{\mu_o\epsilon_o}}$

$\epsilon_o=\dfrac{1}{c^2u_o}$

$\epsilon_o=\dfrac{1}{(299792458\ m/s)^2\times 4\pi\times 10^{-7}\ N.s^2/C^2}$

$\epsilon_o=8.85\times 10^{-12}\ C^2/N.m^2$

Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

A particle’s position is ~r = (ct2 − 2dt3 )iˆ+ (2ct2 − dt2 )jˆ, where c and d are positive constants. Find the expressions for t

Mars2501 [29]

The velocity of the particle is given by the derivative of the position vector:

$\vec v = \dfrac{\mathrm d\vec r}{\mathrm dt} = (2ct-6dt^2)\,\vec\imath + (4ct-2dt)\,\vec\jmath$

(a) The particle is moving in the x-direction when the y-component of velocity is zero:

$4ct-2dt = 2t (2c - d) = 0 \implies t=0$

But we want t > 0, so this never happens, unless 2c = d is given, in which case the y-component is always zero.

(b) Similarly, the particle moves in the y-direction when the x-component vanishes:

$2ct-6dt^2 = 2t (c - 3dt) = 0 \implies t=0 \text{ or }c-3dt = 0$

We drop the zero solution, and we're left with

$c-3dt = 0 \implies c=3dt \implies \boxed{t = \dfrac c{3d}}$

In the case of 2c = d, this times reduces to t = c/(6c) = 1/6.

7 0

2 years ago