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3 years ago

What happens when a magenta light is shone on a green surface?

Physics

1 answer:

Sedbober [7]3 years ago

5 0

Answer:

When Magenta light is shown on a green surface, it looks black.

Explanation:

It absorbs the Magenta light and also reflects none of the light.

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Instantaneous speed is measured

VMariaS [17]

Answer:

C. At a particular instant

Explanation:

Speed is the defined as the ratio between the distance covered by an object and the time taken:

$v=\frac{d}{t}$

where d is the distance and t the time.

However, there are two possible measurements of speed:

- Average speed: this is the speed measured over a non-zero time interval (for example: a car moving 100 metres in 5 seconds; its average speed is

$v=\frac{100 m}{5 s}=20 m/s$

- Instantaneous speed: this is the speed of an object measured at a particular instant in time, so for a time interval that tends to zero. So, in the previous example, the average speed is 20 m/s but the instantaneous speed of the car at various instants of time can be different from that value.

7 0

3 years ago

A wire is wrapped 200 times around a hollow cylinder of cross-sectional area 12 cm2 and length 10 cm. Assuming the turns are spr

Sonja [21]

Answer:

The answer is 10.857mJ

Explanation:

The energy stored in this solenoid is given by the below mentioned equation,

$U = L*I^{2}/2$

where L the inductance of this solenoid is given by the below mentioned equation,

$L = u_{0} *N^{2}*A/L$

Plugging this into the energy equation you obtain the equation for the total energy stored in the magnetic field of the solenoid, given by,

$U = u_{0} *N^{2}*A*I^{2}/(2*L)$

where $u_{0}$ is the permeability of free space which equals to $u_{0} = 4*\pi *10^{-7} T.m/A$ . Plugging all the quantities into the above equation from the data in the question after converting to standard units. of meters instead of centimeters, we get for the energy stored in the coil,

$U = 10.857mJ$

7 0

4 years ago

2. A bus drove 8 meters East, then turned to drive 8 meters North, then 2 meters

Sphinxa [80]

Answer:

Long question good luck:)

Explanation:

3 0

3 years ago

A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .

Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}$

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
$B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}$

Taking out constants from the integral:
$B =\frac{\mu_0 i}{4\pi R^2} \int ds$

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

$B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds$

Evaluate:
$B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}$

Plugging in our givens to solve for the magnetic field strength of one loop:

$B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T$

Multiply by the number of loops to find the total magnetic field:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}$

Using the diagram, if 'z' is the point's height from the center:

$r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}$

Substituting this into our expression:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }$

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds$

Evaluate:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Multiplying by the number of loops:
$B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Plug in the given values:
$B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}$

5 0

2 years ago

Read 2 more answers

Where is there potential energy throughout the loading, cocking, and releasing if the trebuchet?

pshichka [43]

Answer:

adapted from NOVA, a team of historians, engineers, and trade experts recreate a medieval throwing machine called a trebuchet. To launch a projectile, a trebuchet utilizes the transfer of gravitational potential energy into kinetic energy. A massive counterweight at one end of a lever falls because of gravity, causing the other end of the lever to rise and release a projectile from a sling. As part of their design process, the engineers use models to help evaluate how well their designs will work.

Explanation:

4 0

4 years ago