All categories

4 years ago

What two parts of the eye focus light onto the retina

Physics

1 answer:

melamori03 [73]4 years ago

3 0

Answer: the "cornea" ; and the "lens" .
____________________________________________

You might be interested in

Help will mark brain list

Hatshy [7]

A jagged line represents a resistor .

3 0

2 years ago

Mars has a mass of about 6.4 x 1023 kg, and its moon Phobos has a mass of about 9.6 x 1015 kg. If the magnitude of the gravitati

Sliva [168]

Answer:9.55×10^-4 metre

Explanation:

8 0

3 years ago

Name two kinds of longitudinal waves and explain how you know they are longitudinal

jasenka [17]

First of all Longitudinal waves is a matter in the medium that moves parallel to the direction of the wave travels.
1st example: sound travels parallel.
2nd example: when you talk you will here your voice again. because all the frequency bounce back to you.

In the other hand Transverse wave matter in the medium moves perpendicular to direction the wave travels.
For example: light is a good example of transverse wave.

6 0

3 years ago

2) Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another

worty [1.4K]

Answer:

m_1 / m_2 = sqrt (1 / 2)

Explanation:

Given:

- Initial velocity of both skaters V_i = 0

- Velocity of skater 1 after push = V_1

- Velocity of skater after push = V_2

- Distance traveled by skater 1 = s_1

- Distance traveled by skater 2 = s_2

- s_1 = 2*s_2

- Accelerations of both skaters to halt is equal

Find:

What is the ratio m1/m2 of their masses

Solution:

- Apply conservation of momentum for two skaters just before and after the push as follows:

P_i = P_f

0 = m_1*V_1 - m_2*V_2

- Evaluate: m_1 / m_2 = ( V_2 / V_1 )

- Apply Conservation of Energy on both skaters as follows:

- Skater 1:

0.5*m_1*V_1^2 = u_k*m_1*g*s_1

-Simplify: 0.5*V_1^2 = u_k*g*(2*s_2)

- Skater 2:

0.5*m_2*V_2^2 = u_k*m_2*g*s_2

-Simplify: 0.5*V_2^2 = u_k*g*s_2

- Divide the two energy equations for skaters:

(V_1 / V_2)^2 = 2

(V_2 / V_1)^2 = 1 / 2

- simplify: (V_2 / V_1) = sqrt (1 / 2)

-Hence from earlier momentum conservation results:

m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)

6 0

3 years ago

A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b

nekit [7.7K]

<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

$p(t)=v(t)i(t)$

So the average power is:

$P=\frac{1}{T}\intop_{0}^{T}p(t)dt$

But:

$v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)$

So:

$P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}$

$P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}$

In terms of RMS values:

$V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI$

7 0

3 years ago