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3 years ago

why doesn't light bend (refract) when it initially enters a glass? what actually happens to the light inside the semi-circle?

1 answer:

7 0

The light does not bend when it initially enters the glass because it travels from a less dense medium such as air to a dense medium such as glass. The refraction of light happens only on the flat side of the semi-circle, while on the circular side, it strikes on the interface.

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a large sphere is on a horizontal field on a sunny day. at a certain time the shadow reaches out a distance of 10 m from the poi

Mars2501 [29]

The radius of the sphere in meters is ,r = $10\sqrt{5} -20$

Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,

r/10 = tan [arctan(1/2)] = tan (1/2)

Since, tan (θ/2) = 1-cos(θ) / sin(θ)

we find that,

r/10 = $\sqrt{5} -2$

Hence, r = $10\sqrt{5} -20$

So, the radius of the sphere in meters is ,r = $10\sqrt{5} -20$

Learn more about radius (r) of the sphere here;

brainly.com/question/14100787

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5 0

1 year ago

How does Doppler ultrasound technology differ from ultrasound technology

patriot [66]

Answer:

. Doppler ultrasound is based on absorption of sound, and other

ultrasound technology is based on reflection.D.

Explanation:

6 0

2 years ago

Read 2 more answers

Which are characteristics of mammals? Check all that apply.

Neporo4naja [7]

Answer:

one of the characteristics of a mammal is their several hollow bones another is their three chambered heart and the last is highly developed nervous system

Explanation:

the reason i picked those three is because not all mammals live their life on land and also mammals font have internal fertillization when they are done they take care of their babies and when they grow up they live their own life

6 0

2 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

During the experiment if you could double the breakaway magnetic force with all other quantities left unchanged, what is the new

sergiy2304 [10]

There are some missing information in the question.
However, since you are talking about magnetic force, I think you refer to the Lorentz force. When a particle of charge q and velocity v is immersed in a magnetic field of intensity B, the force acting on the particle is:
$F=qvBsin\theta$
where $\theta$ is the angle between the magnetic field and the direction of the particle.
Therefore, if force F is doubled, then also the velocity v must be double of its initial value:
$v=2v_0$

6 0

3 years ago