Ask question

Ask question

All categories

2 years ago

7

An underwater scuba diver looks up toward the water's surface. She sees a buoy floating on the surface and notes its location. S

he then realizes that the buoy may actually be in a different location because she is looking at it from underwater. What is the wave property that best explains this phenomenon?

1 answer:

Fofino [41]2 years ago

8 0

Answer:

Refraction

Explanation:

Whenever there is a change in media through which light travels there is change in the speed of light due to which there is bending of light when there is change in the medium.

When a ray of light passes from and optically denser medium to and optically rarer medium then it bends away from the normal to the interface and vice-versa.

As here the scuba driver is looking from an optically denser medium to an optically rarer medium the boy may appear nearer to the diver in position than is actually in the air.

Snell's law gives the relation between the angle of incidence and the angle of refraction as:

$n=\frac{sin\ i}{sin\ r}$

where

n = refractive index

i = angle of incidence

r = angle of refraction

You might be interested in

This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a

Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = time taken for decomposition = 3 days

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{58}{100}\times 100=58g$

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

$k=\frac{2.303}{3}\log\frac{100}{58}$

$k=0.18days^{-1}$

a) Half-life of radon-222:

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days$

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant = $0.18days^{-1}$

t = time taken for decomposition = ?

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{15}{100}\times 100=15g$

$t=\frac{2.303}{0.18}\log\frac{100}{15}$

$t=10.54days$

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0

3 years ago

A student standing on a knoll throws a snowball horizontally 4.5 meters above the level ground toward a smokestack 15 meters awa

Elan Coil [88]

Answer:

2.4 m

Explanation:

Consider the motion along the vertical direction

$y_{o}$ = initial position of ball above the ground = 4.5 m

$t$ = time taken by the ball to hit the smokestack = 0.65 s

$v_{oy}$ = initial velocity of the ball along vertical direction

$a_{y}$ = acceleration due to gravity = - 9.8 m/s²

$y$ = position of ball at the time of hitting the smokestack

Using the kinematics equation

$y = y_{o} + v_{oy} t + (0.5) a_{y} t^{2}$

inserting the above values

$y = 4.5 + (0) (0.65) + (0.5) (- 9.8) (0.65)^{2} \\y = 2.4 m$

6 0

3 years ago

A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between

Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

$a = \dfrac{F}{m}$

Where, F = frictional force

m = mass

Put the value into the formula

$a=\dfrac{\mu mg}{m}$

$a=\mu g$

$a=0.229\times9.8$

$a=2.244\ m/s^2$

We need to calculate the speed of the sled

Using equation of motion

$v^2=u^2-2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

$v ^2=(5.49)^2-2\times2.244\times3.89$

$v=\sqrt{(5.49)^2-2\times2.244\times3.89}$

$v=3.56\ m/s$

Hence, The speed of the sled is 3.56 m/s.

8 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 0.05-kg car starts from rest at a height of 0.95 m. Assuming no friction, what is the kinetic energy of the car when it reache

statuscvo [17]

The kinetic energy of the car is A.) 0.466 J

6 0

2 years ago

Read 2 more answers