All categories

3 years ago

Which is an example of reflection

1 answer:

Elena-2011 [213]3 years ago

5 0

Common examples include the reflection of light, sound and water waves. The law of reflection says that for specular reflection the angle at which the wave is incident on the surface equals the angle at which it is reflected. Mirrors exhibit specular reflection.

You might be interested in

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What

Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

$\alpha = 6261 \ rev/minutes^2$

$\theta = 613 \ revolutions$

Explanation:

From the question we are told that

The initial angular speed is $w_i = 1120 \ rev/minutes$

The angular speed after $t = 13.8 s = \frac{13.8}{60 } = 0.23 \ minutes$ is $w_f = 2560 \ rev/minutes$

The time for revolution considered is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$

Generally the angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{2560 - 1120 }{0.23}$

=> $\alpha = 6261 \ rev/minutes^2$

Generally the number of revolution made is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$ is mathematically represented as

$\theta = \frac{1}{2} * (w_i + w_f)* t$

=> $\theta = \frac{1}{2} * (1120+ 2560 )* 0.333$

=> $\theta = 613 \ revolutions$

5 0

3 years ago

A planet's temperature depends on its distance from the Sun as well as the strength of its greenhouse effect. Let's remove the e

pishuonlain [190]

Answer:

Smallest Effect

Mars

Earth

Venus

Largest Effect

5 0

3 years ago

A soap box derby car at the top of a ramp has more _________ energy and at the finish line at the bottom of the ramp it has more

astra-53 [7]

Kinetic, potential because, at the top of the ramp it’s going faster. Potential at the bottom of the ramp is potential because, it’s not doing any motion.

5 0

3 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

3 years ago

A steel pot has a bottom with a cross sectional area of .1m2 and has a thickness of 1cm. The pot is filled with boiling water an

jenyasd209 [6]

Answer:

840000 J/min

Explanation:

Area = A = 0.1 m²

Bottom of pot temperature = 200 °C

Thermal conductivity = k = 14 J/sm°C

Thickness = L = 1 cm = 0.01 m

Temperature of boiling water = 100 °C

From the law of heat conduction

Q = kAΔT/L

⇒ Q = 14×0.1×(200-100)/0.01

⇒ Q = 14000 J/s

Converting to J/minute

Q = 14000×60 = 840000 J/min

∴ Heat being conducted through the pot is 840000 J/min

3 0

3 years ago