All categories

3 years ago

What do light waves NOT do when intereacting with matter

Physics

1 answer:

RSB [31]3 years ago

3 0

Answer:

D. Dissolve

Explanation:

A light wave is not a soluble substance, so it cannot dissolve. But it can totally do A, B, and C.

You might be interested in

How much heat is needed to raise the temperature of an empty 2.0 x 101 kg vat made of

scZoUnD [109]

Answer:

Q = 7272 Kilojoules.

Explanation:

<u>Given the following data;</u>

Mass = 2.0*101kg = 202kg

Initial temperature, T1 = 10°C

Final temperature, T2 = 90°C

We know that the specific heat capacity of iron = 450J/kg°C

*To find the quantity of heat*

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 90 - 10

dt = 80°C

Substituting the values into the equation, we have;

$Q = 202*450*80$

Q = 7272KJ or 7272000 Joules.

6 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

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