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3 years ago

Which of the following best models the changes in energy of a blackbody radiator?

2 answers:

8 0

We have that a blackbody radiator either constantly absorbs energy or constantly emits energy, depending on its surroundings. In this case, the energy is continuously and smoothly decreasing, thus it cannot be like B and C.
The energy loss or gain is also monotonous, it has the same direction; a radiator cannot gain energy at some point and then lose some. Hence, it does not resemble a wave either. The most appropriate model is the ramp. Energy is constantly emitted to surroundings and it decreases monotonically.

Katena32 [7]3 years ago

6 0

Answer:

Explanation:

summary of other answer

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A 0.877-kg ball has an elastic, head-on collision with a second ball that is initially at rest. The second ball moves off with h

Art [367]

Answer: the mass of the second ball is 2.631 kg

Explanation:

Given that;

m1 = 0.877 kg

Initial velocity = V0

Initial momentum = m1 × V0

final velocity of m1 is u1, final velocity of m2 is u2 = v0/2

now final momentum = m1 × u1 + m2 × u2

using momentum conservation;

m1×V0 = m1×u1 + m2×v0/2

m1×(v0 - u1) = m2×V0/2 ----- let this be equation 1

Now, for elastic collision;

m1×v0²/2 = m1×u1²/2 + m2×(v0/2)²/2

m1×(v0² - u1²) = m2×(v0/2)² --------- let this be equation 2

now; equation 2 / equation 1

: V0 + u1 = v0/2

2V0 + 2u1 = V0

2u1 = V0 - 2V0

u1 = -V0/2

now we insert in equ 1

m1×3V0/2= m2×V0/2

m1 × 3 = m2

m2 = 0.877 × 3

m2 = 2.631 kg

Therefore, the mass of the second ball is 2.631 kg

7 0

3 years ago

A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end

Dominik [7]

Answer:

wavelength = 0.8989 m

Explanation:

Given data:

weight of wire is 8.25 g

length of wire is 65 cm

speed of sound in room is 344 m/s

time of returning of pulse is 7.84 ms

There are three time period

Time period $= \frac{7.84\times 10^{-3}}{3} = 2.6133\times 10^{-3} s$

$frequency = \frac{1}{T}$

$=\frac{ 1}{2.6133\times 10^{-3})} = 382.6579 Hz$

velocity = wavelength × frequency

wavelength $= \frac{velocity}{frequency}$

$= \frac{344}{382.6579}$

wavelength = 0.8989 m

6 0

3 years ago

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

Schach [20]

Complete Question

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 30.5 m/s (about 68 mph ) around the turn.

Required:

a. What is the race car's centripetal (radial) acceleration?

b. What is the force responsible for the centripetal acceleration in this case?

O normal

O gravity

O friction

O weight

Answer:

question a

$a = 16.32 \ m/s^2$

question b

correct option is option 3

Explanation:

From the question we are told that

The radius is $r = 57.0 \ m \$

The constant speed at which the race car is travelling is $v = 30 .5 \ m/s$

Generally from the question we are told that the track is completely flat so the only force pulling the car to the middle is the frictional force hence the centripetal force is due to the frictional force

Generally the centripetal acceleration is mathematically represented as

$a = \frac{v^2}{r}$