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2 years ago

The emissivity of an ideal reflector has which of the following values?

1 answer:

4 0

Emissivityis a measure of how much thermal radiation a body emits to its environment. On the other hand we have that reflectivity is a measure of how much is reflected, and transmissivity is a measure of how much passes through the object. If a body is required to be ideally reflective to its maximum efficiency, the body should NOT have the property of transmissivity or emissivity. Therefore it should be 0 its emittivity.

Correct answer would be A : ZERO.

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The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha

saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

Δm = 4.9 x 10⁻¹³ kg

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

d. The scale's resolution is too low to read the change in mass

8 0

2 years ago

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ ...Newtons second law Fnet = ma

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ ...f is the friction which is zero here.

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0

2 years ago

Transverse waves on a string have wave speed v=8.00 m/s, amplitude A=0.0700m, and direction, and at t=0 the x-0 end of the wavel

Vilka [71]

Answer:

a. frequency = 25 Hz, period = 0.04 s , wave number = 19.63 rad/m

b. y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

c. 0.0496 m

d. 0.03 s

Explanation:

a. Frequency, f = v/λ where v = wave speed = 8.00 m/s and λ = wavelength = 0.320 m

f = v/λ = 8.00 m/s ÷ 0.320 m = 25 Hz

Period, T = 1/f = 1/25 = 0.04 s

Wave number k = 2π/λ = 2π/0.320 m = 19.63 rad-m⁻¹

b. Using y = Asin(kx - ωt) the equation of a wave

where y = displacement of the wave, A = amplitude of wave = 0.0700 m and ω = angular speed of wave = 2π/T = 2π/0.04 s = 157.08 rad/s

Substituting the variables into y, we have

y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

c. When x = 0.360 m and t = 0.150 s, we substitute these into y to obtain

y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

y = (0.0700 m)sin[(19.63 rad/m × 0.360 m) - (157.08 rad/s × 0.150 s)]

y = (0.0700 m)sin[(7.0668 rad) - (23.562 rad)]

y = (0.0700 m)sin[-16.4952 rad]

y = (0.0700 m) × 0.7084

y = 0.0496 m

d. For the particle at x = 0.360 m to reach its next maximum displacement, y = 0.0700 m at time t. So,

y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

0.0700 m = (0.0700 m)sin[(19.63 rad/m × 0.360 m) - (157.08 rad/s)t]

0.0700 m = (0.0700 m)sin[(7.0668 rad - (157.08 rad/s)t]

Dividing through by 0.0700 m, we have

1 = sin[(7.0668 rad - (157.08 rad/s)t]

sin⁻¹(1) = 7.0668 rad - (157.08 rad/s)t

π/2 = 7.0668 rad - (157.08 rad/s)t

π/2 - 7.0668 rad = - (157.08 rad/s)t

-5.496 rad = - (157.08 rad/s)t

t = -5.496 rad/(-157.08 rad/s) = 0.03 s

6 0

2 years ago

Two rocks are tied to massless strings and whirled in nearly horizontal circles so that the time to travel around the circle onc

Fynjy0 [20]

Answer: $m_1=m_2$

Explanation:

Given

Time period for both string is same

$\frac{2\pi r}{v_1}=\frac{2\pi 2r}{v_2}$

$2v_1=v_2$

and tension in string 2 is twice the first string

$2T_1=T_2$

Tension will provide centripetal acceleration

$2\frac{m_1v_1^2}{r}=\frac{m_2v_2^2}{2r}$

$2\frac{m_1v_1^2}{r}=\frac{m_2\times 4v_1}{2r}$

thus $m_1=m_2$

3 0

2 years ago

Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He m

Oduvanchick [21]

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, $t_t$ of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + $t_t$ + t₁ =6 + 4 + 1.6 = 11.6 seconds.

6 0

2 years ago