Answer:
Explanation:
Given
Em wave is in the form of
where
Wave constant for EM wave k is
Wavelength of wave
A 100 kg cart goes around the inside of a vertical loop of a roller coaster. The radius of the loop is 3 m and the cart moves at
1 answer:
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Answer:
number of electrons = 2.18*10^18 e
Explanation:
In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.
Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:
(1)
i1 = 0.40 A
i2 = 0.75 A
you solve the equation i3 from the equation (1):
Next, you take into account that 1A = 1C/s = 6.24*10^18
Then, you have:
The number of electrons that trough the wire 3 is 2.18*10^18 e/s
Answer:
T₂ = 5646 K
Explanation:
Let's start by finding the power received by the first disc, for this we use Stefan's law
P = σ. A e T⁴
Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of the disk, T the absolute temperature and e the emissivity that for a black body is 1
The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is
I = P / A
Writing this expression for both discs
I₁ A₁ = I₂ A₂
I₂ = I₁ A₁ / A₂
The area of a sphere is
A = 4π r²
I₂ = I₁ (r₁ / r₂)²
r₂ = r₁ ± 5
I₁ = I₂ ( (r₁ ± 5)/r₁)²
.
Let's write the Stefan equation
P / A = σ e T⁴
I = σ e T⁴
This is the intensity that affects the disk, substitute in the intensity equation
σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²
The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white
e₁ T₁⁴ = T₂⁴ (r1 + 5)²/r₁²
T₁ = T₂ {(e₂/e₁)}^{1/4} √(1 ± 1/ r₁)
If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term
(1 ±x) n = 1 ± n x
Where x = 5 / r₁ << 1
We replace
T₁ = T₂ {(e₂/e₁)}^{1/4} (1 ± ½ 5/r1)
T₁ = T₂ {(e₂)}^{1/4} (1 ± 5/2 1/r1)
If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach
T₁ = T₂ {(e₂)}^{1/4} ¼
T<u>₁</u> = T₂ 0.9
5500 = T₂ 0.974
T₂ = 5646 K
Answer:
The cannonball fly horizontally before it strikes the ground, S = 323.25 m
Explanation:
Given data,
The height of the cliff, h = 80 m
The horizontal velocity of the cannonball, Vₓ = 80 m/s
The range of the cannon ball with initial vertical velocity is zero is given by the formula,
S = 323.25 m
Hence, the cannonball fly horizontally before it strikes the ground, S = 323.25 m
Answer:
(C) 4 beats per second.
Explanation:
As we know that the no of beats can be calculated as.
No. of beats is equal to difference in the tuning forks frequencies.
So,
.
Substitute the values of frequencies of 2 tuning forks in the above equation.
Therefore the number of beats per second will be hear by the observer is 4 beats per second.